Content deleted Content added

Inline

Revision as of 20:35, 24 October 2006

Science	Mathematics	Computing/IT	Humanities
Language	Entertainment	Miscellaneous	Archives

How to ask a question

Search first. It's quicker, because you can find the answer in our online encyclopedia instead of waiting for a volunteer to respond. Search Wikipedia using the searchbox. A web search could help too. Common questions about Wikipedia itself, such as how to cite Wikipedia and who owns Wikipedia, are answered in Wikipedia:FAQ.

Sign your question. Type ~~~~ at its end.
Be specific. Explain your question in detail if necessary, addressing exactly what you'd like answered. For information that changes from country to country (or from state to state), such as legal, fiscal or institutional matters, please specify the jurisdiction you're interested in.
Include both a title and a question. The title (top box) should specify the topic of your question. The complete details should be in the bottom box.

Do your own homework. If you need help with a specific part or concept of your homework, feel free to ask, but please don't post entire homework questions and expect us to give you the answers.

Be patient. Questions are answered by other users, and a user who can answer may not be reading the page immediately. A complete answer to your question may be developed over a period of up to seven days.

Do not include your e-mail address. Questions aren't normally answered by e-mail. Be aware that the content on Wikipedia is extensively copied to many websites; making your e-mail address public here may make it very public throughout the Internet.
Edit your question for more discussion. Click the [edit] link on right side of its header line. Please do not start multiple sections about the same topic.
Archived questions If you cannot find your question on the reference desks, please see the Archives.
Unanswered questions If you find that your question has been archived before being answered, you may copy your question from the Archives into a new section on the reference desk.
Do not request medical or legal advice.
Ask a doctor or lawyer instead.

After reading the above, you may
ask a new question by clicking here.
Your question will be added at the bottom of the page.

How to answer a question

Be thorough. Please provide as much of the answer as you are able to.
Be concise, not terse. Please write in a clear and easily understood manner. Keep your answer within the scope of the question as stated.

Link to articles which may have further information relevant to the question.
Be polite to users, especially ones new to Wikipedia. A little fun is fine, but don't be rude.
The reference desk is not a soapbox. Please avoid debating about politics, religion, or other sensitive issues.

October 18

How many seagulls...

...would I need to get to stand very close together in order to create sufficiant mass to form a black hole? --Kurt Shaped Box 00:36, 18 October 2006 (UTC)[reply]

Posting on behalf of User:Ziyi_cai841117: let me count...one, two three, four, five...two hundred billion one thousand and one...two hundred billion one thousand and two...four hundred billion...four hundred trillion...four billion billion billion... --Bowlhover 03:28, 18 October 2006 (UTC)[reply]

How heavy is a seagull? 500 grams? If so, you'd need about 1.2 x 10^31 seagulls (see Tolman-Oppenheimer-Volkoff limit). By the way, why are satirical questions about seagulls often asked? How did this joke originate? --Bowlhover 00:57, 18 October 2006 (UTC)[reply]

Your average gull weighs about a kilo - the biggest gulls weigh up to 3kg. As for the questions about gulls - I think I started that one off. I have an interest in gulls and started asking a few questions about them. A few people saw this and started taking the mickey... :) --Kurt Shaped Box 01:01, 18 October 2006 (UTC)[reply]

Ah I remember that --⁪froth^{T C} 16:34, 18 October 2006 (UTC)[reply]

But would you have to compress them as well? There are plenty of stars more massive than 6 x 10^30 kg (~3 solar masses) which are not black holes. There are stars around 20 times as massive. How about an upper bound for the problem. A rough back of the envelope calculation shows that a sphere of radius ~ 2.68 AU at the same density of water would be roughly the same size as its Schwarzschild Radius. That's a mass of about 2.7×10³⁸ kg.Richard B 22:17, 18 October 2006 (UTC)[reply]

I think you could make a black hole with just a few. But to be on the safe side, I would gather the complete worlds population, and pack them together rather tightly. THen you need to surround the gull mass with a very powerful exploding (or imploding) mechanism. Say an H bomb. Under the massive compressive forces, its just possible you could get a very small BH. On the positive side tho, that would be the end of ANY future gull questions on WP 8-)).--Light current 01:28, 18 October 2006 (UTC)[reply]

A H bomb is not at all powerful enough. You'd just be creating a very big mess. In order to create a black hole, you'd have to overcome the strong nuclear force, which is very strong. --Bowlhover 02:09, 18 October 2006 (UTC)[reply]

You're not related to L. Ron Hubbard by any chance, are you? ;) --Kurt Shaped Box 02:04, 18 October 2006 (UTC)[reply]

You gather a bunch of gulls together. Then surround them with a flock of swallows (African or European - your choice) and get them to throw the coconuts they're carrying at them at close to the speed of light. Voila! Clarityfiend 02:40, 18 October 2006 (UTC)[reply]

Yeah but you forgot that the swallows have swallowed the coconuts thus rendering them useless as missiles 8-) --Light current 15:28, 19 October 2006 (UTC)[reply]

Interesting question. I've often likened the "big crunch" scenario to the mannerisms of a large group of seagulls converging wildly on a stray piece of bread on the ground. freshofftheufo ΓΛĿЌ 05:07, 19 October 2006 (UTC)[reply]

What kind of bird is this?

This bird flew out of a bush in the middle of Manhattan. What is it? grendel|khan 00:46, 18 October 2006 (UTC)[reply]

It's a common starling. In my homeland we have many. --Kurt Shaped Box 00:50, 18 October 2006 (UTC)[reply]

Blame Eugene Schieffelin. MeltBanana 02:38, 18 October 2006 (UTC)[reply]

You can send some of yours back over here if you like - they've started to decline in the UK. Also, has anyone ever considered trapping for the pet trade? Check out those talking starling videos on YouTube... ;) --Kurt Shaped Box 05:55, 18 October 2006 (UTC)[reply]

Nice photograph. —Bromskloss 09:15, 18 October 2006 (UTC)[reply]

Time machine

How can two teenagers build a time machine? --Bowlhover 03:24, 18 October 2006 (UTC)[reply]

You'll need some paper clips, a piece of string, and tachyon-enriched uranium. —Centrx→talk • 03:29, 18 October 2006 (UTC)[reply]

It's easier when you find a guy named Rufus who will lend it to you, preferably disguised as a phone booth. ---Sluzzelin 03:43, 18 October 2006 (UTC)[reply]

It's really easy [1]. Clarityfiend 04:37, 18 October 2006 (UTC)[reply]

A police box would do nicely as well. – Clockwork Soul 04:38, 18 October 2006 (UTC)[reply]

Eat some mushrooms and all will become clear. Vespine 05:01, 18 October 2006 (UTC)[reply]

Forget the paper clips and uranium. You can find out all about time machines here. B00P 05:46, 18 October 2006 (UTC)[reply]

Or, alternatively, you could buy the plans on the Internet. Just be careful not to injure your vital organs in the process :) --Robert Merkel 07:07, 18 October 2006 (UTC)[reply]

It's quite easy ready... 1. Get a cardboard box, old circuit boards, egg time and some glowing fuel (tip: Paint rock with glow-in-the-dark paint). 2. Set up box and glue on circuit boards and 'fuel' 3. get inside cardboard box and set egg timer for the amount of time you wish to travel into the future. 4. Wait for egg timer to countdown, and only then exit your time travel box. 5. You've now sucessfully travelled the set time into the future. Benbread 08:30, 18 October 2006 (UTC)[reply]

Don't forget to write "TIM MASHIN" on the side! Confusing Manifestation 12:04, 18 October 2006 (UTC)[reply]

How about deciding that when you build the machine you will come back to today from the future and give yourselves the plans. If you don't want to risk meeting yourselves then you can work out where you can leave the plans so that you will find them. Write it on your calendar now so that you won't forget ! Now go and look for those plans ! Gandalf61 08:35, 18 October 2006 (UTC)[reply]

Make sure you bring it to the 24 hour all you can eat car wash. --Russoc4 18:13, 18 October 2006 (UTC)[reply]

Ask yourself. The yourself that already built it and went back into the past to tell you about it, that is. freshofftheufo ΓΛĿЌ 05:02, 19 October 2006 (UTC)[reply]

Funny.. Im starting to get deja vu on this thread! Does that mean Im already in the future, or does it mean this is a FAQ?--Light current 15:36, 19 October 2006 (UTC)[reply]

You guys have all got it wrong. Just reach the Planck heat and you've got everything all set! — X [Mac Davis] (SUPERDESK|Help me improve)05:52, 20 October 2006 (UTC)[reply]

dirigibles

designs for commercial use; design principles; performance and capacity. Thanks. SKalmutz--Skalmutz 03:52, 18 October 2006 (UTC)Bold text[reply]

See WP's article on airships, and also the external links mentioned there. Have a look at the P-791 for a recent example of design.---Sluzzelin 04:18, 18 October 2006 (UTC)[reply]

Regarding Earth rotation and revolution

If gravity is a force towards centre of sun, why doesn't the earth fall straight into sun instead of rotating around its own axis and revolving around sun? One explaination given by stephen hawkins in brief history of time says that actually earth is travelling strainght in space-time but in 3 dimensional space it looks curved as a result of removal of time dimension.Does somebody have any simpler answer to understand?

If the earth were to stop moving through space for an instant, it would indeed start moving in a straight line directly towards the sun. And if the sun's gravity stopped for an instant, the earth would continue moving in a straight line instead of orbitting. But you know that neither of those "if"s happen...the earth is in orbit, always tending to move "sideways" relative to "directly towards the sun" and the sun's gravity keeps pulling it closer. These effects are balanced--the gravitational pull is strong enough to keep the earth curving around instead of zooming off but not strong enough to overcome the earth's inertia and pull it any closer. See centripetal force and planetary orbit, and remember that actually the earth and sun orbit each other (but the sun is so much more massive, that result is as if just the earth is orbitting a virtually stationary sun). DMacks 09:05, 18 October 2006 (UTC)[reply]

(edit conflict) Hi! Please sign your comments with "~~~~" at the end. We can forget about the earth rotating around its own axis since that is not related (or its impact is at least very small). An object doesn't have to move in the direction of the force that is acting on it if the object is already moving in another direction. Think of a car that rolls past you. If you give it a nudge, it doesn't immediately change its movement and start moving in the direction you pushed, right? If you continue to push it in the same direction for a long time, however, with nothing holding against, it would more an more in that direction. But the sun does actually not pull the earth in the same direction all the time because toward the sun is not the same direction all the time – it depends on where the earth is at the moment. A more familiar example is when you have an object in a string that you swing around in a circle. Let's say you do it in empty space so we don't have to think about air resistance. The only force on the object is the string pulling it and that force works toward the center, where your hand is. Still, the object don't end up in your hand, but continues in a circle. —Bromskloss 09:08, 18 October 2006 (UTC)[reply]

Just to expand on things a little, there are two answers to your question. The first is based on Newtonian gravitation - that gravity is a force, which pulls the Earth to the Sun. But, as pointed out above, the Earth is travelling in another direction, so the force acts to change that direction. Imagine you've got a big cannon on top of a big hill. If you fire the cannon with a small charge, the cannonball will go forwards a bit and fall at the same time, making a parabolic path. If you use a lot more charge, it will go further and further. If, theoretically speaking, you used a massive amount of charge, you could shoot it so fast that by the time it's fallen a certain distance, the curvature of the Earth means it's still at roughly the same height. As long as it keeps falling down as fast as it's moving forwards, it will track out a circular path around the Earth. This is a (very much) simplified way of making something orbit the Earth, like a satellite. Now, imagine that on a bigger scale, and you have the Earth orbiting the Sun.

Unfortunately, it turns out Newton didn't get things perfectly right, and when Einstein came along and invented General Relativity he made the universe look a whole lot different. In particular, in GR gravity isn't a "force", it's a distortion in the shape of spacetime. Think about standing on top of a sphere, like the Earth. From our scale, it looks like the Earth is flat. But if you tried walking straight forwards, eventually you'd come back to where you started, because in fact the Earth is curved. Now imagine space is like that - curved, but in more complicated ways. Anything moving in space that doesn't have a force acting on it (and remember, gravity isn't a force any more) will move forwards in its little bit of spacetime, which will always look flat on a small scale (and in the universe, the scales get pretty big), but whenever you have something massive, like a star, that makes the overall shape of spacetime curved, so even though you may think you're going straight forward, someone standing far away will watch you trace out a curved path - in particular, spacetime around the Sun is shaped so that the path the planets make is roughly circular - i.e. an orbit. Confusing Manifestation 12:03, 18 October 2006 (UTC)[reply]

I could have sworn this was only a theory... --Wirbelwind 02:32, 20 October 2006 (UTC)[reply]

Well yes, Einstein's theory of general relativity is only a theory in the same way that most all of science is only a theory. It's a convenient way to explain why gravity can act at a distance, seemingly instantaneously. It also happens to hold up to the empirical evidence we've been able to gather so far.sthomson 19:15, 24 October 2006 (UTC)[reply]

.

sinisoidal RC phase shift oscillators

how does a sinusoidal RC oscillator prodduce a sinusoidal output??? I know that an oscillator produces a out put signal even withought a input signal bcause it amplifies the noise signals but noise signals are not always sinosoidal are they?

An oscillator consists of gain more than unity to compensate for losses.So the noise automatically gets amplified,and due to resonant nature of circuit only a particular frequency gets amplified causing suistained oscillations.

The RC produces 60 deg phase shift, so Supposing the output of the amp is 180 out of phase, three such RC networks would eventually produce zero phase shift or in phase oscillations for feedback,thus positive feedback in short.

See Wien bridge oscillator for one answer. Another is the "Twin-T" oscillator, described briefly at RC oscillator.

Atlant 17:03, 18 October 2006 (UTC)[reply]

How can I study theoretical physics in one of the US universities?

I want to study theoretical physics (Quantum mechanics) in one of the US universities. I am nineteen years old. I live in Egypt. What shall I do? How much will this cost me? Is there a way I can get a free scholarship? I have an account under the name Meno25 talk. I sent this message from a public computer because my computer is currently broken down. By the way, I am currently having a wikiholiday.

--196.218.105.36 11:32, 18 October 2006 (UTC)[reply]

First of all, you cannot study just theoretical physics. The canon of undergraduate physics is more or less the same everywhere, and usually contains, within in the first three years (in many conuntries, that is up to the Bachelor of Science degree) courses in: calculus, linear algebra, general physics, mechanics, electrodynamics, lab practices, a bit of quantum mechanics and relativity. If you are then thrilled by it, you'd contnue your studies up to an Honours degree, a Master of Science, or even a PhD, depending on the country and your stamina. But I'm afraid I am not an expert for your main question, because I studied physics in Germany and currently work as physicist in Austria, so I do not know much about scholarships in the USA. But I could tell you about Germany or Austrai, of course, in case you are interested. Simon A. 11:56, 18 October 2006 (UTC)[reply]

Here are some types of scholarships:

Academic scholarships for exceptional students. If you are at the top of your class and score extremely well on tests, you might be able to get one of these.

Athletic scholarships for excellent athletes. Somehow I doubt if many physics majors have athletic scholarships, but there may be a few.

Need-based scholarships. These provide money to those who are poor. Generally speaking, they are only for those of the same nationality as the scholarship fund.

Special group scholarships. You get these by being a member of some class (blind, deaf, black, etc.) or organization.

As for the cost, that will vary widely. However, I would say US$10,000-$20,000/year would be a bare minimum for tuition and expenses. Some are far more expensive. One option is to take the first two years at a community college, which can be much less expensive.

As for getting into the US, you would apply for a student visa, at the American embassy. So long as nobody in your family is a member of a terrorist organization like the Muslim Brotherhood, you can probably get one. However, if you ever overstay your student visa, you are likely to be deported and banned from ever entering the US again.

StuRat 14:02, 18 October 2006 (UTC)[reply]

I don't know about the US, but in the UK, plenty of universities, including some prestigious ones such as Imperial College London and University of St Andrews, offer theoretical physics as a possible course[2]. It tends to overlap normal physics but with some modules changed (more maths, less lab work). It's probably too late to apply to a UK university this year (the 2006 UCAS application season is now in full swing), but registration for 2008 should start up soon, and with physics application rates so low, universities will be competing to accept you! As far as scholarships and bursaries are concerned, unfortunately UK universities tend to ask that overseas students sustain themselves (for more info, see International Students on the UCAS website). La ï ka 15:27, 18 October 2006 (UTC)[reply]

Well I'm paying 28000 USD per year here in america. I was a geek child so no scholarships for me I'm afraid :) If you do decide to study in america you'll be paying it off for many many years. IIRC britain has free higher education (or was that canada?) but you might have to be a citizen. --⁪froth^{T C} 16:24, 18 October 2006 (UTC)[reply]

Canada does not have free higher education (it has severely reduced higher education but international students have to pay twice the amount domestic students do. However becoming a citizen is easier in Canada.) Sweden and other socialist countries do have free higher education but I forget the requirements. ColourBurst 16:43, 18 October 2006 (UTC)[reply]

The UK did have free Uni education. The universities can now charge up to £3000 (~$5000) per year in Top-up fees, but that still sounds a heck of a lot cheaper than a lot of countries. But an international student would have to pay full fees (Cambridge, Leeds and Imperial all charge around the £10,000 (~$18,000) mark for scientific courses to international students). La ï ka 19:02, 18 October 2006 (UTC)[reply]

Doesn't cambridge have like individual dorms and fireplaces and leather furniture in the rooms? At 18 grand? Geez what am I doing in the US? --⁪froth^{T C} 19:16, 18 October 2006 (UTC)[reply]

Some of the rooms are pretty sweet; the best room at Corpus Christi College, Cambridge has ensuite and a private living room! Of course, you need to get at least a first to even be entered into the lottery to earn the room... La ï ka 20:39, 18 October 2006 (UTC)[reply]

As you asked on my Talk page, some infor on German universities: University education used to be free, but this year, most Länder have introduced fees. Their amount is not too clear yet but it should settle below €1000 per year. But note that, as you are not a EU citizen you will AFAIK be required to prove that you own enough money to afford living in Germany in order to get a student visa. Maybe I point out a few differences between German and US universities. Many other countries in northern continental Europe (Austria, Switzerland, Benelux, Scandinavia, ...) are comparable to what I say about Germany.

German universities have no entrance exams, and besides the most popular subjects (which does not include physics) anybody with an Abitur may start studying at any university. Foreigners have to show that they finished secondary education (i.e. high school) at a level comparable to a German Gymnasium.
To compensate for the lack of entrance exams, in physics, the exams after the first and second year are designed to sieve, i.e. to let only those pass who are good enough to study physics. More than a third of the freshmen leave physics after failing these exams!
In US universities most undergrads start with one or two terms of general studies before deciding on a major. German universities require you to decide for your major at the beginning. Also, the starting level is higher due to the longer high school time and our three-tier school system.
While the US has a few elite universities and many medicore ones, the quality of education differs not that much from one German university to another one, and lies below the top US universities but clearly above the average US university. Still, the physics departments of some German universities are much better than average.
Undergrad courses are taught in German! Most foreigners need a year of full time language courses before thay can really start to study. But as every German physicist is usually more or less fluent in English, some universities are considering to offer undergard courses in English. I don't know whether this has already left planning stage.
German freshmen are older than US freshmen and treated more as adults. You are expected to organize your live and your studies yourself. While there is counseling offers you have to actively seek them -- nobody will notice if you are left behind. And few universities have a campus with on-campus housing,
Traditinally, physics studies take five to six years and finish with the Diplom degree, which is comparable to the Master of Science of British or US universities. However, due to the Bologna process, this is now being changed to a two stage scheme: three years to a Bachelor of Sciene, than two years to a Master of Science degree. The PhD is considered not as a part of the studies but as the first employment of a physiscist at university.

If you have more questions or need pointers to information, just ask. And have a look at the web site of the German Academic Exchange Service (DAAD), which represents German universities abroad: [3] Simon A. 21:19, 18 October 2006 (UTC)[reply]

In the U.S. top private colleges it will cost $40,000 to $45,000 per year for tuition, housing, food, books, fees and other expenses, not including airfare. There may be restrictions on you working if you have a student visa, but perhaps some part time work is allowed. You might check on that. Many foreign governments have sponsored their top students or those related to the people in power to study in the U.S. Some foreign students get the general education courses at a cheap community college, whole they also polish their English. There are tests of English proficiency (TOELF) required by many colleges. Check with your government for possible sponsorship. Good luck! Edison 23:24, 18 October 2006 (UTC)[reply]

I believe the guys on Wikiversity can probably help you out to, (more once the site gets running better). — X [Mac Davis] (SUPERDESK|Help me improve)05:49, 20 October 2006 (UTC)[reply]

To learn before you get to the university, try "Opencourseware at MIT". Lecture notes, videos of classes, etc. Free. http://ocw.mit.edu/ --GangofOne 20:08, 20 October 2006 (UTC)[reply]

Science, I think

I am a Marine currently in Iraq. I found a Phillips PYE Unicam PU 4023 Refractive Index Detector, a PYE Unicam LCXP gradient Programmer, and a Hitachi 150-20 Data Processor. I have researched the 3 items and found that they are most likely used together. What I am trying to figure out is what would the owner use the items for.

Sounds like HPLC. DMacks 15:36, 18 October 2006 (UTC)[reply]

Kill you and your mates most likely if he had a chance. DirkvdM 18:32, 18 October 2006 (UTC)[reply]

orbital plane

Are planets in other solar systems on a parallel orbital plane to our planets? Also why is it that pictures of multiple deep space objects (like that hubble deep field scan) seem to show everything scattered about helter-skelter? If planets and moons form in the same plane then why don't universe-level objects like galaxies? --⁪froth^{T C} 16:19, 18 October 2006 (UTC)[reply]

1. No.
2. Because they are scattered about.
3. Your question betrays a misunderstanding. While the planets of the Solar System revolve around the Sun in (pretty much) the same plane, their axes of rotation are all different (Earth's being tilted by 23½°). So why shouldn't galaxies be oriented in any direction?
B00P 07:04, 19 October 2006 (UTC)[reply]

Formation and evolution of the Solar System should clear this up. Essentially, the process that forced everything onto the same plane at the formation of the Solar System was a completely local process, neither affected by nor affecting other star systems, and not controlled by any larger order. In the absence of a reason for things to line up, they don't. Melchoir 16:49, 18 October 2006 (UTC)[reply]

I'm not sure, but I assume it's because the planets' rotation around the Sun and the moons' rotation around the planets forming at the same time, in the same protoplanetary disk, caused them to be in the same plane. Why that disk is not in the same plane as the milky way, I don't know, but what I said suggests that the two disks did not form at the same time. Just the other night I saw in a documentary that the Sun bobs up and down through the Milky Way's disk, almost as if it rotates around something else in the disk (at least, that's what the visualisation loked like). As for larger structures, it's still a mystery why they formed in the first place. What caused structure in the Universe after the Big Bang that resulted in the accretion that resulted in galaxies and such. Sorry to have asked more questions than given answers. :) DirkvdM 18:50, 18 October 2006 (UTC)[reply]

It has been noted that the arrangement of galaxies resembles the arrangement of a pile of bubbles, with clumps of galaxies where the "bubbles" come together, more sparsely placed galaxies along the "skins" of the bubbles, and inter-galactic voids in the center of the bubbles. To me, this implies that something at the center of the bubbles repels normal matter. This would also help to explain why the expansion rate of the universe is accelerating. StuRat 19:09, 18 October 2006 (UTC)[reply]

Or that there's something near the skins that pulls other bubbles to it, like the Great Attractor or something.. hmmm --⁪froth^{T C} 19:18, 18 October 2006 (UTC)[reply]

But that wouldn't explain why the expansion rate of the universe is accelerating. StuRat 20:33, 18 October 2006 (UTC)[reply]

I think the OP has asked an interesting question. Planetary systems and spiral galaxies (the most common type of observed galaxy) exhibit preferred orbital planes, due to the interaction between conservation of angular momentum and gravtitational attraction. Are there any signs of preferred orbital planes in larger scale structures, such as groups, clusters and superclusters ? If not, why not ? Gandalf61 08:34, 19 October 2006 (UTC)[reply]

Flu vaccine

The university I work for is providing flu vaccines for all employees, with no foreseen shortages. Should I as a healthy, fit, not-very-flu-susceptible, 24-year-old male get the vaccine?

My concerns are:

People in more at-risk groups ought to go ahead of me, even though the uni says they have enough (if they have spare, would the spares go back to the hospital?)
If I'm not at risk, I don't want to put unnecessary medicine in my body
I believe in letting my body's own immune system do the work, and hopefully become stronger.

So obviously I'm leaning towards not getting one (I've never gotten one before), but I am wondering if

My beliefs are mistaken or out-dated
I am selfishly allowing flu to spread by not vaccinating myself.

What do people think?

Thanks! -Sam

You don't seem to be a member of any priority group defined by the ACIP. So basically do as you wish. Don't do so thinking that getting the flu will make your immune system "stronger", though, that's just silly. And yes, by not getting the vaccine you are making it ever-so-slightly more likely for others, especially those you live with, to get flu, but it's a marginal difference. - Nunh-huh 17:39, 18 October 2006 (UTC)[reply]

The defined priority groups are:

People at high risk for complications from the flu, including:

Children aged 6 months until their 5th birthday,

Pregnant women,

People 50 years of age and older, and

People of any age with certain chronic medical conditions;

People who live in nursing homes and other long term care facilities.

People who live with or care for those at high risk for complications from flu, including:

Household contacts of persons at high risk for complications from the flu (see above)

Household contacts and out of home caregivers of children less than 6 months of age (these children are too young to be vaccinated)

Healthcare workers. - Nunh-huh 17:39, 18 October 2006 (UTC)[reply]

Right: my question was based on the fact that the university said that they would have no shortages, and so wondered whether non-priority groups ought to also get the vaccine. --Sam

Yes. You should if you don't want to get the flu, with the risks that entails, risks which are greater than those caused by the vaccine. If you don't care whether you get the flu, or spread it to others, you needn't get the vaccine. - Nunh-huh 17:48, 18 October 2006 (UTC)[reply]

There is a shortage of flu vaccine for the nation, (the US, but probably in other countries as well), and it is a real problem. Companies and individuals with the necessary funds purchase huge amounts of vaccine and then resell it to anyone who has money, regardless of risk factors, in "flu clinics" all over the place. This year, the entire stock of flu vaccine made by Sanofi Pasteur was purchased within the first 14 hours it was on sale. Out of the many units, only about 10% went to actual medical facilities (I'm sorry I can't give exact numbers here). This year, in the city I live in in the US, we have 7,500 DPH-appropriated units for the entire priority population (about 20,000 individuals). Some will seek immunization outside of DPH-funded clinics, others will get sick, some of them will die... and their care will cost the city much more than a few vials of vaccine. That said, your university has said that they have a surplus, and they probably do. If I were you, I would ask them where the extra units will go if they are not used at the school. If the answer is "the trash" then I would get one. Otherwise, I would probaly not. And if you do happen to get sick, stay home if you can. In any case, I would tell them that it is not cool that they purchased so much vaccine. It is certainly unfortunate, Sam, that more people are not as thoughtful as you when they are asked if they want a flu shot, then we might just have enough for the people that need it.Tuckerekcut 22:18, 18 October 2006 (UTC)[reply]

There is no flu vaccine shortage. 40 million doses have been shipped to date, the most ever at this point in the season. [4] [5]. There may be distribution problems, which will be corrected. But just like eating all the food on your plate won't do anything to alleviate the suffering of children starving in Africa, foregoing a flu shot when it's available won't do anything to make it available to others elsewhere. - Nunh-huh 00:54, 19 October 2006 (UTC)[reply]

I regularly get flu shots and I'd suggest you do the same. People who have to interact with lots of people on a regular basis (like at your uni) have a vested interest in doing everything they can to prevent the flu from climbing aboard their body. On the one hand, you'll be in contact with people from over a wide area in a somewhat cramped environment, making it very very likely that you will come in contact with some bugs. On the other hand, lots of people from all over will be coming in contact with you - while you're shedding viruses, you'll potentially infect a lot of the folks around you. My understanding is that the one good reason not to get the shot is if you're allergic to eggs, which are used to incubate the vaccine. See our article for more details. To answer your specific question, yes, I think your views are a little out-dated, but asking questions is the best way to reverse that! If it's the shot that's really bothering you, there's also a nasal mist, which is apparently more effective. Matt Deres 01:24, 19 October 2006 (UTC)[reply]

There is no flu vaccine shortage this year. I just got the annual brochure from the DOH and it makes it very clear that there is no shortage and that it should be given to everybody who wants it. -THB 02:22, 19 October 2006 (UTC)[reply]

membrane capacitive current?

Hello. I can´t understand exactly what is meant by a capacitive outward current across a cell membrane, apparently in the opposite direction of an ionic inward current. Is it because of the departure of positive ions from the extracellular space into the intracellular space, that one defines the "lack" of positive charges outside the cell as a current "across" the membrane? Or is capacitive current a real, physical current from the intracellular space to the extracellular one? Thank you for any help. Christian, medical student, Denmark

Take a look at this. If you still have questions, please return. --JWSchmidt 01:20, 20 October 2006 (UTC)[reply]

Thanks a lot. Anyway, I understand that the membrane has the ability to store charges on its surface area, but in my textbook it is written as if there is a specific "current" in opposite direction of the ionic one (ex. in opposite direction of the inward Na+ current in nervecells about to unleash an actionpotential). It is the nature of this "current" that puzzles me.

Any biological membrane has many charges that part of the molecules of the membrane. Under most physiological conditions, a biological membrane is a "fluid mosaic" and some of the charges that are in the structure of the membrane can move in response to a change in the electric field that exists across the membrane. This shifting of charge is very fast and fairly symetrical, as shown here (compare panels A and B). --JWSchmidt 02:20, 24 October 2006 (UTC)[reply]

What bs. Christian is absolutely right in his assumptions, and you're just puzzling him with reference to diagrams that show action potentials, not the fluid mosaic model you brazenly say you know of. Tag radet fran en svensk och ignorera herr Schmidt. Han ar ingen hjalp.

 Hello,I  had the same question and it even led me here now I think I can shed some light.You must first know that  a capacitor after charging has the same voltage of the source that charged it without  decreasing the voltage of that source,so the batteris in the RC circuit of the memrane must first charge the memrane by actually moving charge from one side of the memrane to the other and this charging doesn't drain these batteries(the electro chemical gradients of the ions)this is the capacitive current .Then the batteries (e.g the Na ion gradient)can start the ionc current that will eventually drain the battery when the membrane potiental approaches the equilibrium potential of Na(reversal potential)

DFT bins

I often read terms like transform bin or DFT bin in papers concerning DFT and other transforms.. Can you explain me the meaning?

I think you want to see discrete fourier transform. --Kainaw ^(talk) 20:03, 18 October 2006 (UTC)[reply]

I did, but on en.wiiki I found that therm in Goertzel Algorithm only, without definiton

When you're doing a transform from the time domain to the frequency domain, you need a way to represent the output data. One conventional way is to define a series of "frequency bins" into which you allocate the spectral energy of the waveform being analyzed. For example, you might define 256 frequency bins that are centered on 0 Hz, 1 Hz, 2Hz, ... 254 Hz, and 255 Hz. So when you do the transform, any energy that the transform produces that occurs from >=0.5Hz to <1.5Hz is dumped into the "1 Hz" frequency bin, >=1.5Hz through <2.5Hz is dumped into the 2Hz frequency bin, and so on.

Naturally, the more samples you have, the more finely you can resolve the frequency of the spectral components and the narrower you can make your bins (at least, if you wish).

Atlant 13:10, 19 October 2006 (UTC)[reply]

More specifically, the "bins" are because you're using the DFT rather than the analogue fourier transform - by its very nature, the DFT requires that the output data comes in discrete packages, which are your bins. (And as Atlant pointed out, the size and number of the bins is mostly based on what you want them to be.) Confusing Manifestation 00:38, 20 October 2006 (UTC)[reply]

Thank you! --Ulisse0 08:55, 20 October 2006 (UTC)[reply]

Look at [Z-transform] as well. That's a sampling transform that has discrete time points as well. A key point of the bins is that the summation of the bins over the same frequency range should equal the analog continuous transform. This makes it look like you can lower the noise floor by increasing the number of bins but that is incorrect. --Tbeatty 06:16, 24 October 2006 (UTC)[reply]

Spacetime

One interesting interpretation I heard about relativity is that as you move faster through space you move slower through time. So photons dont actually move through time at all, and if you are stationary in space you are moving at maximum rate through time. Another way to look at, albeit from my laymans viewpoint, is that you have to apply energy to 'brake' your progress through time. Anyway, does this mean that everything travels at the same rate through spacetime, and if so what is that rate mmatt 20:09, 18 October 2006 (UTC)[reply]

Yes: in a very real sense, everything travels at the speed of light through spacetime. (The proper dimensions of spacetime are somewhat hard to define; it really makes just as much sense to say that time and space are the same thing and call the speed of light 1. This makes sense, because for a stationary object, the only speed is "time-speed", and it seems only fair to call that "one second per second" or just 1.) Of course, the very notion of "speed" when time is relative to the observer is somewhat odd; perhaps you'd like to read about proper time, which is one sort of realization of the "constant speed" idea. --Tardis 20:14, 18 October 2006 (UTC)[reply]

According to Einstein, everything is always going at the speed of light it just happens to be that most of that speed is through the fourth dimension (time). When one travels at significant portions of the speed of light through space (say in a fast rocket ship), it's simply transfering some of this speed away from time to space so that you're still going at the speed of light. No faster, no slower. Æµ§œš¹ _{[aɪm ˈfɻɛ̃ⁿdˡi]} 21:01, 18 October 2006 (UTC)[reply]

I'm not sure Einstein would have subscribed to this version of his theory. --Lambiam ^Talk 11:30, 19 October 2006 (UTC)[reply]

I am truly confused at what you two have said. Rest mass never moves at c. Photons experience no time, tardyons and luxons, together, travel at many different speeds. — X [Mac Davis] (SUPERDESK|Help me improve)05:44, 20 October 2006 (UTC)[reply]

Male marsupial nipples ?

Male placental mammals have nonfunctional nipples, so do male marsupials have them, as well ? In the females, those nipples are inside the pouch. Do males also have a partial pouch which contains the nipples ? StuRat 20:18, 18 October 2006 (UTC)[reply]

Better ask User:JackofOz from marsupial land! 8-)--Light current 02:41, 19 October 2006 (UTC)[reply]

I'm flattered, but I have no idea, sorry. What I do know is that marsupials are also found in New Guinea and some adjacent islands; South America; and North America. JackofOz 07:16, 19 October 2006 (UTC)[reply]

I would be very flattered too if another man asked me a question about nipples. — X [Mac Davis] (SUPERDESK|Help me improve)05:40, 20 October 2006 (UTC)[reply]

Yeah, but you could go and check. Don't you have koalas hanging around in your back yard? Or are you afraid that groping in their pouches to feel their nipples would be seen as sexual harrassment? DirkvdM 08:32, 19 October 2006 (UTC)[reply]

You've been here, mate, so you know that isn't true. Maybe they do things differently in Bundaberg, but the last time I saw a koala was years ago, in a zoo. Mind you, I have plenty of possums in my garden. And in a couple of weeks I'm moving to a place in the country, where kangaroos and wombats are frequently seen, so I'll be sure to keep an eye out. :) JackofOz 08:35, 20 October 2006 (UTC)[reply]

If nobody can find out, I know some people in the marsupial lab at uni and can ask them. I may see them tomorrow, but otherwise next Wednesday. So remember to check this post then. Thanks. --liquidGhoul 08:14, 19 October 2006 (UTC)[reply]

Darwin had something to say about this:

[...] long after the progenitors of the whole mammalian class had ceased to be androgynous, both sexes yielded milk, and thus nourished their young; and in the case of marsupials, that both sexes carried their young in marsupial sacks.

Everything I can find on the internets tell me that male marsupials should have nips too, but I can't find anyone saying "YEP, IT'S TRUE, MR. WALLABY HAS LITTLE BOOBS JUST LIKE YOU AND ME", so I guess you're going to have to take this one on faith. freshofftheufo ΓΛĿЌ 09:59, 19 October 2006 (UTC)[reply]

Do I get a prize for having stumped the professors ? :-) StuRat 14:41, 19 October 2006 (UTC)[reply]

For god's sake, all marsupials are mammals. Mammals are called mammals because they have mammary glands. All of them. Even the males. All of them. End of story. -THB 21:03, 19 October 2006 (UTC)[reply]

The question remains, where are the nipples on a male marsupial ? Are they in a pouch or not ? StuRat 22:22, 19 October 2006 (UTC)[reply]

Point of order, THB. Monotremes are mammals and possess mammary glands, but they don't have nipples (either gender). Whether marsupial males have nipples is still a legitimate question. I can't think of a single reason why they wouldn't have them (since the females do), but nature is under no constraints to meet my expectations. Matt Deres 01:46, 20 October 2006 (UTC)[reply]

Why the hell would they have mammary glands but no nipples? — X [Mac Davis] (SUPERDESK|Help me improve)05:40, 20 October 2006 (UTC)[reply]

Marsupials are very strange, in many parts of their biology (have you seen the shape of their sperm?). The female nipple changes in size during the development of the offspring, and, depending on the species, each nipple can be at a different state. I wouldn't be surprised if the males don't have nipples. Having said that, my guess is that they will have small ones in the same place, just without the pouch. --liquidGhoul 12:46, 20 October 2006 (UTC)[reply]

Monotremes are the branch of mammals most like the earliest proto-mammals. The mammary gland/milk/nipple situation had to evolve somehow and I guess monotremes show that the nipple was the last part to develop. The milk just kind of seeps out (I believe milk glands are homologous to sweat glands) and the young suck it from the fur. Matt Deres 01:06, 21 October 2006 (UTC)[reply]

electromagnetic interaction and contact for force fields?

I have read recently in a book called "a brief history of pretty much everything" it calimed that all physical interactions are actually electrons repeling each other, would it be at all possible at least in theory if not in practice at this stage in technology to create a field or beam of electrons with the same properties as matter to move or repel matter like a force field? is this why we cant make one coz we cant make a powerfull enough or dense enough electron cloud? hopefull

I think you would be more likely to melt it than move it.mmatt 21:00, 18 October 2006 (UTC)[reply]

In the absense of atomic nucleii, the electrons tend to fly apart from one another. They can be kept together temporarily, as in a television with a cathod ray tube. --Gerry Ashton 21:04, 18 October 2006 (UTC)[reply]

but if you could make them group them together in the right way they would form a solid force, like a stream or solid force? hopefull

cathode ray tube. Xcomradex 21:16, 18 October 2006 (UTC)[reply]

That didnt answer my question about if electrons could be arranged to form a force that is solid like normal things are if electron repulsion causes things to be solid? Hopefull

The usual way that electrons are "arranged to form a force that is solid like normal things are" is to, well, use normal things, like, say, a forklift. If you could somehow arrange electrons to make a smooth surface, you could perhaps push on objects with it. But the objects would push back, so you'd have to hold the electrons in their formation somehow. What better way to do that than to stick a bunch of nuclei in with them (to bind them to a spot) and let the resulting atoms bond together to hold a shape? (This, of course, is once again "use normal matter".) It is not unreasonable to charge a piece of metal and then manipulate it with an electric field, or run a current through it and then manipulate it with a magnetic field. The latter is more common since you don't need to keep anything charged, and is the principle behind all sorts of things like rail guns and magnetic levitation. Does that help? --Tardis 22:22, 18 October 2006 (UTC)[reply]

I think so , so the answer is "yes you could make a solid electron field if you could keep them charged and stationary" this would move stuff if it touched it, or stop things moving towards it? is that much correct? would it be possible to make a stronger repulsion than normal, so instead of just pushing it repulses away, like magnets? sorry to ask another question on top of that Hopefull

I think you're right -- but realize that tissue paper has the same qualities. If you can keep it stationary (or move it as you like), you can move things around with it or stop objects from passing through it. The problem is that there's no way (short of backing it with steel) to keep it "stationary" (including intact, since parts of it must move if it breaks) when it's called upon to stop, say, a bullet. The same thing is true of the electron sheet: you've no way of holding it in place, except reinforcing it with something stronger (here, the rest of the steel). In fact, it's worse with the electron sheet because they repel each other, so can't hold up against even 0 force without assistance (fortunately, the same rest of the steel is also positively charged and serves to cancel this effect.) As far as increasing the repulsion, no: in fact, uncharged objects will be attracted to a big sheet of electrons (see electrostatic induction, and think about where the separated charges would go). It's only at very short range (this is the definition of "contact") that you can repel anything, and even then the force falls off so strongly that it is again like a wall instead of like a "flow of force" that could accelerate anything away from the electrons. Once again, since normal matter interacts through the very mechanism you're describing, you can't expect (although it's not inherently impossible for) applying this mechanism to be unlike using normal matter in any way. --Tardis 15:50, 19 October 2006 (UTC)[reply]

Ok then so if one could find a method for groupiing electrons together, without a nucleus, and keeping them together [stop them repelling each other , depending on how many and how closely grouped the electron sheet would act like normal matter. with the more electrons per space in the sheet the stronger it will be, so you could vary from tissue paper consistancy? would it be possible to make it stronger than this? wioth more electrons or closer together, also would it be possible to create the charge the electrons build up that repel each other some other way? from a source other than electrons? Hopefull

If you can suppress electrostatic repulsion ("stop them repelling each other"), you can probably do a lot better than simply making an "electronic broom"; see Niven's Wunderland Treatymaker. As far as lots of electrons, remember the bit about normal matter being attracted to charge; with enough of them, you'd be pulling strongly on everything nearby that wasn't actually in contact with the broom. Your remark about "tissue paper consistency" is confused: the whole point of this discussion is the equivalence of all normal matter and your electron sheets. In both cases you can push on things, but need some variety of material-like strength to maintain your pusher's integrity (more in the case of self-repelling electrons). As for other sources of negative charge, there are plenty: antiprotons, some pions, down quarks, etc. — but they tend to be harder to deal with than electrons. --Tardis 18:52, 20 October 2006 (UTC)[reply]

Domestic Poultry

Why do they ~~loose~~ lose their pigments in captivity and turn white? (ie: chickens, turkeys, ducks, flamingos, etc). --Russoc4 22:39, 18 October 2006 (UTC)[reply]

Their loose pigments fall off ? :-) StuRat 23:36, 18 October 2006 (UTC)[reply]

Ah, you got me. --Russoc4 23:41, 18 October 2006 (UTC)[reply]

I think there are multiple answers. Flamingos lose their colour because they get their colour from the food they eat; when they eat different micro-organisms, they lose their pink-ness. ISTR that zoos and other places actually introduce a reddish dye to their food so that they look 'right'. For the others, I'm not sure if there's an actual trend to whiteness, if it's just happenstance that white happened to be bred in barnfowl, or if some of the same answer applies as with flamingos. Matt Deres 01:32, 19 October 2006 (UTC)[reply]

Flamingos need carotenoids in their diet to develop their pink/red color. This is mentioned in the Flamingo article under "color". It is possible that there are similar effects for other birds. Though, I would consider this more likely to be an effect of breeding. I.e. I've seen brown chicken in captivity. There are different chicken types/races. Lukas 01:47, 19 October 2006 (UTC)[reply]

Perhaps white poultry has been intentionally bred because they are easier to find if they get loose. StuRat 01:52, 19 October 2006 (UTC)[reply]

Right. Anyway, the question is still unanswered. Also, is there a reason why lab rats are white? --Russoc4 02:11, 19 October 2006 (UTC)[reply]

If StuRats are white, why shouldn't lab rats be white ? :-) StuRat 23:33, 20 October 2006 (UTC)[reply]

I also don't think they 'lose' their pigment, I think they've been bread that way. Maybe because brown chickens camouflage better, so are harder to find if they escape the coop, as opposed to white birds which would stand out more. Also, white feathers are more aesthetically pleasing to use as stuffing for pillows and duvets. There are probably more reasons. Vespine 02:14, 19 October 2006 (UTC)[reply]

Is the poultry bred to be breaded poultry ? :-) StuRat 02:25, 19 October 2006 (UTC)[reply]

Follow on: As for lab rats my guess is to tell them apart from wild rats? Vespine 02:15, 19 October 2006 (UTC)[reply]

So...they're white "bread". So I guess the brown ones are whole wheat? Clarityfiend 02:27, 19 October 2006 (UTC)[reply]

haha actually, can I say the pun was intended? No, it was a typo;) read:bred ;) Vespine 02:34, 19 October 2006 (UTC)[reply]

Aren't lab rats albinos? Lab animals are often bred to have specific deficiencies. About poultry, there are plenty non-white domesticated ones and there are white wild birds. But there do seem to be more white domesticated ones, so you still have a point. DirkvdM 08:46, 19 October 2006 (UTC)[reply]

With the exception of flamingos and some amphibia (which obtain their colouring from their food), the reason that we have albino or leucistic) domestic animals is simply from selective breeding. There are naturally found albino and leucistic animals of pretty much every species, although they are pretty rare. For whatever reason - probably aesthetic - humans have selected these and bred them so the are more common in domestic species (check out the number of white dogs around, for example). Lab animals are a slightly different story. Contrary to popular belief not all lab rats and mice are white, plenty of different strains are used that are brown (or "agouti"), black and white. There are good technical reasons people used different strains and different colours in experiments, and the genes that cause the differences in pigmentation are a very good model for studying genetics. Rockpocket 18:09, 19 October 2006 (UTC)[reply]

Alas! My question has finally be answered! --Russoc4 22:21, 19 October 2006 (UTC)[reply]

Glad i could help. Rockpocket 05:57, 20 October 2006 (UTC)[reply]

To celebrate, go and rent Pink Flamingos. Actually, no, don't, you would regret it. Seriously. JackofOz 08:27, 20 October 2006 (UTC)[reply]

Electromagnetic Pulse

How does an electromagnetic pulse short-circuit electronic devices? Note: I am not planning on building an E-bomb. There is just something about being arrested for terrorism that scares me. T H L 23:38, 18 October 2006 (UTC)[reply]

See induction. Basically, the EM field induces currents in the circuitry, which can cause a number of different problems, particularly when components are exposed to a greater-than-intended voltage, reverse-biased voltage, or an SCR is switched on (see latch-up).

Over voltages and excessive currents basically destroy the electronic components and devices.--Light current 02:02, 19 October 2006 (UTC)[reply]

EMP rapidly releases the magic smoke from electronic devices over a very wide area.

Atlant 13:13, 19 October 2006 (UTC)[reply]

I should really have said that in all cases electromagnetic energy causes failure by dissipating itself in the component and causing all the magic fumes to escape. Just like letting the genie out of the bottle: you can never get it back in 8-(--Light current 15:42, 19 October 2006 (UTC)[reply]

There is of course another, maybe more satisfying way to destroy a component that has (partially) failed. THis technique is called Widlarizing named after the semiconductor circuit genius and father of the op amp (I thikn) Bob Widlar.--Light current 15:46, 19 October 2006 (UTC)[reply]

October 19

How to Read the Periodic Table of the Elements

How do you read the Periodic table of the Elements? From a grade 6 learning the table in science class

The periodic table is quite a complex representation of the fundamental building blocks of matter. This is a pretty decent one online. I have no idea what level of knowledge grade 6 kids have on this topic so I'll explain the VERY basic. The 'ATOMIC NUMBER' is the primary determining factor of what an element is. The atomic number is the number of protons within the atomic nucleus of the element's atoms. The periodic table starts with Hydrogen which has 1 proton, then helium which has 2, lithium has 3 etc... The reason why they are not always together, i.e. 4 is not near 5, is because the elements are not only listed in order of atomic number but also they are grouped with elements which share common properties, i.e. 2 Helium, 10 neon, 18 Argon, 36 Krypton are all noble gasses and form the right hand column, 3 lithium, 11 sodium, 19 potassium are alkali metals and form the left hand column... There is plently more information packed into the table, like atomic weight, radius, density, boiling point, which you can learn about more later, but the above should be more then enough to get you started. Vespine 04:38, 19 October 2006 (UTC)[reply]

The simple Wikipedia article on this really fails at being simple and is probably just as difficult for a 6th grader to understand. I'll see if I can fix that up in a little bit. freshofftheufo ΓΛĿЌ 04:46, 19 October 2006 (UTC)[reply]

I think you can simplify it all you want to "the sixth grade level" but it is an oversimplification. There is a lot more stuff going on than it would be liked to have said, because then it gets too complicated, and most sixth-grade teachers wouldn't know it. Here's my try:

In it's true form it looks like this.

The periodic table of elements is an arrangement to put a general order to the elements. It follows a system of listing the elements in order of atomic number (how many protons are in the nucleus of the atom) going from left to right, and up to down. Each horizontal row is called a period, and each vertical column is called a group. There are seven periods, and eighteen groups, because the pink part that is usually under it is not counted (they don't fit in very well). The periodic table also has many trends that it follows. Obviously, inherently, the lower and farther to the right you go in the table, the bigger the atomic number. For instance, electronegativity (how reactive the element is) generally increases to higher or farther right the element is. The atomic radius (how fat the atom is) generally increases the lower and farther to the left the element is on the table.

I forgot where I was going with that. Oh well, sounds good to me. — X [Mac Davis] (SUPERDESK|Help me improve)05:29, 20 October 2006 (UTC)[reply]

I don't pretend to be fluent in Simple ("Layman's terms? I don't know any layman's terms."), but someone needs to do something about that article. It confuses the atomic number and period number, and then says that the least metallic elements are found at the bottom-right (where the metalloids, the usual dividers, of course suggest the top-right instead). --Tardis 16:21, 20 October 2006 (UTC)[reply]

Producing Manganese (III)

I'm wondering how its possible to prepare a compound with manganese in a 3+ oxidation state. I've read on the internet of 1. either acidifying manganese sulfate in solution with sulfuric acid and then adding permaganate or 2. adding base to a manganese sulfate solution and waiting while air oxides the precipitate formed (or then one can add sulfuric acid to dissolve the precipitate, leaving the ion in solution). I also read that the first method produces a orangey colour solution and the second produces a dark, brown/black solid.
Does anyone know how well these methods work? And how stable would my product be, i.e. could the resultant solution or solid be put into a bottle for use later in other reactions?
Thanks in Advance. and P.S. Just in Case, no this is not a homework question, nor does it particularly, I think, sound like say a textbook question. 72.56.169.205 00:46, 19 October 2006 (UTC)[reply]

i would worry you would get manganese dioxide (a Mn(IV) compound) from permanganate, but it might be controllable i guess. typically Mn(III)-salen complexes are made by bubbling air through a solution of the corresponding Mn(II)-salen complex in ethanol. i imagine this procedure would be broadly applicable (it's also used to prepare some Co(III) complexes in the same way). an example is in J. Org. Chem. (2006) 1449-57; there are many more out there. Xcomradex 01:08, 19 October 2006 (UTC)[reply]

Traditionally unstable oxidation states of any transition metal (like manganese(III)) can be stabilised by precipitation or by complexing. Check the electrode potentials, but I am sure that you will find that, in aqueous solution, manganese(III) is unstable with respect to disproportionation. --G N Frykman 07:47, 19 October 2006 (UTC)[reply]

Time Dilation Confusion

I don't understand what a 1G acceleration means in the time dilation article. Here it is: Indeed, a constant 1G acceleration would permit humans to circumnavigate the known universe (with a radius of some 13.7 billion light years) in one human lifetime.

1 g is the acceleration due to gravity at Earth's surface: 9.8 m/s². —Keenan Pepper 02:05, 19 October 2006 (UTC)[reply]

I don't really understand either. Let's say you're travelling at 299 792 457 m/s. Since it's impossible to gain more than 1 m/s of speed, what does it mean to accelerate at 1 g? How can you accelerate at a constant 9.8 m/s when doing so would mean travelling faster than light? --Bowlhover 03:20, 19 October 2006 (UTC)[reply]

It sounds to me like a weak statment making a lot of assumptions and ifs, i.e. "if it was somehow possible to accelerate infinitely and if there was enough energy to propel you and if you didn't have to worry about relativistic effects and if you could somehow turn at relatively extreme angles without worrying about dying, then it could be possible". freshofftheufo ΓΛĿЌ 04:43, 19 October 2006 (UTC)[reply]

If you kept relativistic effects then you only need to scoff at the idea of enough energy and the idea of turning at relativistic speeds. But if you accelerated fast enough, you could travel the length of the universe's circumference several times over before you died. Æµ§œš¹ _{[aɪm ˈfɻɛ̃ⁿdˡi]} 05:45, 19 October 2006 (UTC)[reply]

Bowlhover might think of it in terms of Einstein's thought-experiment of an elevator in an infinitly long shaft. A steady acceleration of 1G is what would make it feel as if the elevator was were not moving and you were on the surface of the Earth.
Now, here's the difference between Newtonian and Eisteinian physics. In Classical physics, a steady acceleration results in a straight-line growth of your velocity; Relativistic physics has it that as you approach the speed of light, a steady acceleration increases your velocity by smaller and smaller increments, and so you never get to go faster than light.
B00P 07:28, 19 October 2006 (UTC)[reply]

Doesn't a "steady" acceleration mean that your speed is always increasing at the same rate? Or does it mean that you're always applying the same force to the accelerating object? --Bowlhover 16:16, 19 October 2006 (UTC)[reply]

Of course, a 1G acceleration would never get you off the ground. Assuming you start from the surface of the Earth. But even if you start from space, you'd have to navigate such that you always follow a line that lies where the gravity of surrounding objects cancels out (sort of follow a gravitational ridge), because else you'd eventually get caught by some star. DirkvdM 08:57, 19 October 2006 (UTC)[reply]

Of course a 1G acceleration would get me off the ground. If I'm moving relative to the Earth, I can get off-planet. If I'm accelerating, that's even better. (Objects only accelerate downloads at 9.8 m/s if they're in free-fall. A rocket firing its engines is not falling.) --Bowlhover 16:16, 19 October 2006 (UTC)[reply]

So long as your kinetic energy exceeds your (negative) gravitational potential energy with respect to some object (like your capturing star), you won't be "caught". (See the virial theorem, or just think about reversibility, for why direction doesn't matter and you won't accidentally end up in orbit.) You'll be deflected, certainly, but you're always being deflected. So you can pick any course which (with a margin of safety for the deflections) doesn't intersect something you'd rather not encounter. --Tardis 15:07, 19 October 2006 (UTC)[reply]

Also, Space is pretty empty - I'd say you'd be unlucky to have your path intersect a star - if we assume an average density of 0.1 stars per cubic parsec - typical in our neck of the woods, then perhaps we need to be within 2 solar radii to collide with the star (the deflection is minimal near to c - light passing the Sun is deflected only by about 1 arcsecond). So to get on average 1 star collision, we'd need to have a disc with radius of 2 solar radii to sweep out a volume of roughly 10 cubic parsecs - roughly a path length of 1×10¹⁵ parsecs. Now that's only rough - but even assuming that we need to be within 1 AU of a star to collide, we're still talking of the order of a path length of 1×10¹¹ parsecs. And to even come within 100 AU of a star, you'd have to travel on average 13 Mpc - and that's assuming constant star density - outside of galaxies, star densities are extremely low. Richard B 20:00, 19 October 2006 (UTC)[reply]

To come back to the original question: I suppose the article means that the accelartion is such that the astronauts in the space ship feel a force pulling them towards the stern of their vessel which is the same as they would experience o Earth due to gravitation, i.e. 9.8 Newton per kilogram. It is better to think of it as N/kg than m/s², because we avoid the question whether the spaceship accelerates with 9.8 m/s². And, from the point of view of an observer on the Earth, it indeed does so only at the beginning, until it gets near light speed. Now, you might think that once the ship has reached 99% of light speed not much more will happen. However, note that even though the ship does not get much faster any longer from the Earth observer's point of view, it still gets more and more massive and time on it gets more and more dilated. The latter is the reason why the astronauts can cross the whole (visible) universe: In the Earth observer's reference frame, the time on the space ship passes so slowly, that within the astronauts' livetime of a few decades, several billion year have passed on Earth: enough time to cross the visible universe if one is travelling with nearly the speed of light. Simon A. 22:16, 19 October 2006 (UTC)[reply]

Thanks That was an Enormus help!

And another thing, I don't see how if you're accelerating at 9.8 m/s away from the center of the earth, how you'd get off the ground. I would have thought it's like the bouncing ball paradox, where you drop a rubber ball and it bounces back to 0.6 times the original height, falls again, the 0.6 times that height, and so on, but it never hits 0 because 0.6^infinity is still a number, no matter how infinitely small, but for all pratical purposes, the ball does stop bouncing. So something accelerating at 9.8 m/s might go up a distance of 1/infinity, then the opposing force (gravity) would bring it back down, and the process would repeat, causing (in practicallity) the object to not move. Eh it's been awhile since I took physics. I probably am missing something as simple as -9.8 m/s + 9.8 m/s = 0.0 m/s which made my whole paragraph nonsense =P --Wirbelwind 02:21, 20 October 2006 (UTC)[reply]

If you're really accelerating at 1g away from the center of the earth (as measured by someone on the earth), you're obviously getting off the ground. If you weren't, your acceleration would be 0 instead (witness all the rocks sitting around, not accelerating anywhere). Now, it's true that if we hang a rocket from a string and light it, only to discover its "acceleration" (that is, its thrust divided by its mass) is 1g, we will find that it simply releases all the tension in the string (since the string isn't having to support its weight) and doesn't actually go anywhere. (Of course, presumably as the rocket's mass decreased its "acceleration" would increase and it would eventually start climbing.) This is the confusion: the "acceleration capacity" of an engine must be added to any other forces on it (the string, gravity, wind, etc.) in order to discover the actual acceleration of the object. Further confusion is the "perceived gravity" within an object: for that you have to consider the actual acceleration of the object and any real gravitational fields in the area. It has nothing whatsoever to do with infinitessimals or geometric series. (But do recall that since, in the normal real number system, "infinity" is not a number, neither is

0.6^{\infty }

. And things that "never happen" can still be said to be the limiting state of a system!) Finally, since this was originally a relativistic question (even if only an SR one), I'll note that in GR gravity isn't even a force: instead, we say that whether it's the ground or a pathetic rocket engine that's keeping you from falling, you really are being accelerated (relative to not rest but rather the natural state of motion which is "falling freely toward the earth"). So then we simply count up all the accelerations due to real forces acting on you and that's your "perceived gravity", or acceleration in your own frame. (This is the acceleration that can still be 1g when you're already just a snail's pace below c.) Your acceleration takes on a different value, equivalent in concept to the classical notion (where sitting on the ground is 0, not 1g), if measured by an observer outside any gravitational fields. --Tardis 16:03, 20 October 2006 (UTC)[reply]

Shape of the Universe...

In a previous post, there was mention of circumnnavigation of the Universe. I suppose the "poster" is assuming the Universe is spherical? Could the Universe in fact be a near-perfect sphere? If so, wouldn`t that make it a CLOSED Universe? I tend to think, that with all the evidence about the Universe so far, that it is in fact an OPEN Universe. Anyone care to comment? Thanks Dave 205.188.116.74 02:44, 19 October 2006 (UTC)[reply]

The expression "the known universe" usually (and also here) refers to the observable universe. The part we can observe is spherical – which has nothing to do with a possible curvature of space: it is spherical in the same sense in which a billiard ball is spherical. There is a lot about "the known universe" that we don't know. --Lambiam ^Talk 11:40, 19 October 2006 (UTC)[reply]

See [6] and [7], as well as our article on "shape of the universe". Basically, although the CMB data does not rule out the possibility of a near-spherical universe, it does not support it either. The evidence also doesn't strongly support any other model of the Universe. --Bowlhover 03:38, 19 October 2006 (UTC)[reply]

Fairly interesting stuff Bowlhover. I guess we`ll just have to wait for new evidence promised in the next decade. TYVM Dave 205.188.116.74 04:02, 19 October 2006 (UTC)[reply]

If the "true" curvature of space has an average magnitude of, say 1 in 10⁹⁰ m, no observation ever may establish the difference with global absolute flatness. So even if the universe is finite, we may conceivably never see evidence supporting that. The term "closed" is in my opinion less felicitous, because in the terminology of topology also the infinite flat plane and space are closed. --Lambiam ^Talk 11:14, 19 October 2006 (UTC)[reply]

It is fairly well accepted that the Universe is flat (BOOMERANG). This would rule out the possibility of the Universe being spherical. There are however other topologies in which the Universe can be an unbounded closed system.

PHYSICS,Mechanical Advantage&related topics pertaining to Pulleys

In a single Pulley system where the Pulley with the Load(L) is hung by a string whose one end is tied to a fixed rigid support and the other end is being pulled upward. Now,it is observed that when the Load is pulled vertically up by a distance d,the string is pulled up by the Effort(E) to a distance 2d upward.

Kndly help me understand mathematically how the 2d vertcal shift can be proved as against d vertical shift of the Load L.

Regards,

S Ghosh India

Have you read this: "pulley"? I'm not sure what you mean by mathematically proved? E x 2d = L x d, or 2Ed = Ld therefore E = L/2 perhaps? Vespine 04:45, 19 October 2006 (UTC)[reply]

The amount of work done is Force x Distance. The amount of energy stored as potential energy by lifting the weight must be the same as the work put in (law of conservation of energy). Therefore, if the weight goes up force F x distance D raised, and the work going in which is only F/2 force then it must move twice as far: F x D = F/2 x (D x 2). RJFJR 13:44, 19 October 2006 (UTC)[reply]

Lightning Protection

Any idea on how UPS such as APC UPS provide Lightning Protection to PC'S through Electric lines and telephone lines???

Have you read our article on surge protector? --Shantavira 08:24, 19 October 2006 (UTC)[reply]

Varistor, Transient voltage suppression diode, and Transil will also be helpful. And if you want the ultimate in surge protection, see motor-generator.

Seriously, the biggest data processing shops have an interesting drill: When a lightning storm is approaching, they fire up the auxiliary power (diesel generators, gas turbines, or what-have-you) ahead of time, switch the input side of their UPSs from the utility-purchased mains power to the diesels, and in that way, isolate themselves from most of the possible effects of the lightning. Remember, very-high speed telecommunications circuits are mostly optical fibers anyway, so there's no concern there about lightning-induced transients. And by routinely running the auxiliary power system, they (almost) never get surprised by it failing to start in a real emergency.

Atlant 13:21, 19 October 2006 (UTC)[reply]

Bug Rivet Paradox

This is a paradox in special relativity. It's about trying to kill a bug using a short rivet.

Looking at this url http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/bugrivet.html#c1

My question is this: If I'm that bug, will I be killed? 211.28.178.86 13:15, 19 October 2006 (UTC)[reply]

The way " I " see it, you`re dead either way. If YOU, the bug, does the accelerating, then, that should be clear:your 'hole' isn`t deep enough. On the other hand, if the rivet does the accelerating, you`re 'safe', but only until the rivet slows-down to non-relavistic speeds, and re-lengthens to it`s 'regular' length...ouch! Dave 205.188.116.74 14:06, 19 October 2006 (UTC)[reply]

Yes. These paradoxes are quite popular in introductions to relativity theory, but surprisingly, most text book never spell out the lesson the student is supposed to learn from them, namely: The standard mechanics idealization to assume perfectly rigid bodies breaks down in special relativity. The solution goes as follows: The bug should be worried even if it only considers its own frame of reference. On first sight, the bug might do the following faulty reasoning: At time t=12.9 ps the rivet head hits the wall and the rivet is stopped instantly. Hence it does not have to worry, because the rivet end stops dead before reaching the bottom of the pit, leaving a gap of ??? cm. (I leave out the numbers to leave some work for you to do. ;->) However, there are no rigid bodies. The bottom of the rivet does not notice immediatly that the head has stopped. It will fly on until the signal from the head (as mediated by the electrostatic forces that hold the material of the rivet together) reaches the end. How long does this take? Well, the length of the rivet in the bug' s frame of reference is 0.9 cm / 2.29, and the signal, travelling at light speed, takes ??? ps to reach the end of the rivet. Until the, the end flies on without breaking at its speed of 0.9 c. So, is this enough time to reach and squash the poor bug? Finish the math and you'll know. Simon A. 14:15, 19 October 2006 (UTC)[reply]

Dave: You overlooked that the "regular" length (the rest length) of the rivet is too short to reach the bug. Simon A. 14:19, 19 October 2006 (UTC)[reply]

You`re absolutely correct Simon. I tried to correct but received an 'editing conflict' before I could post. Either way though, the momentum of the bug would surely squash itself. I know, not the point. Very interesting stuff just the same. Thanks Dave 205.188.116.74 14:24, 19 October 2006 (UTC)[reply]

Simon got toward this point, but it's possible to ignore frames of reference altogether here. It is impossible (due to causality restrictions) to arrest an object from the back (as in, grabbing the rivet head) before its front has reached a distance of

L \over c/v-1

from the back end. In this case, that's

{\frac {0.8\,\mathrm {cm} }{2.29(1/0.9-1)}}=3.14\,\mathrm {cm}

. The bug isn't the only thing that's getting squished; the hole itself is going to be tripled in length! (Assuming that its substrate, like most materials, is not in the position to stop a slug of metal going 0.9c...) --Tardis 21:00, 19 October 2006 (UTC)[reply]

Doesn't information propagate through the rivet at the speed of sound instead of the speed of light? --Bowlhover 04:45, 20 October 2006 (UTC)[reply]

If the body was perfectly rigid as it says above, what would the speed of sound be?--Light current 10:25, 20 October 2006 (UTC)[reply]

It will be infinite. Therefore, there's no such thing as a perfectly rigid body. --Bowlhover 12:29, 20 October 2006 (UTC)[reply]

OR.. it will be limited to 'c' in the case of the infinitely rigid body. Take your choice. THey probably work out to the same answer.--Light current 16:36, 21 October 2006 (UTC)[reply]

One can probably postulate some sort of "relativistic rigidity" where objects maintain their shapes to the degree that causality allows. As for the speed of sound, it wasn't actually relevant; I didn't say that you could stop an object at the length I specified, just that you couldn't stop it at any shorter length. --Tardis 15:47, 20 October 2006 (UTC)[reply]

This link says the bug doesnt get squashed ( I think !) http://www.rdrop.com/~half/Creations/Puzzles/pole.and.barn/index.html . Anyway, it gives a more satisying answer than the maleability of a lighspeed rivet, interesting as that approach may be. mmatt

On the other hand, see http://math.ucr.edu/~jdp/Relativity/Bug_Rivet.html for a nice interactive Flash animation by John de Pillis that shows the bug is necessarily skewered. In fact, the animation shows that after collision with the plate, the rivet must "stretch" beyond its at-rest length due to the time delay for the collision signal to travel from the impact point at the rivet head on one end to the leading point at the other end.

air pollution

What is the major pollutants that the epa would like to get rid of in the air?

What is the chemical composition of the pollutants?

Is there any movement to rid the air of these pollutants?

The US Environmental Protection Agency, under control by Republicans, has had it's original mission (to protect the environment) subverted. Their new purpose is to protect industry from environmentalists. For example, they are working to prevent California from enforcing tough environmental legislation. Some common pollutants are ozone (O3), carbon monoxide (CO), carbon dioxide (CO2), and particulates (various compositions). Many of these also occur naturally, but in smaller quantities (CO2) or at higher altitudes (O3). StuRat 14:34, 19 October 2006 (UTC)[reply]

I generally hate the EPA because it's history makes me angry. e.g. DDT and recycling. It is too much to say everything to say what I think and why. I hate that last sentance. But too much. The major pollutants they hate would have to be CO, SO₂, CFCs, and nitrogen oxides. O₃ and CO₂ I would have to say are not in the actionable effective brunt of many initiatives, and much further down on the list, because their acuteness is much lower. Something tells me I'm going to get a disagreement on this. :) Oh well, all for the better anyway. — X [Mac Davis] (SUPERDESK|Help me improve)05:07, 20 October 2006 (UTC)[reply]

Do you have any idea how much CO₂ is released into the atmosphere. The ones you mentioned are certainly a problem, but the amounts of CO₂ are easier to take on first. Unfortunately, the worst polluter (United States) didn't sign the Kyoto Protocol... - Mgm|^(talk) 10:08, 20 October 2006 (UTC)[reply]

Proof of an Open Universe?...

The bug/rivet paradox reminded me of this...A photon travels at the speed of light, always. If it had a wristwatch on, the watch would never tick a single second. Since no time elapses, wouldn`t that place said photon EVERYWHERE in the Universe at once? If so, wouldn`t that make the Universe infinitely bright? Since the Universe is NOT infinitely bright, can one conclude that the Universe is OPEN? Am I missing something obvious here? Thanks, Dave 205.188.116.74 14:40, 19 October 2006 (UTC)[reply]

Where did you go from the constant speed of a photon to time freezing? 130.179.252.41 14:43, 19 October 2006 (UTC)[reply]

Sorry for delay. I got 'blocked'. AOL. If photon is speeding-away from my frame of reference, its 'watch' wouldn`t tic. If watch isn`t 'ticking', wouldn`t the photon go an infinite distance, in ZERO time? Dave 172.130.33.102 15:01, 19 October 2006 (UTC)[reply]

Yeah, you are. It's true that in the "photon's rest frame" (a physically dubious concept, but historically interesting and useful for some thought experiments) that no time passes. However, in everyone else's reference frame, it is moving at the speed of light, a large but finite speed, so from your perspective the photon is not everywhere in the universe at once. I'd also like to note that it's not that photons give off light: they are light. In order to be seen, they have to be absorbed by your eye. Photon is now a featured article, so you can learn a lot more there. -- SCZenz 15:53, 19 October 2006 (UTC)[reply]

Its true; as the photon travels at the speed of light, in one year it travels precisely one light year from the obeservers perspective. La ï ka 18:37, 19 October 2006 (UTC)[reply]

I agree, but, how much 'older' is that photon umm, La ka? 205.188.116.74 18:45, 19 October 2006 (UTC) Dave[reply]

I does not age because it experiences no time. Hence, it also cannot decay. This gives, by the way, a neat explanation why all those fundamental interactions that are mediated by massless gauge bosons, i.e. electromagnetic force and gravitation, have infinite range. But that is another story. Simon A. 21:30, 19 October 2006 (UTC)[reply]

Very interesting, since a few years ago, I took a physics course which dealt with the fundamentals of it, where we had to calculate relative time differences. The example was probably something like, two watches were synchronized on earth, and one was place inside a space shuttle while the other was left on earth. The shuttle went to space, traveling at some great speed (nowhere near the speed of light). The shuttle returns to space and the clocks put side by side, and the one from the shuttle is a minute behind the one on earth. This is more background info in case someone doesn't know what the topic meant. But from what I understand, time still elapses for photons, just at a much slower rate, else it would create the paradox that one light photon is everywhere in space at once, but that's obviously not true, else you can never have a dark room. So in that sense, you can't prove with photons and this theory that the universe has no boundries. --Wirbelwind 01:55, 20 October 2006 (UTC)[reply]

If the photon were wearing a watch, it would see time passing at the same rate as you or I see our watches (while they are still attached to us). However, if it were possible for an observer to see the photons watch he/she would not see it ticking. What the photon would see is everything that ever happened happen at exactly the same time (and at exactly the same place). It would not perceive time to pass more slowley for itself, only that time for everything else will speed up. The Universe being bright in every direction is, however, and interesting problem. It is known as Olbers Paradox. If the article is there it should explain how it is resolved.

I think your thought experiment breaks down at the very beginning. I believe all the Lorentz style equations require a rest mass or a rest frame. I don't think it makes sense to talk about the photon frame of reference with respect to other frames that are related by invariant mass. Also wouldn't the uncertainty pinciple take over and your error would go to infinity if you tried to narrow down the frame of reference? In other words, at inifintely bright (a point) it's infinitely large in which case it's meaningless.

300 channels and nothing on Static?

Hi, all! Two questions: a) WHAT is the nature of TV static? (Is the 'fuzz' a natural behavior for a TV, or are external electromagnetic radiation factors more important?) b) More importantly, WHY is there no article that talks about it in Wikipedia? If there is, then by all means do say so! And if there is not... well, I don't think I have to say more, now do I? Kreachure 17:45, 19 October 2006 (UTC)[reply]

It's a combination of a variety of interfering sources, such as EM, in the cabling, and other forms of interference. I think the dominant one is low Signal-to-noise ratio as well as some electronic noise.

There's an article on White noise, which is what the "static" is usually called. That or "snow". --Charlene.fic 18:28, 19 October 2006 (UTC)[reply]

Believe it or not, some of those 'specks' you see, when tuned to an 'off' channel, are cosmic waves! Dave 205.188.116.74 18:39, 19 October 2006 (UTC)[reply]

That would be a cosmic ray, which isn't exactly the same as a wave. But it's true cosmic rays can cause some noise (of almost-negligible effect, though).

That's peachy! Question a) answered. Now, how come none of the articles you mention even have the word television in them? Upon further examination, there doesn't seem to be any explanation of how the many sources of radiation from the universe affect TV reception, and thus create the famous TV static... are we supposed to conclude the fuzz effect from the random stimulus of all signal frequencies of vibration on our own? Do any of those articles explain why is it black and white fuzz dancing around? I think not, and if a simple question like this isn't answered in Wikipedia, then (I guess I did need to say it) there has to be an article on that, or at least a mention on any article related to this subject. So, I recommend to anyone who is familiar with this subject to Be bold and create TV static (or whatever name you want to put it)! I'm sure people who are more interested in data compression stochastic processes or telecommunications in general than me, would be very thankful to have an encyclopedia that talks about this common subject. (Otherwise, why do you answer questions like this for? So that it reflects here only, and not in the improvement of Wikipedia?) Kreachure 22:08, 19 October 2006 (UTC)[reply]

We have the stubby article Noise (video). --Lambiam ^Talk 13:31, 20 October 2006 (UTC)[reply]

dimensions of time

Can anybody get their head around what it would be like to exist in a universe with more than one dimension of time. What would it be like ?mmatt 19:08, 19 October 2006 (UTC)[reply]

You'll have to first meaningfully quantify what a "second dimension" of time would be. Otherwise it's just gobbledygook. — Lomn 20:17, 19 October 2006 (UTC)[reply]

Same as a second dimension of space would be. You would have two dimensions of time to travel through. So it would be more like a plane of time. Maybe you could 'go around' a point in time same way as you could go around a point in space if you have two dimensions? What would it mean to be able to see and even travel in two directions in time as you can do in two dimensions of space? I dont know what the implications are or is it even possible to comprehend - Thats why I asked. Maybe the answer is that its gobbledygook. Is that what you think ? mmatt 20:33, 19 October 2006 (UTC)[reply]

Objects would still navigate linearly through the timeplane (how boring, nothing exciting) however, it is possible to fit an infinitely long line of time into a finite are of time plane. Be we do not how objects would navigate the timplane, wether it would random, or controllable, or set. Philc _T^E_C^I 20:50, 19 October 2006 (UTC)[reply]

Hindu time is circular. -THB 21:05, 19 October 2006 (UTC)[reply]

Yeh, but thats religion, not science. Philc _T^E_C^I 21:33, 19 October 2006 (UTC)[reply]

See F-theory for a bit of info about extra time dimensions. JMiall 21:41, 19 October 2006 (UTC)[reply]

With issues like "what is the nature of time?" science too has fundamentally metaphysical and axiomatic answers. --Fastfission 21:40, 19 October 2006 (UTC)[reply]

Correct me if I'm wrong, but I do not believe "What is the nature of time?" is a question to be answered by physics. Perhaps metaphysics. — X [Mac Davis] (SUPERDESK|Help me improve)04:59, 20 October 2006 (UTC)[reply]

I think taking the "dimension" metaphor of time too seriously is misleading here — it is not a "road" or a "channel" or a "plane" that one can just imagine having multiples ones of. Our time and fourth dimension articles have a good description of the two major philosophical views of time, neither of which take the "time is something one walks on" metaphor too seriously. Time as a dimension governs the activities of the spatial dimensions (or is governed by them, depending on how you look at it). "Another dimension" of time might simply mean that under certain conditions these spatial dimensions would exhibit different effects (the increase of entropy outlined in the second law of thermodynamics, for example, might happen faster or slower than it would in the reference frame governed by the other dimension of time, or the second law of thermodynamics might not apply at all, for example). I think getting too wrapped up in spatial metaphors is a sure-fire way to coming up with bad science-fiction approximations, if not gobbledygook. --Fastfission 21:40, 19 October 2006 (UTC)[reply]

Flatland -THB 22:41, 19 October 2006 (UTC)[reply]

Inside a black hole, t becomes a spacelike dimension and the exterior space dimensions become timelike, but that has more to do with what kind of paths you can travel - outside a black hole, you can sit somewhere so that your position doesn't change but your time co-ordinate does, but inside the black hole you've got to travel to the center. The other problem is that it's because you're trying to measure with co-ordinates based on space outside the black hole, so you wouldn't be able to observe it yourself - the time co-ordinate you take with you would stay timelike. Confusing Manifestation 00:32, 20 October 2006 (UTC)[reply]

The following quote from the article F-theory is of interest: "Max Tegmark has written a paper arguing that life cannot exist in a universe with more than one macroscopic temporal dimension, because differential equations would not be hyperbolic in such a universe, rendering prediction of "future" states impossible". For the same reason, it is not possible to have a notion of "what it is like" to exist in such a universe. --Lambiam ^Talk 13:52, 20 October 2006 (UTC)[reply]

Isn't this related to the notion of parallel universes? At every point in time you can take different directions, splitting reality up in many copies that are ever so slightly different, but then diverge ever more (or possibly come back together again?). We might actually live in such a reality, but we can only see one timeline. Just like a flatlander can see something moving in three dimensions instantly diappear. Except that that happens at the same moment they see it appear - so they can't really see it. Does this make sense? DirkvdM 08:40, 22 October 2006 (UTC)[reply]

Current/Voltage

If i compared electricity in a wire to water in a tube, and voltage was the amount of water in the tube, what would current/amperage be? would it be something like pressure, or am I just not understanding this. Ilikefood 21:15, 19 October 2006 (UTC)[reply]

Voltage is the speed of the water. Current would be the amount of water flowing through the tube, per second. --Bowlhover 22:14, 19 October 2006 (UTC)[reply]

Voltage is probably most analogous to pressure, and current is essentially volume of flow. TenOfAllTrades(talk) 22:22, 19 October 2006 (UTC)[reply]

Ilikefood...It might be interesting to note that other analogies can be made, such as: a bucket with a hole in it, capacitor; a check-valve, diode/rectifier; a neoprene 'balloon', somewhere along a pipe, might be akin to a coil, etc...Dave 172.130.33.102 22:52, 19 October 2006 (UTC)[reply]

Since voltage is also known as potential, it's similar to the water pressure that can force water through a pipe. Current is often compared to the flow of water in a pipe, in many many electrical engineering books, and I've seen a few of them. --Wirbelwind 01:38, 20 October 2006 (UTC)[reply]

Thanks. Ilikefood 18:17, 20 October 2006 (UTC)[reply]

Fusion

Could someone create a fusion reaction and just keep feeding it with fuel to create a stable fusion reaction?

Sure...it`s called a STAR. Sorry. Despite the bad joke, I`ll still sign. Wish you would too. Dave 205.188.116.74 23:59, 19 October 2006 (UTC)[reply]

That is the goal, but so far it's not been so easy. See tokamak for one possible design of such a reactor. StuRat 01:41, 20 October 2006 (UTC)[reply]

Sorry and thanks for that.68.120.225.121 06:28, 21 October 2006 (UTC)[reply]

October 20

The fourth dimension?

I've read the article and I still don't understand it. Is it actually physically possible for the human brain to directly visualize a 4-dimensional object, or is it one of those things like directly visualizing the concept of literal 'nothingness' that is beyond our capabilities? --Kurt Shaped Box 00:55, 20 October 2006 (UTC)[reply]

It sounds pretty much beyond our capabilities to me, and I'd be pretty suspicious of anyone who claimed they could visualize otherwise (though, of course, it would be impossible to really know if they were telling the truth). We have brains hard-wired for three spatial dimensions + a time dimension. Anything beyond that is not really in our "intuitive" hardware, I don't think. We can create analogies and simulations and calculations and etc. but all of that are poor substitutes to any sort of real way to directly visualize a fourth spatial dimension. --Fastfission 01:10, 20 October 2006 (UTC)[reply]

No, it's not possible because we live in a 3-dimensional world. See Flatland and Spaceland (book). -THB

The theory goes, you can't see past your own dimension. Like, a point can't see a line, a line can't see a 3D shape, 3D (us) can't see 4D. --Wirbelwind 01:28, 20 October 2006 (UTC)[reply]

Or like somebody who is completely colorblind would find it impossible to visualize what color looks like, even though they could understand it theoretically as light at certain wavelengths. StuRat 01:35, 20 October 2006 (UTC)[reply]

I've read somewhere that blind mathematicians actually had a pretty good intuition of 4th dimension and topology. If I recall correctly, they mentioned Bernard Morin among others. ☢ Ҡi∊ff⌇↯ 01:38, 20 October 2006 (UTC)[reply]

But can you even say they even know what the third dimension is supposed to be like, much less the 4th? --Wirbelwind 02:24, 20 October 2006 (UTC)[reply]

Since when is the third dimension dependant of vision? The point here is that they have a more developed spatial intuition because they have to build 3-D objects in their minds. They already visualize (aw crap) the third dimension in a more abstract way than we do, so perhaps they are better prepared to visualize higher dimensions. ☢ Ҡi∊ff⌇↯ 02:30, 20 October 2006 (UTC)[reply]

What I mean is how do you know they visualize 3D correctly, because if they don't, they probably didn't visualize 4D correctly. Their way of seeing 4D is probably similar to understanding 4 dimensional arrays in programming, which is hardly a visual. --Wirbelwind 02:38, 20 October 2006 (UTC)[reply]

In the end, I find it somewhat questionable about whether we really see in 3-D ourselves. I mean, our eyes individually give 2-D images of the world which are compiled into a 3-D understanding (sometimes quite a wrong one, as optical illusions point out) of the world. But perhaps we'd be dropping down a rabbit hole here. (Honestly seeing in 1-D makes almost no sense to me. 2-D makes a lot of sense, because you open up a plane. 3-D makes conceptual sense but I haven't seen any way to visualize it that didn't involve building it up from composite 3-D images. You can, however, perceive the 3-D directly if you open up perception to include non-visual sources, I suppose. But by this point I think I'm falling down that rabbit hole I mentioned...) --Fastfission 02:54, 20 October 2006 (UTC)[reply]

Actually, that a very good point. That reminded me of some religious argument where they said we live in 3D so we can see below us, or 2D and 1D. God or whoever is in 4D and thus can see 3D, or everything in the physical world. But that really is another subject, and one I'm much less interested in =P --Wirbelwind 02:59, 20 October 2006 (UTC)[reply]

Eh, that is assumed for sure that a four dimensional "being" would have great more power over us. Imagine a three dimensional being seeing over a two dimensional world. No safe or prision is safe, because it is just a square. If you get what I mean. The possibilities are endless, since the 4D being could see and manipulate anything physical in the third dimension without restraint. It's tough to explain. Hyperspace does it well. — X [Mac Davis] (SUPERDESK|Help me improve)04:55, 20 October 2006 (UTC)[reply]

Wirbelwind, what did you mean by "visualising 3D correctly"? What would be the correct perception of it, and who gets to decide that? Also, I don't think it's possible to be categorical about the dimensionality of God. In fact, I'm sure it's not. JackofOz 08:17, 20 October 2006 (UTC)[reply]

Some people claim to be able to visualise a closed universe. I simply can't do that, because the concept of space/time/space-time coming to an end somewhere/when, offends my notion of an all-ness that is somehow not all. But that's not to say others can't conceptualise these things. They say they can, so who am I to deny that? Same with a 4D object. For me, it's just too strange to understand visually, but I'm sure some people can do it, or they know what it would "look" like if they could, which amounts to the same thing. JackofOz 08:17, 20 October 2006 (UTC)[reply]

I've always believed that in reality we simply won't be able to interact with a 4 dimensional being. It can't affect us and we can't affect it. Nil Einne 12:26, 20 October 2006 (UTC)[reply]

If the n dimensions of a being intersect with the m dimensions of a different being and they have l dimensions in common, they should be able to interact in those l dimensions..... Paul venter 15:31, 26 October 2006 (UTC)[reply]

While it's difficult to say what the human mind is capable of, I would say no assuming you're talking about a fourth spatial dimension. However I would say humans are capable of visualising four dimensions in terms of the 3 'spatial' dimensions and 'time' as in Eistein's theory or relativity... Nil Einne 12:12, 20 October 2006 (UTC)[reply]

I actually saw The Fifth Dimension , close up and in person.Edison 23:30, 20 October 2006 (UTC)[reply]

I think it's possible. 'Visualize' as in 'imagine' and not 'visualize' as in 'see'. The concept of "see" in itself is a 3D concept - our eyes and the way our mind processes visual information is in a 3D 'template.' But i think it's possible to imagine what a fourth dimention would be like. I think shows it nicely. You can't really imagine how the 4D cube would look like - because the way we percieve objects is competely 3D. But if you look at the 4D cube carefully, it's easy to understand it. Each point has four perpendicular lines coming off it, one for each spatial dimention. you can't 'see' it, but you could 'imagine' how it would exist if the rest of the world was also 4D.

Just another thought - if we can accurately use a 2D plane to draw a convincing representation of a 3D object, shouldn't it be very possible to use a 3D plane to construct a convincing representation of a 4D object? --`/aksha 13:01, 21 October 2006 (UTC)[reply]

It is a bit more complicated than that. We being able to draw a convincing 3D object on a 2D plane is a form of optical illusion; one made possible by our way of perceiving 3 dimensions outside of that 2D plain. Seeing as we are only able to see 3 dimensions through shading, itself a kind of optical illusion, that would extremely difficult, if not impossible, to do. We would have to create an optical illusion inside of an optical illusion. It may not be impossible, but... T H L 02:27, 22 October 2006 (UTC)[reply]

Just like a cube's 'shadow' on paper is two squares with the corners connected, a third dimensional 'shadow' of a hypercube would be two cubes with their corners connected. A nice visualisation-aid is Image:Hcube fold.gif DirkvdM 08:56, 22 October 2006 (UTC)[reply]

Yes, but the line would have to be connected at a 90 degree angle to the others; which we would not be able to perceive. Representing one is possible, as the image you linked to and the one above shows, but actually creating a convincing recreation of one in 3D is another story. T H L 15:51, 22 October 2006 (UTC)[reply]

What, build a real model of a hypercube in 3D, complete with right angles? That would of course be as impossible as drawing a 2D version of a cube, with all right angles. DirkvdM 19:11, 22 October 2006 (UTC)[reply]

Sorry, I lost myself for a second. I hate it when I do that. Yes, that could be done; I'm not sure how convincing it would be, but it could be done with supplies as simple as straws and marshmallows. T H L 21:22, 22 October 2006 (UTC)[reply]

ossification and calcification

what is the relationship between the process of ossification and calcification?

See ossification and calcification. And sign your posts!!! -THB 02:46, 20 October 2006 (UTC)[reply]

Combustion of BIO Gas...(Didn't get an answer yet!!)

I'm working on an anaerobic digester design and i found that the CH4:CO2 ratio is between 1:1 and 3:2. In addition it may contain Hydrogen Sulphide, Nitrogen, Amonia and Hydrogen in small amounts.

I'm planning to utilize this gas to produce CO2 to be used in another industry. So i would like to know further information about following issues.

1. will it be effective to combust methane+CO2+other trace gasses mixture to produce CO2 (will the excess CO2 affect the complete combustion?)

Do you mean you want to combust methane and O2 ? StuRat 03:49, 20 October 2006 (UTC)[reply]

2. What will be the products after combustion of the above gas mixture?

3. what will happen to NH3 and H2S after combustion?

It is planned to combust the mixture in lower temperature (Less than 200 C)

Appreciate any suggestion to success my effort and any links regarding this process will be really helpful to me...!! Thank you. --192.248.8.100 03:41, 20 October 2006 (UTC)Sithara, SL[reply]

rechargeable batteries

Why are the voltages lower on rechargeable batteries than on non-rechargeables? ie, AA alkaline is 1.5 volts, while AA NiH rechargeable is 1.2 volts

Please sign your posts. Have you tried reading our page on rechargable batteries? I don't know if it answers your question but it's a start. However this is a very simple and basic topic in chemistry. I suggest you start at Electrochemical cell. Unfortunately, it doesn't look like we cover this that well at the moment but any basic chemistry book will cover this and there must be a ton of websites as well Nil Einne 12:15, 20 October 2006 (UTC)[reply]

Cell voltage depends entirely on the particular chemical reactions that the battery designer chooses. And while the NiCd and NiMH reactions are pretty low voltage, this isn't true of all rechargeable battery chemistries. Lead acid cells tend to run about 2.0 V and lithium cells tend to run about 3.6 V.

Atlant 18:45, 23 October 2006 (UTC)[reply]

The Sun Core

Hi, the question is how many years does it take for heat generated in the Sun's core to reach its outer layers, the photosphere? Thank you.

--202.184.216.70 07:46, 20 October 2006 (UTC)[reply]

Our article on the Sun covers it. ☢ Ҡi∊ff⌇↯ 07:59, 20 October 2006 (UTC)[reply]

Puzzling physics question

This is for a student I am helping.

An iron cube floats on mercury at 0°C. The temperature is increased to 30°C. a) Does the cube float higher or lower in the mercury? b) What is the percentage change in the volume of submerged iron?

I assume that the iron would sink somewhat, intuitively I suppose. But I'm not sure how to calculate the % change in submurged volume. I assume that you would use, in part, the change-in-volume equation, i.e., ΔV = β V₀ ΔT. The β values for iron and mercury are known. BenC7 10:18, 20 October 2006 (UTC)[reply]

THe upthrust equals the weight of fluid displaced (Archies principle)
THe weight of liquid displaced is the depth times csa of cube times its density.
does the density of mercury change between 0 and 30 celscius?

OTOH, is this a surface tension problem?--Light current 10:39, 20 October 2006 (UTC)[reply]

No. It is part of the physics curriculum dealing with heat. And the mass and size of both metals are unknown. BenC7 10:42, 20 October 2006 (UTC)[reply]

Ah, so its just the temp of the cube thats raised (not the mercury)?--Light current 10:45, 20 October 2006 (UTC)[reply]

Remember that both the iron cube and the mercury will expand, and that will change their densities. The expansion rates will be enough to tell if it will float more or sink when heated up. ☢ Ҡi∊ff⌇↯ 10:46, 20 October 2006 (UTC)[reply]

The question isn't terribly clear on whether just the mercury, or the whole system changes in temperature. I assume that it will be the whole system. OK, so let's say, for argument's sake, that the β value for iron is 3 times as much as that for mercury. So it will come up higher in the mercury. What does this mean for the percentage change? Is it possible to work the question out without knowing the density? BenC7 10:51, 20 October 2006 (UTC)[reply]

Yes, I think it is. If you find out the iron expands 3 times more than the mercury, its density will drop proportionally more than the mercury, therefore it will float even better than before. You can use the variation of the proportion of densities to work the variation of the percentage of immersion. ☢ Ҡi∊ff⌇↯ 10:54, 20 October 2006 (UTC)[reply]

OK - the β for mercury is 20; for iron it is 4. What is the answer, then? BenC7 10:59, 20 October 2006 (UTC)[reply]

Consider that an object with density

D_{1}

is immersed in a liquid with density

D_{2}

, floating with a certain N% of it submerged. If I halve

D_{1}

, N will also be halved (it floats twice as much). If I make the liquid twice as dense (double

D_{2}

) N will also be halved. Therefore, we have a multiplier for N given by

{\frac {D_{1}}{D_{2}}}

.

Now, given the above coefficients of expansion, we have the relative expanded volumes for 30° C as:

V_{mercury}=1+30\times 20\times 10^{-6}=1+6\times 10^{-4}

V_{iron}=1+30\times 4\times 10^{-6}=1+1.2\times 10^{-4}

Since densities are inversely proportional to the volumes, their relative densities at 30° C are then:

D_{mercury}={\frac {1}{V_{mercury}}}={\frac {1}{1+6\times 10^{-4}}}

D_{iron}={\frac {1}{V_{iron}}}={\frac {1}{1+1.2\times 10^{-4}}}

(these values are the multipliers that give us the density of each material at 30° when multiplied by their original density at 0° - we don't have that, so we work with proportions)

Throwing that in the first equation, we have:

N=N_{0}{\frac {D_{iron}}{D_{mercury}}}\approx 1.00047994N_{0}

Which means a 100.048% ~~change in~~ of the submersion amount from before, that is, it will sink (higher submersion than before) by 0.048%.

I'm not too confident about this result, though (I'm a bit rusty with these things, and it's been a bad week... so sorry about that) but that's basically the sort of thing you'll have to do here. Hope it was any help. ☢ Ҡi∊ff⌇↯ 12:09, 20 October 2006 (UTC)[reply]

Correction, that means the cube has 100.048% OF the submersion it had before, or a 0.048% change in the amount of submersion from before. StuRat 13:02, 20 October 2006 (UTC)[reply]

Good one! Thanks for the correction. I often get confused with some of the use of "from"\"of" "in"\"on" in english. ☢ Ҡi∊ff⌇↯ 13:58, 20 October 2006 (UTC)[reply]

No problem, lots of ads seem to get this wrong, too, along with saying "get 10 items ALL for $1" when they really mean "get 10 items EACH for $1". StuRat 14:38, 20 October 2006 (UTC)[reply]

Hmm, I accidentally divided the α values by three to get the β values instead of multiplying them. So in actual fact they are 180 x 10^-6 for Hg and 36 x 10^-6 for Fe. I think this makes the answer a 0.4% change. Sounds plausible. Thanks guys. BenC7 05:16, 22 October 2006 (UTC)[reply]

Metabolism etc.

It says in a number of health-related books that "If your energy input exceeds your energy output, you will gain weight. If your energy output exceeds your energy input, you will lose weight." I can understand that in many cases this would be true. But what about for skinny people (like myself), who seem to be able to eat large amounts of food and never put on weight? BenC7 10:46, 20 October 2006 (UTC)[reply]

Your enegry output equals your energy input? (Or you have a tapeworm- 8-)-Light current 11:02, 20 October 2006 (UTC)[reply]

Try reading a bit on metabolism and stuff. Especially Basal metabolic rate (which coincidently looks like it might need some improvements to make it more encylopaedic) Nil Einne 12:22, 20 October 2006 (UTC)[reply]

Some possible mechanisms for being able to eat a large amount without gaining weight:

You expend more energy than others. Some people seem to be always moving. Is this you ?

The food passes through you largely undigested. StuRat 12:56, 20 October 2006 (UTC)[reply]

Or: The food is digested just as for most people, but your liver does not convert excess energy into triglycerides that are stored in adipose tissue. --Lambiam ^Talk 14:17, 20 October 2006 (UTC)[reply]

"If your energy input exceeds your energy output, you will gain weight. If your energy output exceeds your energy input, you will lose weight." that statement is actually a very simplistic explaination of metabolism and weight. Whether you put on weight or gain weight is VERY complicated. Genetics, age, health, and a lot of other factors which are not directly related to metabolism affects weight. Also, being skinny doesn't mean you have a low energy output. You may want to take a look at Obesity, which does discuss how weight lose/weight gain works. --`/aksha 12:49, 21 October 2006 (UTC)[reply]

OK, so if the liver does not convert the excess energy into fats (to simplify), but digestion occurs the same as in most people, where does the excess energy go? BenC7 05:31, 22 October 2006 (UTC)[reply]

Down the drain when you flush the toilet. --Lambiam ^Talk 19:08, 22 October 2006 (UTC)[reply]

Beta particles

It's been a while since I've done physics so forgive me if this is a silly question. But someone was claiming that CRT monitors emit beta particles. I know how CRT monitors work and I also think I know what beta particles are. However I don't know if he is correct or not. Are high speed electrons only considered beta particles when emitted by beta decay? If so the answer would be they're not. But what is the difference between beta particles and electrons from an 'electron gun'. My guess is from the CRT, the velocity is very different and the number/concentration is a lot less but I don't know. Don't get me wrong, I'm not worried that I'm going to get cancer or anything I'm just wondering about the technicalities... Nil Einne 12:22, 20 October 2006 (UTC)[reply]

No need to worry. The electrons are completely absorbed by the thick glass in front. The guy your were talking with might have been worrying about X rays. As you might know, these are produced in an X-ray tube by accelerating electrons and smashing them into the anode. The energies involved are comparable, and hence, a CRT tube in fact produces X rays. As the glass of the tube should not only shield you from the cathode ray (i.e. the elctrons) but also from these X rays, it is very thick glass which contains lead. Nevertheless, customers are concerned by it, especially those working a lot in front of a CRT monitor, and hence, manufacturers of CRT monitors like to proudly put a sticker on their product that they comply with the TCO standard. Simon A. 13:43, 20 October 2006 (UTC)[reply]

And the term "beta particle" for an electron (or positron) is indeed normally only used in the context of radioactivity in the form of beta decay. The only relevant thing that may be physically different from other electrons is the high energy, which is fully determined by the speed of the particle. --Lambiam ^Talk 14:28, 20 October 2006 (UTC)[reply]

Human Body

Where is the actual centre of gravity to be found in the human body

I would think somewhere around the small of the back, along the centre line, but its giong to depend on the distribution of fat on your body.--Light current 17:10, 20 October 2006 (UTC)[reply]

It's somewhere in the lower abdomen, just above the pubic area. If you've seen dancers, ice skaters, and acrobats lifting their partners, you might notice the place where they have to hold their partners to balance them (the CG point), is a bit uncomfortably close to the pubic area. StuRat 17:13, 20 October 2006 (UTC)[reply]

Just as is the case in other physical characteristics, the center of mass varies from person to person. In general, the center is higher in the male adult than the female adult. Also, there is also a correlation between higer center of mass and poor health-- see Metabolic syndrome.--Mark Bornfeld DDS 18:11, 20 October 2006 (UTC)[reply]

The higher the muscle mass (relatively), the closer the center of gravity is to the chest area. The lower the muscle mass, the closer the CoG is to the abdomen. Note how skinny the figure skaters being held usually are. A popular demonstration of this fact used in biology/anatomy/physics classes (depending on which class the school teaches about the CoG in) is: To have a boy, usually a football player, put his head against a wall bending his back to an 80 degree angle. Then, have the boy grab a box weighing ~30 pounds (~13.6 kg), and try to stand up without moving his feet. He shouldn't be able to do it. After that, have a weak-looking girl try; she should be able to do it with ease. After the class finishes laughing, explain why this happened. T H L 22:50, 20 October 2006 (UTC)[reply]

Depends on where those muscles are. Speaking of football players, I know one who is all muscles below the waist and skinny above that.(Then again, you were probably not talking about the original football. :) ) Also, he's Dutch, so the cycling will also help in strong leg muscles. Something similar will be the case with skaters, which would explain why they have to be held up near the groin, as StuRat pointed out. DirkvdM 09:11, 22 October 2006 (UTC)[reply]

Right, I forgot that everyone else calls a different sport football ;). In the general population, meaning people who do not do athletics professionally, muscle mass tends to be distributed in such a way that the higher the muscle mass compared to the overall mass of the body, the closer the center of gravity is to the chest, and vice versa. Professional athletes probably aren't the best examples, seeing as many train one area of the body harder than the others; creating more exceptions. If every area of the body is trained equally, and they can lift the same amount pound/kilo for pound/kilo, then the center of gravity will move closer to the chest as the muscle mass increases. The greater amount of testosterone in men tends to cause the center of gravity to be closer to their chest; the exception being men like me who sit at the computer eating Tostitos and drinking Coke all day; so basically American men age 40-death are the exception. Anyway, muscle distribution tends to be equal enough for this to be the norm. T H L 21:45, 22 October 2006 (UTC)[reply]

THL: Middle-aged men might feel a little less creeped out by being categorized as "40 and up", as opposed to "age 40-death". :-) StuRat 00:36, 23 October 2006 (UTC)[reply]

I'll keep that in mind ;) T H L 01:19, 23 October 2006 (UTC)[reply]

The Past

Does the past exist. If it does is it frozen or always changing. 64.119.104.205 16:13, 20 October 2006 (UTC)bgoldie[reply]

Well, if the future exists now, then (as the present also exists-- obviously), it follows that the past also exists. This is becuase the future's past includes the present, and the present is the future as seen from the past! We believe the future can be changed becuase we have not reached it yet and the present is changeable. So to change the past , you would have to make it at least the present by going back there. Do I make myself clear? 8-)--Light current 17:19, 20 October 2006 (UTC)[reply]

But the future doesn't exist now. "Exist" is a present tense verb. The past is that which once "existed", but no longer exists. Definitions of existence vary, but in normal usage things "exist" only in the present. --Shantavira 17:28, 20 October 2006 (UTC)[reply]

)edcon)

I dont think its proven that the future does not actually exist now. Anyway the future will exist (probably) and at that time the present will be the past. If the future doent exist now, how can we hope to travel there other than at the same old rate?--Light current 17:59, 20 October 2006 (UTC)[reply]

(This is one perception of the universe, enabling me to answer your question)

We here touch upon the fourth dimension, that in which objects no longer simply exist as lengths, but (seen from our view) start to move. The past you are looking for is a point in this dimension which has certain qualities, so yes, it has existed. In the fourth dimension, the progress from that point in time to right now, has become a line - you may say that this line has snaked through a landscape, touching upon different places as it went along. Each second you choose to measure between the past (thirty minutes ago) and now (now) will see the line/snake having progressed a bit beyond what it was at. It changed coordinates each time. These coordinates you may understand as data about what the universe and everything else looked like right then. This is because "past" and "future" are the victims to diffuse semantics, and that your question is completely unanswerable to anyone without a proper language for it. I pray to god some alien civilization comes upon us to resolve these things soon.

Your point along this line of time, or rather just your line in the fourth dimension, will keep progressing. Because of that it is correct to say - at least I would claim so - that the past is frozen. This demands that the universe is as intuitive as I want it to be. Don't count on it. 81.93.102.3 17:56, 20 October 2006 (UTC)[reply]

NOW lasts forever, but NOW has just passed. :) Dave 205.188.116.74 18:06, 20 October 2006 (UTC)[reply]

Seriously though, Stephen Hawking declares that a very tiny fraction of a second, after the Big Bang, The future was 'carved in stone', so to speak...if one could possibly know the position and momentum of every particle, one could conceivably calculate every future event. I`m paraphrasing, of course. If one could suppose that the Universe was 'pre-determined' to be a 'quantum' Universe, even quantum mechanics would be 'covered'. So, if this is all true, then the future already "is"...one cannot change the future! Dave 205.188.116.74 18:23, 20 October 2006 (UTC)[reply]

In some civilisations, the past is in front of us, as we can imagine that we see it clearly ; future stays behinds ... -- DLL ^{.. T} 18:47, 20 October 2006 (UTC)[reply]

What if the past exists and is constantly changing. We're not aware of it because it already happened. The future also exists and is constantly changing. The future has an equal effect on the present as the past. Time, if we could observe it as a 4 dimensional being would be like a guitar string stretching from the big bang till the end of time. It would be constantly changing and vibrating. Einstein said time was an illusion. The relationship between cause and effect is not as we perceive it. bgoldie

I think that is a very interesting theory that bgoldie has just outlined. It tends to make sense to me. I dont know why it should, but it does--Light current 20:16, 20 October 2006 (UTC)[reply]

I have 'proof' that the future already exists. I just made a post AFTER the page was last modified. : ) Dave 205.188.116.74 20:12, 20 October 2006 (UTC)[reply]

If the past is changing, does it have a future? And is that future also changing? Is it perhaps the past of our changing future? Will the past of the future that will have been changed be part of an everchanging present, or is the present frozen? And if the present is changing, when is it changing? Is it changing now or in the future? Can we visualize these changes as the vibrations of the guitar strings being strung/having been strung/willing be strung from the big bang till the gnab gib? Should these vibrations be quantized and unified with string theory? I hope this makes sense to Light current. --Lambiam ^Talk 23:04, 20 October 2006 (UTC)[reply]

Per Philip Gosse's Omphalos (book) we can have no assurance that the world was not created 1 minute ago, so the past might be fake.Edison 23:37, 20 October 2006 (UTC)[reply]

But is it possible that the world has not yet been created and that what we are experiencing is the changing past of that future creation? Also, if the Omphalos (book) was written in the past and not created just one minute ago, it may be changing in the past and have said something very different. We can have no assurance that the world will end in 1 minute. The future may be real. If the past is fake, can it still change or is it frozen? Also, I wonder, does time always flow in the same direction? How do we know it does not occasionally flow backwards or sideways. Or stops flowing. Does the time in the past also flow, or only in the present? Will the questions change when the answers change? These are just a few of the questions I have. --Lambiam ^Talk 00:56, 21 October 2006 (UTC)[reply]

It doesn't matter to us if time stops or not, because we wouldn't notice. It wouldn't matter if it went backwards or "to the side" if possible, or time went faster or slower. Time is the reference, and the only way to know or observe any change in time would have to be from "outside" of time, wouldn't it?X [Mac Davis] (SUPERDESK|Help me improve) 22:33, 21 October 2006 (UTC)[reply]

Subatomic particles have been observed to disappear and then reappear. I believe that these particles are travelling to the past or the future. They become part of another atom in a different time. They are linked (entangled) with this particle. From the point of view of the particle the two times are the same. It is our illusion of "The Present" that makes us think that they are seperate. bgoldie

Tough pool leak...

I know this isn`t 'rocket' science but I`m really at a loss here. I have a particular type of pool. We`re not supposed to use copyright names so let`s just call it a "not very difficult at all" pool. You know, the type that has a ballon, which I`ll refer to here on in as, the bladder, which, once filled with air, simply fill pool with water, and the bladder 'floats' the walls up. Well, my bladder has developed several leaks. With the pool-repair kits that are sold won`t work...when I place a patch on a leak, the internal pressure pushes the patch away. I cannot deflate the bladder as that would empty the pool. i.e. walls would 'fall'. Can anyone here suggest a method that might work, without having to completely deflate bladder? I really don`t want to empty pool. Thanks, Dave 205.188.116.74 17:03, 20 October 2006 (UTC)[reply]

First, the rule about not using copyrighted names doesn't apply to the Ref Desk. Second, I think you will indeed need to deflate the bladders and put the patches on the inside, so the pressure holds them in place. You will then need to let them fully dry and cure before putting any air or water pressure on them. StuRat 17:08, 20 October 2006 (UTC)[reply]

I was afraid of that. Thanks for the heads-up about copyright. Dave 205.188.116.74 17:36, 20 October 2006 (UTC)[reply]

Copyright != trademark. —Keenan Pepper 18:05, 21 October 2006 (UTC)[reply]

Dave,I think I have a Potential Solution to your delmna.Why not try to saturate Q-Tip with glue.Poke Q-tip through hole.Inside pressure should hold in place,repairingleak.That way you won't cut up bladder,creating another hole to patch.Andrea216.218.126.83 17:51, 20 October 2006 (UTC)[reply]

Just so people aren't confused: you can't copyright names. You trademark them. (See intellectual property for a discussion of this confusion.) And it's actually relatively hard to infringe on a trademark, particularly accidentally. (The most common case is treating a trademark as a common noun, a la "Legos" as opposed to "Lego bricks" or even "Lego brand bricks".) See the manual of style; as far as I know that's the only trademark policy on Wikipedia, and it doesn't even dissuade their use. --Tardis 18:08, 20 October 2006 (UTC)[reply]

Oh My! Andrea...I tried your trick. I just went and checked...it WORKED! What a brilliant solution. You`ve saved me a HUGE headache. Thank you so very much. Dave 205.188.116.74 18:31, 20 October 2006 (UTC)[reply]

Nasal mucus

Having a bad cold ATM, this is a question that has just come to mind. What determines the color of nasal mucus. I mean sometimes its clearish, sometimes greyish, and just recently greenish.--Light current 17:08, 20 October 2006 (UTC)[reply]

My understanding is that it is, at least in part, colored by bacterial colonization. I'm not 100% sure of this: a physician would probably be better prepared than I to answer this fully. – Clockwork Soul 19:00, 20 October 2006 (UTC)[reply]

Also Im getting different colors between nostrils. Is that normal?

I dunno, the type of crayon you recently shoved up your nose? What I really wanted to say is that I think ish is a great word for nasal mucus: "I've got some terrible green ish today." Perhaps I should wipe this on the Language Reference Desk. MeltBanana 18:59, 20 October 2006 (UTC)[reply]

Yeah you could wipe some over there. See who bites the monkey. Actually the correct term is liquid snot. but I wanted to be polite! 8-)--Light current 00:17, 21 October 2006 (UTC)[reply]

If requesting medical or legal advice, please consider asking a doctor or lawyer instead. Doctors sometimes ask what color it is to help determine if it is a common cold or a bacterial infection.Edison 23:39, 20 October 2006 (UTC)[reply]

Doc never asked me that. They just said try some of this, try some of that! None has worked yet 8-(--Light current 00:15, 21 October 2006 (UTC)[reply]

When I had bronchitis, I was told by my doctor that producing green-yellow mucus was a sign that it was bacterial, not viral, but according to the article bronchitis, the colour has no bearing on this... La ï ka 10:29, 21 October 2006 (UTC)[reply]

colored snot means there's an infection. It's like if you cut yourself really bad, and the wound gets infected, and you don't treat it...you'll end up with colored crap oozing out of the wound. Same deal - greyish --> yellow --> green snot means some kind of infection, in that order too (as in green = worst).

To clarify, i'm talking about colored snot, not colored mucus. Which is what you're asking, since you can't really tell the color of the mucus. i believe the thing that contributes to the yellow/green factor in snot is actually pus, not mucus. Colored snot (not colored mucus) is an indication of bacterial infection - since pus is produced by the body during inflammatory responses against bacteria. --`/aksha 12:37, 21 October 2006 (UTC)[reply]

Sorry to hijack the thread, but somehow the fact that Wikipedia has an article on snot just seems hilarious to me this morning. 192.168.1.1 10:00, 21 Rocktober 2006 (PST)

That s'not funny! 8-)--Light current 17:07, 21 October 2006 (UTC)[reply]

Wikipedia has everything. (since when did RD questions become threads anyway...) --`/aksha 02:23, 22 October 2006 (UTC)[reply]

Questions about cholesterol and saturated fat

A couple questions that I am interested in mainly from the point of view of my personal health, fully aware of all the caveats about Wikipedia not offering medical advice. Here goes... (If you choose to inline your replies, please use at least two colons of indentation.)

Question 1: If you switch from a diet high in saturates to a diet which is low in fat and in particular low in saturates, what is a typical timescale for blood (LDL) cholesterol levels to come down?

Motivation for question 1. Until three months ago, I had a diet which was moderately high in saturates (dairy fat and hydrogenated vegetable fats), and also my overall calorie intake was high. In mid-July I decided to address this, and increased my fruit and veg intake while reducing fat intake, particularly saturates, controlling overall calories, and also getting more exercise. The health benefits are fairly obvious in that my weight has already dropped from 97kg to a more appropriate 82kg (I am 1.83m tall), but I was wondering specifically about cholesterol. My cholesterol was never previously measured but I can only assume that it was on the high side. Should I bother to get it measured? Or by what stage can I start to assume that it is unlikely still to be high? (I am male, and in my mid-thirties.)

I wouldn't bother measuring blood cholesterol since, as you've said, you have no base reading to compare with. Also, blood cholesterol levels are a measure of the rate of change of your body cholesterol level more than they measure the amount of stored cholesterol. As a result, blood cholesterol may be high when your body cholesterol is either rapidly increasing or decreasing. StuRat 18:29, 20 October 2006 (UTC)[reply]

Question 2: If you consume saturated fat but burn off the calories, can it still raise your cholesterol?

Motivation for question 2. Although it has become very rare for me to eat chocolate and cookies during my normal routine, there can be days when I am out hiking all day, when I need the extra energy. Such foods are a very portable source of extra calories. Is it reasonable on those days to supplement an otherwise balanced diet with some of these sorts of food for the extra energy, and not worry about cholesterol? Or is it still better to avoid them, and instead carry extra bread, nuts etc? (I am aware there may be dental issues too.)

You should still have a healthy diet, no matter how much you exercise. Athletes who eat huge quantities of bacon and eggs are headed for health problems some day. StuRat 18:22, 20 October 2006 (UTC)[reply]

Many thanks. Slarey 17:14, 20 October 2006 (UTC)[reply]

Can't answer the first question exactly, but your cholesterol level should respond (if it does respond) within a month or so. Keep in mind that cardiovascular risk is correlated with several different blood lipids, as well as other measurable blood constituents. Not all of these parameters respond to dietary or other behavioral means of control; some are genetically determined. If you are unfortunate enough to have a heritable hyperlipidemia, dietary control will not be all that effective. Why should you assume your cholesterol is on the high side? In any case, a blood lipid profile is a routine test, and would certainly be part of a typical diagnostic exam for a male in his mid-thirties. As for the occasional consumption of high-fat foods-- for the average person, this would not cause a lasting change in the blood lipids (although it might cause a short-term dramatic rise in the chylomicron fraction of your triglycerides), as long as it is done in the context of an overall prudent diet. Man does not live by bread alone...--Mark Bornfeld DDS 18:31, 20 October 2006 (UTC)[reply]

Many thanks for your help. To answer your question, I simply assumed that cholesterol would have been high because of all the cheese, cookies etc I was eating. Anyway, I will ask for a blood lipid profile to be done at some stage. (I didn't actually realize the test measured more than just cholesterol, or that it was something considered routine in one's mid-30s.) I am sure at the time there will be appropriate advice given when discussing the results, but based on what you say I will probably assume that provided that levels are generally okay I don't need to worry too much about what forms of high-energy snacks I use on the odd occasions when I need them e.g. hill walking. Slarey 18:09, 23 October 2006 (UTC)[reply]

link between: diabetes, CHF, renal insufficiency, amputation

Hi, Im trying to understand how each of these diseases effect each other. If a person has diabetes, how would this influence CHF and renal insufficiency? Any information that would better my understanding woud be much appreciated. Thank you in advance

Jodie

I'll try to be short-winded: Diabetes -> high blood glucose -> atherosclerosis and vascular damage -> tissue damage. Since the kidneys and retinae have blind arteries (without anastomoses) they usually go first, but the tissue of the heart will be damaged too, leading to CHF. Decreased circulation in distal vessels, such as those in the limbs, causes ischemia, which increases the risk of infection, which necessitates amputation in extreme cases. Note that while Diabetes can lead to CHF, limb ischemia and renal insufficiency, these conditions will not cause diabetes. In a patient with CHF and RI from other causes, diabetes will further exacerbate the problems. Tuckerekcut 19:25, 20 October 2006 (UTC)[reply]

Patched conics & Lunar Orbit Insertion

Hi guys, an astrodynamics question for you all:

I'm doing a research project (final year of Aeronautical Engineering MEng) which involves determining the maximum payload I can get to the moon's surface, for the minimum cost/kg, from a starting point of 20 tonnes in a Low Earth Orbit (say 200 km, conveniently at the same inclination as the moon's orbit). One of the big factors influencing this is the selection of a suitable transfer orbit, but I'm really struggling to get my head around the orbital mechanics involved. I've got a stack of orbital mechanics, mission design and astrodynamics textbooks but I can't find a simple method for calculating the required delta V's to go from LEO to LLO (Low Lunar Orbit). I understand Hohmann transfers, and I think I understand the principle of patched conics, but all the descriptions of the latter method deal with transfers to other planets, which is different from going to the moon because you leave the Earth's sphere of influence. Basically what I'm asking is, what should the calculations look like for a patched-conic determination of an LEO-LLO transfer? Sorry if this is too obscure... generally seems like there's always someone here who knows the answers to things! Thanks --YFB ¿ 20:05, 20 October 2006 (UTC)[reply]

NASA is (or should be) your friend! 8-)?--Light current 20:09, 20 October 2006 (UTC)[reply]

Well, yes and no... where do I go to ask them a (technical) question?

Try here first Jet Propulsion Laboratory. THen follow the links all the way. Youll find a phone number! 8-)--Light current 20:53, 20 October 2006 (UTC)[reply]

Also have you tried the links in Celestial mechanics--Light current 21:15, 20 October 2006 (UTC)[reply]

Also has your reseasch unearthed Astrodynamics#The patched conic approximation ? If not, you just aint been looking 8-)--Light current 21:17, 20 October 2006 (UTC)[reply]

Then of course you have [8] on orbital mechanics.--Light current 21:19, 20 October 2006 (UTC)[reply]

Cheers, Light current; I've tried all the latter three and come up dry - as I said, I've ploughed through a stack of textbooks (Chobotov's Orbital Mechanics; Battin's Introduction to the Mathematics & Methods of Astrodynamics; Interplanetary Mission Analysis & Design etc. etc.) and done a fairly extensive bit of googling, but not found a decent (worked) explanation for a lunar case anywhere. I think I'm having a bad night, I can't seem to figure out what you mean by 'follow the links all the way' on the JPL website - which ones?! Could you paste a URL for me? Cheers, --YFB ¿ 21:30, 20 October 2006 (UTC)[reply]

Go to JPL official site [[9]] then click on public sevices. You should find a contact number there. You could try calling it. 8-)--Light current 21:59, 20 October 2006 (UTC)[reply]

Well, that sort of worked :-) - I didn't get anywhere with that number but I ended up speaking to the Education Outreach office and they passed me over to someone else, who gave me an email address for another someone, who should be able to help but if not I'm welcome to call back... phew! In the meantime, if anyone else understands my question, I'm still open to answers (please!) --YFB ¿ 22:30, 20 October 2006 (UTC)[reply]

Aha! some success then? Thats what research is all about. Asking people who know the answers! 8-)--Light current 22:37, 20 October 2006 (UTC)[reply]

If you want a Hohmann transfer orbit, then the formula for delta-v is on that page. There may be a more energy efficient way, however. Richard B 23:55, 20 October 2006 (UTC)[reply]

Thanks, but as I said in my question, I understand Hohmann transfers already. The problem is that a Hohmann transfer only gives the delta-v to transfer from a circular low-earth orbit to another circular earth orbit with the same 'altitude' as the moon (mean 384400 km). What I need is a method to calculate the delta-v to go from LEO to a circular orbit about the moon (i.e. to be captured by the moon and start orbiting it instead of directly orbiting the earth). To approximate that manoeuvre you need to do a 'patched conics' calculation, which is what I'm struggling with. Out of interest, I've obtained copies of that and other papers on low-energy transfers, but they're extremely difficult to model quickly as they depend on chaotic dynamics for which you need a searching algorithm and numerical back-integration. Also, they work by going a very, very long way away from Earth and then coming back again, which means that they tend to take a very long time (10+ times as long as a Hohmann transfer). Cheers though, --YFB ¿ 00:34, 21 October 2006 (UTC)[reply]

OK, well the patched conic approximation just looks at the most dominant body - rather than taking into account others. So how about we try an orbit with the following;

A transfer orbit to take us from LEO to some point in the vicinity of the Moon - give the object some Delta-V in LEO - assume that only the Earth and object exists in this calculation.
Give the object a Delta-V whilst in the vicinity of the Moon, to effectively place the object into a parabolic trajectory around the Moon, with closest point to the Moon equal to the desired circular orbit radius - calculate this as though only the Moon and the object exists. A parabolic trajectory has V = escape velocity at that radius, so the necessary velocity should be straightforward to calculate.
Once at the closest point to the Moon, give the object a Delta-V to bring it into a circular orbit around the Moon - again, assume that only the Moon and object exists in the approximation.

Now you've got 3 different phases of the trajectory to calculate. How about we work backwards - start by taking the desirable orbit around the Moon, and work out the circular orbit velocity. The last delta-V - from the parabolic trajectory to circular orbit will have a delta-V of (sqrt(2)-1) times the circular orbit velocity. Then project the parabolic trajectory back to a point where the Moon just dominates over the Earth. Then look at what transfer orbit you'll need to minimise the delta-V in LEO, and at the point between the transfer orbit and parabolic trajectory. The transfer orbit needn't be the same as the Hohmann transfer orbit to take you to a circular orbit around the Earth at the Moon's distance. Richard B 18:32, 21 October 2006 (UTC)[reply]

You are talking about "calculating the delta V's". To me, that sounds like you will apply them as short impulses and already know how many they are and when they will take place. Is that how I should read it? (Sorry if I'm asking stupid questions – I just want to learn.) I mean, isn't it possible that the best trajectory results from thrusting in the right way during the whole trip? —Bromskloss 09:54, 21 October 2006 (UTC)[reply]

As far as I know (I'm not claiming to be an expert) that's not the most efficient solution. Taking the 'most direct route' (i.e. a one-burn tangent) is actually less efficient, because effectively you are fighting against your own orbital inertia as well as the gravitational attraction of the central body (here the Earth). The Hohmann transfer is the most efficient 'two-burn' trajectory, because you use one initial burn to move from your initial orbit to an ellipse with its apoapsis at the target altitude, and then a second burn to turn this elliptical orbit into a new circular orbit at apoapsis. By doing that you minimise the energy you expend in overcoming the inertia of your initial orbit. 'Impulsive' (i.e. infinitesimally short) burns are the ideal, because the shorter the burn, the greater the component of thrust that goes in the intended direction. If you make a long burn, you will gradually shift into the desired orbit rather than 'snapping' into it. During the transistion, some of your thrust will be going in a direction other than the intended one, and is wasted.

For my problem, I know that a 'classical transfer' (i.e. not a low-energy transfer based on chaotic dynamics) consists of an initial burn at LEO to enter the transfer trajectory, followed by a further burn (or burns) to enter a circular orbit about the moon. What I'm having trouble with is working out what is required to make the latter insertion to the moon's orbit. Hope that makes a bit of sense, --YFB ¿ 14:02, 21 October 2006 (UTC)[reply]

Looks like you may have to ask your Prof then! 8-)--Light current 17:16, 21 October 2006 (UTC)[reply]

What is the total number of known chemical elements?

I want to know the total number of known chemical elements. I mean the discovered ones. Thanks.

--196.202.91.65 22:15, 20 October 2006 (UTC)[reply]

Your answer is in the second sentence of the very article you've linked to.- Nunh-huh 22:18, 20 October 2006 (UTC)[reply]

Or, if you don't want to go there, the answer is 117: everything from 1 to 118 with the exception of Ununseptium or 117.Phsource 22:23, 20 October 2006 (UTC)[reply]

You'd be better served by actually going there and reading the details. - Nunh-huh 22:40, 20 October 2006 (UTC)[reply]

Thanks to every one who answered my question. I, of course, read the article before posting the question, but the version I read was old and didn't contain the number of articles.

--196.202.92.94 07:04, 21 October 2006 (UTC)[reply]

gee...that must have been a real old version you read. Because the answer to your question has been in the article since June 2004. --`/aksha 12:37, 21 October 2006 (UTC)[reply]

Isn't the number of discovered ones 92? The ones above that are manmade. DirkvdM 09:19, 22 October 2006 (UTC)[reply]

Because of this mention of "discovered", I cannot resist quoting the end of Tom Lehrer's The Elements song:

These [102] are the only ones of which the news has come to Harvard,

and there may be many others but they haven't been discarvard.

That song was from 1959. Even by about the late 1980s the number had only grown to 103 (this is just when I personally happened to first hear the Lehrer song; I remember checking, and all it was missing was lawrencium), so much of the increase since the song is relatively recent. Arbitrary username 18:41, 23 October 2006 (UTC)[reply]

Known compounds

What is the total number of Known chemical compounds? (to the nearest 1000) 8-)--Light current 22:28, 20 October 2006 (UTC)[reply]

By known, are you including all hypothesised compounds as well (such as UuoO₃)? La ï ka 22:54, 20 October 2006 (UTC)[reply]

Actually, on a more useful note, currently there are about 30,100,000 compounds registered under the CAS registry number system, but the number is growing constantly; the current number can be seen here. La ï ka 22:58, 20 October 2006 (UTC)[reply]

I suppose I was asking how many are naturally occuring of have been synthesised. Not how many were theoretically possible. 8-|--Light current 00:08, 21 October 2006 (UTC)[reply]

When I last looked, the CAS site said: 11,962,499 Commercially available chemicals. I guess these have been isolated or synthesized, but perhaps this works like just-in-time synthesis-on-demand :) --Lambiam ^Talk 00:27, 21 October 2006 (UTC)[reply]

I believe all 30,000,000 at CAS are real ( have been synthesized or found). Theorectical possibilites would be many times greater. They say 4,000 new ones a day. --GangofOne 05:58, 21 October 2006 (UTC)[reply]

is this a shockwave affect or just accelerated air?

http://en.wikipedia.org/wiki/Vortex_ring_gun

I asked this before but never got an answer. does the effect used in this [the version that just uses the air in the cylinder anyways] count as an over pressure shockwave, or not? robin

A shockwave is just a pulse of high pressure air, isnt it?--Light current 00:44, 21 October 2006 (UTC)[reply]

I'm pretty sure you did get an answer. Quoting: "I think this is just a more powerful Airzooka. … There is no supersonic air motion, and no shockwave." —Ilmari Karonen (talk) 12:57, 21 October 2006 (UTC)[reply]

oooh i didnt come up with that, stupid browser , cheers and thanks Robin

October 21

James McCollum

Still looking for some biographical info on Canadian engineer, James K McCollum. He was the co-inventor of the Sleeve-Valve mechanism for internal combustion engines along with Scotsman, Peter Burt.

Jerry Wells, Australia.

Weve already tried to answer this question. If these links are red, its pretty pointless asking further. Why dont you Google it?--Light current 00:45, 21 October 2006 (UTC)[reply]

Page updating problems.....

I`m having a terrible time updating this page. I`m usually on AOL,,,get blocked often on account of that. I`m currantly on Mozilla but, even worse...re-fresh/update brings me to about this time, YESTERDAY! Can anyone offer some assistance? TY Dave 172.128.242.194 02:33, 21 October 2006 (UTC)[reply]

Simply creating an account would help. This would also make it easier for us to diagnose your problems by referring to your edit history.--Shantavira 08:53, 21 October 2006 (UTC)[reply]

Reading Wikipedia:Advice to AOL users would be a good idea =) --`/aksha 12:27, 21 October 2006 (UTC)[reply]

Phase-determination pressure

Is the pressure on a phase diagram the vapor pressure of the substance in question, or is it the total pressure of the substance's environment? It seems to be vapor pressure when discussing the equilibrium between liquid and gas, but it seems to be total pressure when discussing the pure phases (e.g., the diamond/graphite transition). Which is it, or is it some third possibility? --Tardis 02:43, 21 October 2006 (UTC)[reply]

It's the external pressure applied, i.e., if you put the substance in a rigid container with a movable piston and apply some pressure to the piston, the phase diagram tells you the phase the subtance will take. It's only equal to the vapor pressure when the gas in equilibrium with another phase. For example, say the temperature is between that of the triple point and the critical point. If the applied pressure is less than the vapor pressure at that temperature, the substance will be completely gaseous. As the pressure is increased, the volume decreases slowly, until the pressure reaches the vapor pressure and the gas begins condensing into a liquid. At this point the piston suddenly slides in without resistance, because the gas condenses into a liquid and the volume changes while the pressure remains at the vapor pressure (moving the piston does no work because dp = 0). Then when the substance is completely liquified, the pressure begins to increase again, now greater than the vapor pressure. —Keenan Pepper 18:25, 21 October 2006 (UTC)[reply]

This is all from Atkins' Physical Chemistry, ISBN 0-7167-8759-8. —Keenan Pepper 18:28, 21 October 2006 (UTC)[reply]

Cheese longevity

I buy packages of shredded cheddar cheese in a sealed zippable bag (as in, initially properly sealed and re-sealable). The bag claims that it should be refrigerated, of course, and finished within "3-5 days" of opening. However, I use only a little at a time and so have seen it last much longer than that in an arid environment without any apparent ill effects. Just how long should I expect it to be safely edible, and what would be the first sign that it's not? --Tardis 03:12, 21 October 2006 (UTC)[reply]

The first sign will be visible specs of mold. How long they take to develop depends mainly on how sterile you keep it when it is opened. If you reach a grubby hand down in the bag, you will likely heavily contaminate it. If you open it upside down, so nothing can fall in, then let some cheese out and quickly reseal it, then drive the air out, you might get lucky and not have any bacteria in there that like cheese. Another suggestion, buy smaller bags or divide your bag up into single use bags, then freeze each until ready to use. StuRat 03:28, 21 October 2006 (UTC)[reply]

no way dude, what's-his-face (keenan pepper???)'s friend worked in the cheese industry!! Wait for his repesonse.

No whey ! :-) StuRat 11:57, 21 October 2006 (UTC)[reply]

Ive heard cheesy jokes before 8-(, but that on just about takes the biscuit. 8-)--Light current 17:12, 21 October 2006 (UTC)[reply]

I have no idea who you're talking about. —Keenan Pepper 17:53, 21 October 2006 (UTC)[reply]

Sorry I cant see the joke here 8-(--Light current 19:27, 21 October 2006 (UTC)[reply]

The cleanliness of the processing and packaging have a large effect on the longevity of shredded cheese, which is inherently shorter than that of a wheel or wedge of cheese. If the cheese gets moldy quickly, and you are sure you did not contaminate it, you might just wonder "Who cut the cheese?"Edison 21:52, 21 October 2006 (UTC)[reply]

Yes, the much greater surface area of shredded cheese allows it all to get moldy, as opposed to just the surface of a large chunk of cheese. StuRat 00:20, 23 October 2006 (UTC)[reply]

Whistling sine waves

A whistling sound is pretty close to a sine wave. Why? What are the processes behind the sound in whistles that make it like that? I understand the mouth and lips work as a resonant chamber, but why, exactly, no timbres are produced? Our article on whistling mentions this fact, but doesn't give any explanation. ☢ Ҡi∊ff⌇↯ 05:59, 21 October 2006 (UTC)[reply]

The mouth and the lips form a high Q resonant chamber. Look at resonance and Q factor. The Q is so high that it filters out frequencies higher and lower than the one you are whistling leaving essentailly only one frequency. This one frequency must be a sine wave. 8-0--Light current 07:31, 21 October 2006 (UTC)[reply]

Ah, I've never thought about it like that. Makes sense. Thanks for the explanation. :) ☢ Ҡi∊ff⌇↯ 07:54, 21 October 2006 (UTC)[reply]

Light Current, are you sure about your explanation? The Q factor of, say, a woodwind instrument is probably higher than that of the mouth cavity, and they do have a tone that is clearly not sinoidal. After all, also a high-Q resonator gets higher harmonics excited alongside of the fundamental, and the amplitude of the harmonics (in units of the fundamental's amplitude) is what defines the tone or timbre. What a high Q factor rids you of is line broadening, but it does not enforce a pure sine. By the way, Kieff initially asks about whistles and Light Current talk about whistling without a whistle. So, of which one do you two now claim that they had an extraordinary pure sine? I doubt both, actually, but as we all three probably have a microphone input bin our computers, we can easily check. Simon A. 22:43, 21 October 2006 (UTC)[reply]

Well the very fact that woodwind instruments have a certain timbre proves that the instrument Q is much less than your mouth/lip system. --Light current 23:11, 21 October 2006 (UTC)[reply]

No, Simon, I was actually talking about whistling with the mouth (that's why I wikilinked to whistling instead of whistle.) The reason I asked is that I've already recorded myself whistling and looked at the wave, and it's very close to a sine wave. ☢ Ҡi∊ff⌇↯ 10:36, 23 October 2006 (UTC)[reply]

IIRC, the difference between whistling and the sound of a woodwind instrument is that the woodwind instrument has a vibrating reed that produces a range of harmonics, whereas whistling is highly filtered noise (caused by air turbulence). Whistling is unvoiced whereas the woodwind is "voiced" in the way that it is analogous to a vibrating vocal fold.

So whilsting is just noise filtered drastically to single out an individual harmonic. There is of course a bit of noise in the "signal" too, since our mouths aren't perfect. Magic Window 17:31, 24 October 2006 (UTC)[reply]

COSTAS loop

In costas loop, the phase of generated carrier must be such that it matches with that of Incomimg carrier so that angle phi is zero.But why is the error signal given to the Voltage Controlled Oscillator instead of a phase shifter???

What do you mean by phase controller? THe freq and phase of the VCO is controlled by the vco i/p voltage

--Light current 09:02, 21 October 2006 (UTC)[reply]

Transmission line theory

We all learn that circuits have a fixed value of resistance,capacitance and inductance.But why sudden change in theory when it comes to transmission lines with topis such as reflectivity?Any lucid and simple explaination?

In transmission lines, resistance, capacitance and inductance are still there, but the elements are not lumped. THey are distributed over the length of the line. Any discrete lumped RLC cct (like ther ones youre used to) will also have parameters like input match, reflection coefficient etc. but at low frequencies, these parameters do not matter so much as they do at high freuencies where transmission lines are commonly used.--Light current 09:08, 21 October 2006 (UTC)[reply]

I'm not sure who taught you that circuits have a "fixed value" of resistance, capacitance, and inductance, but its usually not true. For example, the resistance of a typical physical resistor changes noticeably depending on the temperature. Variable capacitors can be constructed by using moving plates. Also active circuit elements like transistors and diodes are typically nonlinear -- that means the resistance and capacitance vary depending on the applied voltages or currents. In fact a reverse-biased diode is often a more economic way to form a variable capacitor than a physically moving device. We normally just approximate these circuits as linear circuits, with localized values of the elements, determined at some moment of time, because then we can calclulate the behavior. Our article Electrical network touches on this in the section on Piecewise-linear approximation, but I can't say the discussion is complete enough to give a good understanding. -- The Photon 22:38, 21 October 2006 (UTC)[reply]

I dont think the questioner was really asking about nonlinear circuits but just the diff between the circuit theory and em field theory approaches.--Light current 23:01, 21 October 2006 (UTC)[reply]

has to do with frequency and size of the components. Lumped is easier to analyze. But consider this: You have a generator with an impedance and a fixed resistive load. If the resistive load is far from the generator and the impedance is changed far from the generator, how does it know? there is a lag in that information. The answer is that the generator sends out a fixed power value and what cannot be aborbed is returned as a reflection. As the frequency increases, that distance is made smaller and smaller until it makes it difficult to speak in terms of voltage and current and the preferrred parameter is power (S parameters). The bottom line is that the source has no idea what the load is until the power is reflected back. Lumped analysis is adequate when the dimension of the circuit far exceed the wavelength. --Tbeatty 07:33, 22 October 2006 (UTC)[reply]

Reality

I present two situations.During the genesis of time, the universal laws of reality came into being:-

1)The Domain of these laws started spreading at the speed of light(max speed limit) across this current diemension after the beginning of the big bang and that the universe is not defined in the domain beyond this hypothetical diemension.

2)The domain of these laws were applicable instantaneously across the whole diemension once the universe was created.

Which of the above is correct??

Given that information cannot travel faster than the speed of light, the only way for us to arbitrate between them (ie going past the universe "horizon" and doing some experiments) is impossible. Hence the question is, in a very fundamental sense, totally unanswerable. Batmanand | Talk 16:40, 21 October 2006 (UTC)[reply]

Can you equate information with laws? If laws are instead properties of space, then (2) seems to me to be right. Clarityfiend 21:55, 21 October 2006 (UTC)[reply]

The big bang theory gets around this whole question by saying that at the beginning of time, the spacelike diameter of spacetime was infinitesimally small. This means: before time, there also was no space. At te big bang, time and space spread out together. The spreading of time and space was, by the way, not limited by the speed of light, as this limitation applies to motion within spacetime, not to the expansion of spacetime itself. Maybe we have articles on the flatness problem and on cosmic inflation which explain more. This stuff is, however, quite difficult to grasp. Simon A. 22:47, 21 October 2006 (UTC)[reply]

But the expansion of spacetime isn't the issue here, he's asking about the expansion of constants within the domain of spacetime. Whether spacetime was inconceivably small or 50% as big as it is now, if the laws were to substantiate after spacetime itself came into existance, size and distance doesn't seem to logically be a factor. On a hunch I'd reject both options, going instead for a neater "the laws came into existance the same instant that space time did" style argument. freshofftheufo ΓΛĿЌ 04:46, 22 October 2006 (UTC)[reply]

I think what he's trying to say is that when the universe was scrunched up infintesimally small, any fluctuations in what we now think of as constants had the opportunity to settle down to some kind of average value, and then when the universe expanded it carried those average values out. Or something like that. Confusing Manifestation 09:50, 22 October 2006 (UTC)[reply]

Can you have fluctuations in an infinitely small space? freshofftheufo ΓΛĿЌ 12:42, 27 October 2006 (UTC)[reply]

Ic 8038 and Ic 555

Both Ic555 and Ic8038 can be used as free running oscillators.Then why have both of them???

8038 can generate triangle, sine and square [10]. 555 can only generate pulse waveforms [11]--Light current 09:11, 21 October 2006 (UTC)[reply]

And the 555 is cheaper (and older). And it even has a Wikipedia article: 555 timer IC Simon A. 22:53, 21 October 2006 (UTC)[reply]

SATA computer

I purchased a laptop computer with an SATA hard disk + XP Home Edition. I want to install my previous copy of XP Pro on it but failed (my last computer was dead so it's legal to transfer the XP to my new laptop). The XP installer could not see my Compaq Presario's SATA hard disk. How do I solve this problem? -- Toytoy 14:15, 21 October 2006 (UTC)[reply]

Legal may be, but difficult, as you will have already 'activated' the operating system om your old computer and it may not activate on the new one.

You dont say the make of the 2 computers. Is it possible that you had a special version of XP Pro that was designed for (and only recognises) your old computers hardware?--Light current 14:21, 21 October 2006 (UTC)[reply]

Have you downloaded the latest XP SP and SATA drivers? When you say it doesnt see it you mean it doesnt detect it at all ? Sounds duff. Try another SATA port, different drive, different lead, etc. By the way, youre in the wrong section mmatt 15:03, 21 October 2006 (UTC)[reply]

My XP is a site licensed copy. At lest in theory, I can install it on a new computer even multiple computers. -- Toytoy 15:10, 21 October 2006 (UTC)[reply]

It sounds like you have a crippled version of XP Pro that will only install on the computer it was sold with. You need a generic version of XP Pro that will install on any computer - then hope your product key works with it. --Kainaw ^(talk) 16:42, 21 October 2006 (UTC)[reply]

Put the drivers for the SATA disk on a floppy disk and press F6 during install when it says to do so at the bottom. This way you can install SATA/RAID drivers for the disk that Windows will be installed on. - Dammit 16:45, 21 October 2006 (UTC)[reply]

Youngs experiment

In Youngs experiment, why doesnt the wavefunction collapse when it happens on one of the two slits? Whats the difference between the slit and the detector from the photons point of view ? How do you detect a single photon anyway ? Also, if you slightly offset the light source from the centre point between the two slits, would you able to detect which slit it went through depending on the time it took to reach the detector. The problem with this experiment is that its so simple you feel you should be able to outsmart it ! mmatt 14:41, 21 October 2006 (UTC)

Have you read double-slit experiment? What do you mean by "outsmart it"? There's nothing to outsmart. As for detecting a photon - your eyes do it all the time. So do cameras. There are many forms of photon detectors. From the photon's point of view, the slit is an opening to pass through. The detector is a dead-end that it slams into at the speed of light. --Kainaw ^(talk) 16:41, 21 October 2006 (UTC)[reply]

Wait a minute, a lot of the photons do slam into the walls around the slits, don't they? Still, some slip through and give the double-slit effect. —Bromskloss 17:09, 21 October 2006 (UTC)[reply]

By outsmart it I mean that that you feel as if you should be able to somehow infer which slit it went through, but of course you cant. Its a trick of the mind, I know.

How do you detect a single photon? I know it can be done, but how? The eye detects lots of photons, not just one. I dont want to get hung up on this, its not really relevant to the main question.

Are the photons that get cleanly through the slit the only ones that contribute to the effect? I always thought they scattered off the edges, thats how they change direction backwards to contribute to the effect, otherwise you would just get a very narrow band directly in the line-of-site from the light source ? mmatt 19:04, 21 October 2006 (UTC)[reply]

The eye can detect a single photon! From Rod cell:

A rod cell is sensitive enough to respond to a single photon of light, and is about 100 times more sensitive to a single photon than cones.

As for the scattering you mention, I don't want to say too much because I'm not absolutely sure, but you seem to view the photon as a particle, rather than a wave. A wave spreads out after a small passage. Think of waves on the surface of water. —Bromskloss 20:53, 21 October 2006 (UTC)[reply]

Isnt it a particle and a wave ? matt

It's not that we can't detect a single photon, it's that we can't EMIT a single photon. However, we CAN emit a single electron, and detect this. Watch, as the electron experiences the exact same wave-particle duality, pushing the relevance of this experiment into the macroscopic world. :) 81.93.102.3 19:36, 21 October 2006 (UTC)[reply]

but the question still applies to electrons, right ? You have to get some scatter off the grating/slit or the electrons/photons shoot off in the wrong direction to interfere ? So if they interact with the slit in this way, why doesnt the slit ( or the sides of the slit ) collapse the wavefunction ? mmatt

Even before reaching the slits, they pass air molecules without collapsing. (Well, I guess some may collapse, but not all.) —Bromskloss 20:58, 21 October 2006 (UTC)[reply]

Right. I though about that, but I thought that because an air molecule is so small, you wouldnt collapse the wave function - you would have a superposition of two possible paths one of which involved collision with an air molecule. But with the slit ( or the sides of the split ), which is macroscopic, the wavefunction would collapse mmatt 21:09, 21 October 2006 (UTC)[reply]

Hub Gears Wont Change

I had to remove the black thing that controls my gears in order to tighten a loose nut for the wheel. When I removed the black thing (that covers the nut through which some pin things protrude), the pins came out quite far. Now that I put back the covering, having tightened the nut; a) the gears don't change (no matter where I push the gear lever on the handlebars) and it's stuck in some lowish gear and

b) the gear lever on the handlebar wont push into 6th or 7th gear.

Can anyone diagnose the problem and suggest a solution? The manufacture is 'Sachs' but this was insufficient to draw useful information for me to fix my bike. --Username132 (talk) 15:00, 21 October 2006 (UTC)[reply]

I take it that you are talking about a bicycle with derailleur gears, right?

I also assume that it is a standard modern gear shift. Mounting these is not that difficult but also far from trivial. First check by comparing with a working bike that you have remounted the derailleur correctly. Make sure that both its springs operates in the correct way: there is one spring inside the parallelogram shaped lever that moves the derailleurs two little cogwheels over the gears. This spring pulls the derailelleur over the highest gear (i.e. smallest gearwheel) and if you pull with the Bowden cable against this spring, you switch down to lower gears. That your stuck in a low gear seems to indicate that ever the Bowden cable is too tight or the sping is blocked or the derailleur is mounted the wrong way. There is also a chapter on derailleur adjustment in the Wikibook on Bicycle repair. Reading it might be a good idea. Simon A. 23:08, 21 October 2006 (UTC)[reply]

Ok, stupid me, I overlooked that your heading explicitly sais that you are talking about hub gears. These were always a bit of a mystery to me, so I don't have a better suggestion than checking that the Bowden cable is ok. Simon A. 23:10, 21 October 2006 (UTC)[reply]

Now, I got interested in hub gears and had a look at the German Wikipedia, which explains quite a lot in its article de:Nabenschaltung. There, you find the hint that before disassembling a rear wheel you should switch to that gear that relaxes all the springs in the hub, i.e. the one where the Bowden cable is not under tension (and the little chain nearly all the way in I suppose). If you want to adjust the gearshift, you have to set the switch at the handlebar to a gear certain setting and the hub gear itself to the same gear (by pulling at the little chain I suppose). They give a table which gear to use for adjustments for different makes. (How many gears do you have, and how old is the bike? That should be sufficient to identify the model.) Then, there should be a little bolt that allows to finetune the length of the cable in order to match hub gear and handlebar switch markings. Simon A. 23:56, 21 October 2006 (UTC)[reply]

I fixed it by pushing the pin in until it stopped. The "Schaltbox" has the manufacturer 'Sachs' and 'super 7' written on it (it's a 7-gear bike). I don't know the age, cause I bought it second hand but I'd guess that it was made in the last five, maybe even three years. Username132 (talk) 14:09, 22 October 2006 (UTC)[reply]

Congrats for solving the proble. By the way, Fichtel & Sachs has been bought be SRAM a couple of years ago, and on their website you find a link Techical Documents with a manual for a "SRAM S7 hub gear". Maybe that's yours, or at least its successor model? Check it out here. Simon A. 09:44, 23 October 2006 (UTC)[reply]

Thanks, that's perfect! --Username132 (talk) 16:13, 23 October 2006 (UTC)[reply]

Atheroma

We were always told this was fatty deposits. Our article says it is other sorts of debris. Which is it and if its not fatty, why should we stop eating fatty food?--Light current 16:45, 21 October 2006 (UTC)[reply]

So you die later instead of earlier. - Nunh-huh 17:24, 21 October 2006 (UTC)[reply]

New Iraq strategy: open a chain of fast food restaurants selling triple cheeseburgers, all the fries you can eat. Clarityfiend 17:37, 21 October 2006 (UTC)[reply]

Thirty years ago, the diet police were hating on animal-fat based butter, and proclaimed transfat-based margarine to be healthy. That's why my grandma only uses margarine, hehe. It has flipped around, with transfat being bad too. Then fat was generally bad. But mostly it was saturated fat... but also unsaturated too, except for olive oil, and rapeseed. There was the anti-sugar craze, and the idea that carbohydrates were what really got us fat, not fat. Maybe we should detox or go raw, or vegetarian at least. Organic? Cereal used to be great for you, but it has too much sugar. We don't get enough vitamins and minerals. Ah, chloresterol is what causes heartattacks now, not fat. Heavy metals concentrate in fish. The Cato Institute published an article: As Taubes pointed out in his article, the simplistic notion that dietary fat is bad was a political and business judgment, not a scientific one. Despite ambiguous science, in 1977 a Senate committee led by Sen. George McGovern issued a report advising Americans to consume less fat to avoid "killer diseases" supposedly sweeping the country. The politically dutiful National Institutes of Health soon joined the anti-fat bandwagon, a move that spawned the low-fat food industry. What exactly is the basis for trusting the food alarmists again? The best way I believe is to read the journals yourself, not watch the news. The news will tell you that french fries and breastmilk cause cancer. Remember those days? Atherosclerosis is caused soley due to LDL and VLDL cholesterol, according to every single source I have heard in the mainstream media—even though macrophages, and calcification have quite major roles. I bet the media pushes the chloresterol part because it is scary, and there is someone to blame. What you eat is partially based on your decision—some foods have more or less cholesterol. You don't decide if macrophages or ossification will obstruct or harden your arteries, but you can be to blame for eating yourslef to death.X [Mac Davis] (SUPERDESK|Help me improve) 00:12, 22 October 2006 (UTC)[reply]

Here's another free bit of advice: don't get your dietary recommendations from teenagers, and don't get your understanding of the pathogenesis of atherosclerosis from them, either. - Nunh-huh 05:24, 22 October 2006 (UTC)[reply]

How can we tell whos a teenager? Most people here act like one (including me) 8-)--Light current 14:08, 22 October 2006 (UTC)[reply]

Excellent work from mac davis! I think a lot of the confusion with health is disinformation in the media. they take a "phase 1" study of about 30 people and tout it immediatly as fact just to fill in column inches (or webspace) the result for us is now complete confusion and having to use the wikipedia reference desk and wikipedia to seperate fact from fiction! chocolate helps your heart This one from the BBC is a good example, so now I think "hey maybe a bit of chocolate is good for you" and its a headline based on 20 people!!!

Sauropod size

I'm trying to write a book on the largest sauropods. The technical information like the discovery, etc., is easy. However, it's the size that continues to differ. Various references such as different books and websites give different statistics. My main sources are EnchantedLearning.com, DinoRuss' Lair, Wikipedia itself, and the DinoArchive and a few of my own calculations. As you can see below, the estimates vary considerably.

1. Argentinosaurus: 22-45 m, 60-100 tons 2. Paralititan: 20-32 m, 60-80 tons 3. Seismosaurus : 30-54 m, 25-150 tons 4. Supersaurus : 24-45 m, 30-100 tons 5. Bruhathkayosaurus: 28-44 m, 150-220 tons 6. Amphicoelias: 37-62 m, 50-170 tons 7. Brachiosaurus: 21-30 m, 15-80 tons 9. Argyosaurus: 20-40 m, 45-80+ tons 10.Ultrasauros: 25-35 m, 40-180 tons

As you can see, the variation is not negligible. I am also aware that these creatures are known for very little fossil evidence. If you can,please suggest corrections for this table (taking into account that these creatures are known for very little fossil evidence). I appreciate the help.

You're using Wikipedia as a source? It's very unrealiable. I suggest using books as your main source, instead of websites.

As for the question, I think it's better to include the range of size estimates in your book. But if you really need to come up with a single figure, try taking the more reliable size/weight estimates and averaging them, while noting in your book that estimates vary wildly. --Bowlhover 19:47, 21 October 2006 (UTC)[reply]

What does KIBRA stand for?

Kibra is the name of a gene associated with memory performance in humans. Kibra sounds interesting and exotic. What is the origin of this name? It is an acronym? Tavilis 18:40, 21 October 2006 (UTC)[reply]

Sounds like it is is short for kidney and brain MeltBanana 23:07, 21 October 2006 (UTC)[reply]

Surfaces and bonding

Suppose you got a block of some metal, say gold, and cut it with a sharp blade in a good vacuum, so the cut surfaces had no gases adsorbed on then. Then you pressed the two cut surfaces back together. Would they stick together? Would the cut disappear and the metal become one solid block again? If not, why not?

Also, same question for a network covalent solid, like diamond. Would the cut surface be covered with dangling bonds (unpaired electrons)? Would the cut surfaces attract each other and stick?

Any good books about this stuff? —Keenan Pepper 19:14, 21 October 2006 (UTC)[reply]

Yes, I think cut metal parts would fuse together. How would you cut diamond neatly? Diamonds dont stick together for some reason unknown to me. Try welding diamonds!--Light current 19:31, 21 October 2006 (UTC)[reply]

What you described with metals is called a cold weld, and you can even do it with two different metals. I don't think it would work with diamonds unless you had an extremely high pressure and temperature to reform the bonds, however. StuRat 20:12, 21 October 2006 (UTC)[reply]

Yeah thats what I was trying to say. I suppose it must be possible with diamond given enough temp and pressure as Sturat says. After all, we are told that diamond only forms under condtions of extreme temperature and pressure.--Light current 20:21, 21 October 2006 (UTC)[reply]

Thanks, cold weld and asperities (once I figured out the latter was misspelled in the article =P) are much better strings to search for. I don't understand why you say a network covalent solid wouldn't do the same thing, though. There are dangling bonds and no adsorbed gases... why wouldn't they stick together? —Keenan Pepper 23:00, 21 October 2006 (UTC)[reply]

I would assume the 'dangling bonds' of the network covalent diamond would just form bonds across the surface. So one 'dangling bond' forming a proper bond with another 'dangling bond' that you've just cut. Meaning when you stick the two pieces together, the 'dangling bonds' won't be dangling anymore. So instead of having one piece of diamond with a cut through it, you'll end up with two discrete pieces of diamond that won't bond with each other. --`/aksha 02:34, 22 October 2006 (UTC)[reply]

It seems some of the surface atoms do move and form pi bonds (a process called reconstruction), but the most stable environment for an atom is inside the crystal, so there is still an unavoidable surface energy. —Keenan Pepper 20:48, 22 October 2006 (UTC)[reply]

Zero-point energy and substance compatabilty.

I have read on one of the sites on NASA that they want to use Zero-point energy equilibrium state distruptions, It claims that ZPE is almost like atmospheric pressure, and making changes in its density or distribution might move objects. But how can this be if it is EM and some objects are not effected by EM? Are there any kinds of materials ZPE cannot effect? physicsmoron

Could you at least link to the NASA page so we know what you're talking about? —Keenan Pepper 23:17, 21 October 2006 (UTC)[reply]

That is rather speculative. I suggest Casimir effect. I do not believe any objects could be unaffected by the Casimir effect. X [Mac Davis] (SUPERDESK|Help me improve) 00:21, 22 October 2006 (UTC)[reply]

General Relativity

While researching General Relativity, a broad range of theories, mathematic equations, and list of related topics come up. I need to create a project on BASIC ideas of general relativity on a ninth grade level and am wondering what should I include. I'm planning on including Einstien, Gravity, Wormhole, Black holes, and the unification of Special Relativity. Are there any aspects of General Relativity that I am missing that can be explained in layman's terms and pictures? Am I including things that I should not be? Thanks in advance for your speedy replies. I am thankful to have this resource. 72.77.122.83 21:51, 21 October 2006 (UTC) BeBe2[reply]

I don't know if I'd include wormholes. They're strictly hypothetical and certainly not a central part of relativity. Clarityfiend 22:02, 21 October 2006 (UTC)[reply]

I wouldn't give 9th graders any of the math involved. I would include the theory that gravity works like a sheet of rubber, where if you have a small marble and a baseball on it, they would go toward each other, but the small marble will go toward the baseball more. Gravity wells, which do include blackholes/wormholes. How we orbit the sun, and how satelites are launched to stay in orbit. How gyroscopes work and how they're used for navigation (I never understood that back in high school myself). These are really the ones I would include for 9th graders. How long does this have to be? --Wirbelwindヴィルヴェルヴィント (talk) 22:47, 21 October 2006 (UTC)[reply]

Well, I have to be able to disply it on a flat poster board with pictures and captions. 72.77.122.83 23:27, 21 October 2006 (UTC) BeBe2[reply]

I like the research topic. Organizationally, special relativity was developed first, and does not include gravity. It is mathematically fairly simple (compared to other topics), and may be accessible to someone with knowledge of algebra (books like Wheeler and Taylor's Spacetime physics assume a knowledge of about algebra 2). The idea is if you assume two things, some interesting results can be shown. For more on this, see Introduction to special relativity, Postulates of special relativity, and (if you're daring) special relativity. One of the results of this work was to show that the distance (as we ordinarily measure it) is not fixed between two points in space. You might see two things one meter apart, while I see them occur two meters apart.

Gravity, in the pre-relatavistic (called "Newtonian") system, depends on the distance between two objects. For instance, the sun is gravitationally far more attractive than the earth, but we feel the earth considerably more since we are closer to it. But in the relatavistic framework, distance is not fixed between two objects... So dealing with gravity becomes a challenge. General relativity deals with gravitation, and is mathematically quite complicated (requiring college level mathematics). Gravity, under general relativity, is not a force but bending of space. An analogy is riding in an elevator: you feel heavier as the elevator starts moving up. You might (if you had never been in an elevator before) think that you're sitting still and gravity had changed to make you heavier. Equivelently, as I sit in my chair, I don't see myself speeding up (just as you don't see any difference in the elevator), however I am, for the purposes of relativistic calculations, being moved in bent space. See Introduction to general relativity.

Black holes form when enough mass gets packed together that nothing can escape from it. It is a consequence of general relativity, and observations seem to support their existence. Wormholes are somewhat more theoretical, in the sense that they are not observed to exist. Hope this helps, --TeaDrinker 23:59, 21 October 2006 (UTC)[reply]

Quantum mechanics? wave theory?

I have been reading a book on quantum mechanics - and I found one section confusing. I hope that I can phrase the question clearly.

When a series of electrons are sent through a slit - then the resulting pattern formed is a wave. The probability that a particular electron lands at any position on the wave is given by the square of the amplitude of the wave.

My question- Why square the amplitude? (I think this is what is happening -see the Wikipedia article "Normalisable wavefunction" )- I know that squaring makes the mathematics more convenient (gets rid of negatives and also gives a convenient sum of 1 when squaring components sine and cosine)But I'm still left with the basic question - why the square?

any help will be very interesting to me

We know that the probability is a real number between 0 and 1. We can observe, say by doing a one-slit experiment, that the probability doesn't have any phase information itself. Yet we also observe the diffraction occurring in the two-slit experiment. Therefore we (well, some very smart physicists like Schrödinger) invent a model that can predict the results of the two-slit and other experiments. For convenience, we express this model in terms of complex numbers inside the unit circle, which can carry phase information, but when multiplied by their complex conjugates always give a real number between 0 and 1.

So the answer is not that there is a reason to square the wavefunction value to get the probability, but that the wave function was invented as a complex number with magnitude equal to the sqare root of the probability in order to explain and predict the experimental observations. -- The Photon 22:25, 21 October 2006 (UTC)[reply]

Also note that a wave function whose interpretation is the probability of finding a particle at a particular position would not be a correct model of the observed phenomenon — not if we model the combined effect of two waves using superposition, as we normally do. Since probability is a value between 0 and 1, you will never have destructive interference in the model, which is inconsistent with observations. --71.246.5.19 13:38, 22 October 2006 (UTC)[reply]

G Protein Coupled Receptors

Does anybody have any thoughts on how many GPCRs there are in the human genome? I've read papers that give values of 750, 900, >1000. What's the latest, or most accepted value I should use?

Thanks very much for your help.

Aaadddaaammm 23:38, 21 October 2006 (UTC)[reply]

According to this recent article, "the family of G protein–coupled receptors (GPCRs), which in man contains about 791 proteins". --JWSchmidt 23:52, 22 October 2006 (UTC)[reply]

October 22

Craneflies - how have these really crappy insects continued to survive?

I mean, c'mon - the cranefly has got to be the worst-designed insect in all of creation. It's big, slow, mindless, fragile, has no method of defending itself from predators (despite being big and meaty) and is a terrible flier. Every year when they hatch, everything that eats insects feasts upon them. If God exists, the cranefly was either created on one of his off-days, or was some strange practical joke. If reincarnation is real, the people who've really been bad in this life must come back as craneflies as punishment. Seriously, how has this stupid, pointless, annoying creature managed to avoid extinction? --Kurt Shaped Box 01:39, 22 October 2006 (UTC)[reply]

There's many of them. ☢ Ҡi∊ff⌇↯ 02:05, 22 October 2006 (UTC)[reply]

"It's big, slow, mindless, fragile," - as oppossed to the...uh....intelligent-minded insects? =) I guess god really does have off-days after all, because i always thought humans were god's "strange practical joke". --`/aksha 02:30, 22 October 2006 (UTC)[reply]

Most insects at least try to escape if you go for them. I could walk up to a cranefly and slowly squash it with my finger and it wouldn't even see me coming. --Kurt Shaped Box 02:35, 22 October 2006 (UTC)[reply]

Probably insane mating. Lovebugs same thing. But these taste bad. They also more importantly taste bad to birds.X [Mac Davis] (SUPERDESK|Help me improve) 03:32, 22 October 2006 (UTC)[reply]

I wonder if the fact that they don't try to escape makes the females easier to mate with? God I feel dirty thinking like that. freshofftheufo ΓΛĿЌ 05:06, 22 October 2006 (UTC)[reply]

There are a lot of different techniques an organism can use to survive extinction. High fecundity is one of the more common ones. Not to mention, it is impossible to see these things when they aren't flying (camouflage). They are also of very little nutritional value, and it would be a waste of time for an animal to specialise. They only leave the ground once a year to mate, the rest of the time they are larvae. As long as enough mate to regenerate the population (and that wouldn't take much, I assume one female can make thousands of eggs), they will not become extinct. Simple! :) --liquidGhoul 05:20, 22 October 2006 (UTC)[reply]

Add to that a high agility that makes it possible to escape attackers and what do you get? The bloody mosquito. The reason it's not extinct is not for lack of us trying. Rarely has mankind been so united as in its efforts to exterminate mosquitoes. And they still thrive. DirkvdM 09:39, 22 October 2006 (UTC)[reply]

(reading the cranefly article) "However, it is very easy to accidentally break off their delicate legs when catching them, even without direct contact. This may help them to evade the birds who pursue them as prey." well, there's (part of) your answer. They sacrifice their own legs to survive. I guess they don't exactly need legs to mate and reproduce... --`/aksha 12:39, 22 October 2006 (UTC)[reply]

Meaning of TY1CL1

As an example "TT-P-95 TY1CL1 - PAINT,RUBBER", can someone tell me what "TY1CL1" means please? I see it in the title of numerous products.

Thanks

John

Chest rash?

I understand that this question should be directed to a dermatologist, however I just wanted to get a sense of whether or not that would even be necessary. The below pictured rashes (?) in the middle of my chest appeared about three months ago. There are approximately 25 white circular rashes, the largest about a centimeter in diameter, between my pectoral muscles. They are more defined than the resolution of the picture allows. They are not painful, itchy, nor can I feel them at all. They do not protrude from the skin. Does anyone have any idea what they might be? Cduffner 03:58, 22 October 2006 (UTC)[reply]

Have you been tanning? If so, they could be old scars. My various deep scars (from multiple motorcycle accidents, getting shot twice, and stabbed once) do not tan. The surface ones - like scratches and such - do eventually tan over. I have similar circles on the back of my right hand from gravel in one accident that do not protrude, itch, or hurt. When I tan, they stay white while the rest of my hand turns red (note, I tan red, not brown). There are a few spots that I don't remember getting scars and I only see when I tan. I was told about two of my left fingers - burned by an iron when I was a toddler. But, I assume that I had some injury long before my memory kicked in. --Kainaw ^(talk) 04:47, 22 October 2006 (UTC)[reply]

Hard to tell from the picture. Could be vitiligo. A dermatologist could tell you. - Nunh-huh 05:24, 22 October 2006 (UTC)[reply]

Could it really be vitiligo? It doesn't affect any other area of my skin, however I know my grandfather had it. It couldn't be condyloma lata, could it? I'll try to get a clearer picture. Thanks for everyone's help. User:Cduffner 05:35, 22 October 2006 (UTC)[reply]

Vitiligo can be limited to one or a few areas. As for condylomata lata - no. - Nunh-huh 05:48, 22 October 2006 (UTC)[reply]

Maybe they're just blemishes that have always been there. Your chest is usually covered by clothes, so you would be less likely to notice them, and very few people examine their own chest closely on a regular basis. I remember the day I discovered the freckles on my shoulders that had been there my entire life. I thought it was specks of dirt! XD

Blemish is no more specific an answer than "spots" although in an advertising context it is always code for acne. Freckles are usually not present at birth and can appear any time throughout childhood. alteripse 13:13, 22 October 2006 (UTC)[reply]

Not vitiligo. Might be tinea. alteripse 13:13, 22 October 2006 (UTC)[reply]

Maybe some birth mark you haven't noticed until now???BlueLighter 14:15, 22 October 2006 (UTC)[reply]

Thanks alot, you've all been a great source of information and help. I figure I better just go and ask a dermatologist to be safe. --Cduffner

THis looks a very fuzzy picture. Its not a still from the Alien autopsy film is it?--Light current 17:57, 22 October 2006 (UTC)[reply]

So will you report back to us and award a prize to whichever answer was closest? alteripse 21:47, 22 October 2006 (UTC)[reply]

Then let me quickly get my prize-winning entry in: paucibacillary Hansen's disease. --Lambiam ^Talk 22:23, 22 October 2006 (UTC)[reply]

UPDATE: Turns out it's a fungal infection. Treatment? Apply medicated dandruff shampoo (i.e. Selsun Blue) on the affected area daily for 2-3 weeks. I probably got it from the pool. Thanks guys. --Cduffner

Told ya. But what amazes me is how you got in to see a dermatologist that fast. What is your secret? alteripse 23:13, 22 October 2006 (UTC)[reply]

My secret? Have a father who's a dermatologist. :D --Cduffner

Why on earth didn't you ask him first? Also wonder how Kainaw found himself shot and stabbed? --Username132 (talk) 13:50, 17 December 2006 (UTC)[reply]

BONDING IN SOLIDS

Explain the forces between two interacting atoms when they are brought nearer to form a molecule

Start with atom and molecule, they link to other articles that explain things in even greater detail. Weregerbil 07:00, 22 October 2006 (UTC)[reply]

There are many types of chemical bonds. See that link for details. StuRat 09:57, 22 October 2006 (UTC)[reply]

Molecules are formed by Covalent bonds, you should also take a look at that article. --`/aksha 12:26, 22 October 2006 (UTC)[reply]

Software Radio

In software radio,How is it posible to diffrentiate between noise and actual modulated signal???

Unless you have a copy of the message being sent or accutaely know its statistics (same thing really), it is not possible. That is the fundamental problem of communications. If you do have a copy, then no information is being sent! 8-)--Light current 13:39, 22 October 2006 (UTC)[reply]

would enough negative charge from other sources such as pions, down quarks etc react with normal matter?

would the negative charge from pions, downquarks etc , if controlled in sufficient amounts have the same effect as the negative charges that repel electrons from each other? and if so would the reaction be more powerfull, so instead of just stopping the other electrons, pushing them violently away? Hopefull

All charged particles obey the same laws of electromagnetism. The force only depends on the charges and the distance between them; no other properties of the particles have any effect on the electric force. That said, pions rarely last more than a microsecond, and down quarks can never be separated from other quarks because of color confinement, so neither can be "controlled in sufficient amounts". —Keenan Pepper 20:26, 22 October 2006 (UTC)[reply]

can anything other than particles produce charge enough to repel objects instead of just stopping them or are particles always required, can a photon or electron be charged up with more energy to repel stuff? hopefull

I don't understand what you mean by "repel objects instead of just stopping them". Electrons repel other electrons. —Keenan Pepper 04:48, 23 October 2006 (UTC)[reply]

All matter that we know of consists of particles. Photons have no electrical charge and do not repel. Each electron has an unalterable charge of −1. Charge is not a form of energy and is a conserved quantity in all interactions, so it cannot be "produced". Two objects will repel each other if they are both charged and their charges have the same sign. Opposites attract. See Static electricity. --Lambiam ^Talk 09:57, 23 October 2006 (UTC)[reply]

What do you mean by particles? Aren't pions and down quarks particles? Isn't the only thing known to exsist that isn't known to be a particle or collection of them gravity? Isn't even that strongly suspected to be a particle? — Daniel 03:50, 29 October 2006 (UTC)[reply]

Another EM wind question

How powerfull EM waves are required to project air at the speed and directional focus of this weapon, would it be possible to create an EM version of this weapon http://en.wikipedia.org/wiki/Vortex_ring_gun in the future that doesnt need shells, instead accelterates the air useing EM radiation. not now but perhaps in the future. Been trying to figure this out with other facts, but figured i should just ask directly already. Robin

I'm not an expert, but I think the intensity of the radiation needed to result in a notable acceleration of the air would be such that you get a deadly EM-ray gun. The point is that most of the radiation would just pass through. You'd also need rather large amounts of energy. --Lambiam ^Talk 18:47, 22 October 2006 (UTC)[reply]

Affordable Device for Measuring soil water content

Simple and affordable: Is there a simple affordable device for homeowners to use to determine if there is enough water content in the soil. This is for gardening, lawn care, general landscaping. This would work at my home.

more complex but afforable: I have a background in electrical engineering and would like to find/put together an affordable watering system (soil moisture measurement, automatic watering system, feedback loop,etc..). I found some info on Time Domain Reflectometers and have worked with those for another application but suspect that is expensive. Just need some ideas to get started. I tried the cheap (<$50.00) water timers (time of day - battery, amount of time- mechanical) but have not found one that works. This is for a family owned farm. My siblings want a small garden but no one has time to make the 60-90 minute drive to water daily.

Eostdiek

Simple resistive moisture detectors fed to comparators to decide whether to turn water off or on is all thats needed. You could control from a computer-- but whats the point?--Light current 14:31, 22 October 2006 (UTC)[reply]

Controlling it from a computer would have some advantages:

You could turn the water on and off from a remote site, via the internet.

You could base how much to water not only on current soil moisture levels, but also on the weather forecast. This could either be done with a fully automatic system that reads forecast info or by a human who reads a forecast and sets the amount to water (if any), as appropriate. StuRat 20:31, 22 October 2006 (UTC)[reply]

A device called rapid moisture metre is availlable for soil engineering use.The device uses carbide which is mixed with the sample of soil and acetelyne gas is produced.The pressure of the gas is proportional to the amount of water in the sample.The device is easily availlable at soil lab equipment shops.amrahs 17:08, 22 October 2006 (UTC)[reply]

If you read the initial question, youll see that what you propose is not suitable becuase the questioner wants remote automatic control that your scheme does not provide.--Light current 17:54, 22 October 2006 (UTC)[reply]

IR remote

What exactly would you need to build a simple IR remote control. Like a TV remote, only more simple, just an IR on/off switch. Obviously other than just an IR lamp/receiver set.--71.247.247.67 13:50, 22 October 2006 (UTC)[reply]

Basically, what would I need to make a simple one of these? and a matching receiver of course--71.247.247.67 13:53, 22 October 2006 (UTC)[reply]

Try Discover Circuits.[12]. There probably one there. Yes there is! [13] 8-)--Light current 14:21, 22 October 2006 (UTC)[reply]

What's this aprox. called again?

\varepsilon \left[\Psi \right]={\frac {\left\langle \Psi |{\hat {H}}|\Psi \right\rangle }{\left\langle \Psi |\Psi \right\rangle }}

~~It had a specific name, but I just can't place it, anyone?~~--71.247.247.67 14:16, 22 October 2006 (UTC)[reply]

Nevermind, it just came to me, and this after about 20 minutes of playing with <math> markup--71.247.247.67 14:19, 22 October 2006 (UTC)[reply]

Aqueous Compounds

I know that when a metal is added to water a base forms, and when a non-metal is added to water an acid forms, but can someone explain to me the steps in which a metal or non-metal combine with water to form an acid or base? thanks

It's been many years since I took chemestry, but I seem to recall water being both a weak acid and a weak base, depending on what is introduced to it. Ah, I was correct: Water (molecule)#Amphoteric nature of water. --Wirbelwindヴィルヴェルヴィント (talk) 20:08, 22 October 2006 (UTC)[reply]

First of all, what you said isn't true for all metals and all nonmetals. If you add the metal gold or the nonmetal argon to water, nothing happens. Second of all, even those elements that do react with water do it by different mechanisms. However, there are some general trends. If a metal is reactive enough to displace hydrogen from water, then it does so in a redox reaction and forms the hydroxide. For example, alkali metals react like this: 2M + 2H₂O → 2M⁺ + 2OH⁻ + H₂. Nonmetals are usually more complicated; I think all the halogens undergo disproportionation reactions. —Keenan Pepper 20:18, 22 October 2006 (UTC)[reply]

Solar position equations

Every web site I have visited that computes solar position provides a list of the various measurements such as hour angle, declination, azimuth, altitude, etc. The problem is that they all seem to differ even when the exact same time, date, latitude and longitude is entered. I thought that I could resolve this simply by finding the equations from which these values were computed but again there are a variety of equations given to compute each value and non of the results agree with the computations done online. I thought at first the problem might be in conversions from radians to degrees and vise versa but the results are still inacurate or wrong. All I am looking to do is to replace a table of sunsets and sunrizes with an equation or series of equations that compute sunrise and sunset. Is there a web site or article where the equations are given and that will provide consistent and accurate results or am I doomed to using tables instead? Adaptron 20:35, 22 October 2006 (UTC)[reply]

I think that the reason the websites differ is because some compute sunrise/sunset according to the middle of the Sun, and some according to its upper edge. It is possible to calculate the Sun's position by yourself (a few months ago, I wrote a program to do it); however, you'll need a lot of equations, and your results will be nowhere near as accurate as the websites'.

I suggest using NASA's HORIZONS system. When it comes to accuracy, you can't beat NASA. The HORIZONS system can provide all kinds of information, including sunrise and sunset times. --Bowlhover 22:15, 22 October 2006 (UTC)[reply]

Assuming the output data is accurate I can use it for verification. However what I need to to be able to generate sunset and sunrise from within a computer program versus looking up the times. Accuracy is always important but reasonable accuracy is acceptable. While some web sites have reasonably accurate times the other values that are listed do not agree with other web sites or with the equations I have. There must be a web site somewhere that lists reasonably accurate equations with example data that can be verified unless this is considered to be some sort of secret doings that only a trusted few can be told. Adaptron 22:50, 22 October 2006 (UTC)[reply]

Note that the exact time of apparent sunrise would also vary depening on local geography and weather. For example, the Sun will appear to rise far earlier on the top of a mountain than at the bottom of the Grand Canyon. Certain atmospheric conditions will also cause more diffraction of sunlight than others, causing the apparent sunrise to occur a bit earlier. Perhaps some of your calculations are "tweaked" to account for the local phenomena in a particular location ? StuRat 23:57, 22 October 2006 (UTC)[reply]

Actually the irony here is that while the web based sunrise and sunset results displayed on the websites are seem accurate and identical it is the other values such as time offset, equation of time, solar hour angle, azimuth and zenith that are not the same. In addition the equations provided by the web sites are not the same and the results for sunset and sunrise are inaccurate and not the same. How can the web site provide accurate times for sunrise and sunset but not accurate results for the other values of for sunrise and sunset from the equations they provide??? A list of equations to computer sunrise and sunset from latitude and longitude (and elevation if necessary) do not seem that far fetched. I would think they would be located all over the web. When you consider all of the other detailed and deep depth coverage of even esoteric topics on the web it just does not seem to make sense that there is no web site that provides an accurate set of equations with examples and other details. I get the feeling that it is a jealously guarded secret. Adaptron 01:59, 23 October 2006 (UTC)[reply]

There's no secret. Try sunrise equation. For a very detailed tutorial on how to write a program to calculate the Sun's position, try this. Also, from the same website, this webpage gives a tutorial on how to compute sunrise/sunset times. --Bowlhover 02:49, 23 October 2006 (UTC)[reply]

The Earth is not a sphere but more like an ellipsoid. Differences in the model used for approximating the shape, and differences between that model and the actual shape, can make a difference for angles and other relevant quantities. These differences should be quite small, and I don't know if they could possibly account for what you see. --Lambiam ^Talk 10:09, 23 October 2006 (UTC)[reply]

Okay what appears to be the problem is that many of these sites want you to purchase their software including the Naval Observatory. Although the prices are not outrageous ($13 - $29.95) it appears that they are using the tactic of only providing incomplete, crude or outright incorrect equations while at the same time providing accurate results on their web sites. Although the equations they are using to generate results may be accurate, detailed and complete the ones they have published on the web site are not. Adaptron 16:01, 23 October 2006 (UTC)[reply]

Equilibrium Constant, Kp

I am studying equilibrium constants in my college-level chem class and am having trouble understanding reasons behind the directions of the reaction for the equilibrium. When dealing with gases, if the volume is decreased, the pressure increases and the reaction favors the side of the equation with the smaller number of moles of gas and vice versa. It does make logical sense that if there is less volume the reaction will favor less moles, but I am wondering if there is something more concrete to explain this. I think my teacher said the reason is based on the kinetic molecular theory, but I've consulted my text and another text and I don't see how the kinetic molecular theory can explain this. Can anyone shed some light? Thanks.71.240.47.118 21:06, 22 October 2006 (UTC)MS[reply]

In general, these kind of equilbria are determined by entropy considerations, but there's an easy way to talk about this particular case: whichever side of the equation has fewer moles of gas must, on average, be the result of synthesis and not decomposition reactions. Synthesis reactions will happen more frequently when there are more particles per volume, because there are more collisions (more attempts at combination). However, the decomposition reactions are not encouraged by a greater number density (at least, not nearly to the same extent), so when the number density increases it takes more molecules that can decompose to balance the increased number of combination reactions, so the equilibrium shifts. --Tardis 20:07, 24 October 2006 (UTC)[reply]

i have great cofusion in Kp n Kc. it has been observed that sometimes Kc is greater than Kp in some of the reactions while in some cases it is lesser than that???????why is that so i want a logical answer

Nails

Why do we have white part close to the bed of our nails? They look like half moons, and I could swear that at some point in my life they seemed more pronounced.. so what are they?

Svetlana Miljkovic 22:11, 22 October 2006 (UTC)[reply]

It's called the lunula; our article says only a little about its function. -- Finlay McWalter | Talk 22:17, 22 October 2006 (UTC)[reply]

I believe that's where the nail is still growing. After that point, it's just being pushed along. StuRat 23:44, 22 October 2006 (UTC)[reply]

If that is so, if you cut your nails flat then leave them to grow without cutting, the ends should take up the shape of the 1/2 moons-- Yes?.--Light current 02:10, 23 October 2006 (UTC)[reply]

Sort of. The shape of your nails depends on not only how they grow, but how they are cut and torn, bumped etc. But they do tend to grow in that shape which is why nail clippers are curved, and if you peel your nail off (not the whole thing, just the tip) it will make a curved shape. 74.117.47.205 03:24, 23 October 2006 (UTC)[reply]

Ah well, thats funny. I always cut my toe nails in a striaght line. And they continue to grow in a straight line! 8-)--Light current 14:57, 23 October 2006 (UTC)[reply]

Just because they are growing across a curved area doesn't mean they are growing at different rates, which is what would be needed to change a straight cut to a curved edge. StuRat 19:30, 23 October 2006 (UTC)[reply]

Rainbow

I am trying to help my son with his homework and we are stumped, please help. When you stand with your back to the sun, you see the rainbow with a circular arc. Couls you move off to one sidr and then see the rainbow as a segment of an ellipse rather than a segment of a circle? Defend your answer.

Did you read our article Rainbow? You can't move to one side any more than you can jump over your shadow, even if you take a big leap. Promise you won't attack our answer. --Lambiam ^Talk 22:44, 22 October 2006 (UTC)[reply]

In other words, of course you can move to one side from your starting position -- but just as your shadow follows you, so does the rainbow. It keeps its circular shape, which exists because the raindrops that light up with each particular color are the ones that form a specific angle with the sun and your eye. (There is a second angle for each color that also works, but less well, hence the double rainbows when the rainbow is intense enough.) As you move, the position where this happens moves with you.

If you're talking about a natural rainbow in the sky, when you move sideways the movement of the rainbow is usually too small to see. You move sideways 20 feet, the rainbow moves sideways 20 feet, but you're looking at it against a plain sky background or against a background of distant objects and you can't see 20 feet of motion at that distance. So you have the illusion that the rainbow is fixed in place. But try it with a spray of water that you can get close to, like from a fountain, or a garden hose nozzle (held by someone else and set to a fine spray). A lawn sprinkler would also work if it's the kind that makes a continuous fine spray that stays in one place, and still another choice is the spray from a waterfall if you can get close enough to it. Get the sun angle right and you'll easily see a rainbow. Now move and watch the rainbow moving with you.

Even with a natural rainbow you can occasionally see it move with you, if you're watching it from a moving vehicle and you catch it against the right sort of background. For example, I've seen this when driving past a mountain lake where there was a rain shower over the lake and sunlight shining on the rain; the rainbow could be seen moving against the mountains on the far side.

--Anonymous, 05:23 UTC, October 23.

Factual basis in psychology for claims about the power of the imagination

Frequently, we hear talk of how something bad has happened to someone, like a relationship has broken down, or they have lost their job or something, and people say that deep down, the person "wanted it to happen." For example, if someone has low self-esteem, and they get a really great job, they might (allegedly) sabotage themselves, and somehow contrive (subconsciously) to get fired, in spite of making every outward sign of effort. The idea is that the underlying self-image is more powerful than the determination to succeed. Presumably it can also work positively for people with high self-esteem. These kinds of observations are commonplace, but what is the scientific basis for them? Do psychologists believe in this power of the imagination, and do they base this on science, or their general understanding and clinical experience? The Mad Echidna 21:41, 22 October 2006 (UTC)[reply]

Coincidentally, there was a study in my home town just recently. It's not exactly like the examples you gave, but it's close. False stereotypes can affect performance: B.C. study. Anchoress 22:33, 22 October 2006 (UTC)[reply]

People who believe they're going to fail, no matter what, are more likely to fail, no matter what. The explanation is simple: they have no incentive to improve their performance. People who believe they're going to succeed, no matter what, are more likely to fail, no matter what. The explanation is simple: they have no incentive to improve their performance. (But they are more likely to bluster their way out of the mess they create.) Applying this pop psychology insight as a post-hoc explanation to failing individuals has a problem of unfalsifiability. --Lambiam ^Talk 10:25, 23 October 2006 (UTC)[reply]

Is the human brain a chemical machine?

I would like to ask some further questions. Is the human brain a chemical machine? Is thinking the result of chemical processes? 202.168.50.40 21:50, 22 October 2006 (UTC)[reply]

That seems a bit of an oversimplification. At the very least it is a combination of electrochemistry and chemistry. Read our article Brain, and check out the role of neurons. Then see also the arguments from various sides in Type physicalism. --Lambiam ^Talk 22:40, 22 October 2006 (UTC)[reply]

Francis Crick wrote a book called The Astonishing Hypothesis in which he observed, "a person's mental activities are entirely due to the behavior of nerve cells, glial cells, and the atoms, ions, and molecules that make them up and influence them." --JWSchmidt 22:44, 22 October 2006 (UTC)[reply]

This is a question that will seem more and more pointless, the older and more educated you become. The answer of course is "yes and no". The entire human body can be described as a chemical machine, and we learn more details every year. That perspective has yielded enormous understanding of mechanisms of inheritance, function, disease, and even some aspects of behavior and personality. On the other hand, there are many other perspectives about the human mind that also contain truth and wisdom, including some truly valuable insights from religious and artistic perspectives. Another perspective entirely is the interpersonal, the way in which our minds are shaped by our relationships with other people. None of these perspectives can be described well from the "chemical machine" perspective. Intelligent adults recognize that different perspectives are useful for different purposes, and choose the best perspective for a specific purpose. The chemical machine perspective does not tell you what would make your life worth living. alteripse 23:08, 22 October 2006 (UTC)[reply]

Reductionist answers can answer some kinds of questions very well, and some kinds of questions very poorly. Regarding the mind as a chemical machine is useful for answering some questions ("What is the physiological cause of happiness?") but not for others ("What is the ultimate cause of happiness? How can I achieve it with the people I love?"). Answering questions in only physiochemical terms is rarely useful for anyone but the physiologist or the chemist. --Fastfission 00:53, 23 October 2006 (UTC)[reply]

Obviously yes. How else do you thimk it works?--Light current 02:15, 23 October 2006 (UTC)[reply]

Perhaps we should say electro-chemical.--Light current 16:32, 23 October 2006 (UTC)[reply]

lens

What type of corrective lenses are in the eyeglasses of a person who is nearsighted?

~~Convex~~. The things can't start a fire at all. :( Hyenaste ^(tell) 22:36, 22 October 2006 (UTC)[reply]

I think concave, actually. See Lens (optics). A convex lens can be used to focus light to a point, and thus start a fire. A concave lens spreads light out, thus (sans modification) can not be used to start a fire. This error was made in Lord of the flies when the myopic (nearsighted) character Piggy starts a fire with his glasses[14]. --TeaDrinker 23:04, 22 October 2006 (UTC)[reply]

Oh of course. I was thinking mirror instead of lens (a convex mirror focuses light, ya?). Hyenaste ^(tell) 00:05, 23 October 2006 (UTC)[reply]

No, a concave mirror focuses light. A convex lens also focuses light, but you don't use it in a pair of glasses to treat myopia. You use it to treat hyperopia. --Bowlhover 01:56, 23 October 2006 (UTC)[reply]

You should all go see an optician! (if you can) 8-(--Light current 02:03, 23 October 2006 (UTC)[reply]

Lab report!

Hello there! I'm a little stumped with my lab report I have to hand in tomorrow. I'm quite stuck on a part where it asks me the molarity of my acid solution. The lab was a titration lab with HCl and NaOH. Currently, I have the volume of my acid solution and base solution. It asks for the molarity. I understand Molarity = Moles / Volume (Liters). How would i be able to find the moles of my acid? I don't remember correctly but is it possible to find moles from liters? --Agester 22:55, 22 October 2006 (UTC)[reply]

According to the hydrochloric acid article, the density of HCI is 1.18 g/cm^3. Since you know how many litres of the acid you have, you can use this to find out the mass.

Once you have the mass, you can calculate the number of moles it's equivilent to. One mole of HCI is equal to 36.46 g. --Bowlhover 23:20, 22 October 2006 (UTC)[reply]

I'm not sure Bowlhover's answer is correct - the density given is for a 37% solution of HCl, which may not be what you have. I would have thought that, given that the experiment was a titration, the purpose of it was to determine the molarity of your HCl solution by finding out how much was required to neutralise a given volume of NaOH solution of known molarity. What you need to do is to find out the molarity of the NaOH, then check the volume of HCl you'd titrated at the end point, and then using those figures and the equation of the reaction, determine the molarity of your HCl solution. --YFB ¿ 23:31, 22 October 2006 (UTC)[reply]

I know if i find the molarity of the acid OR base i could find the molarity of the other. The main problem is that all i have is the volume of BOTH the base and acid.... wait! i think i have an idea! since it's a one to one mole ratio (for neutralization which is what i did in the experiment!) wouldn't my moles be 1? since the formula is written as HCl_(aq) + NaOH_(aq) ---> NaCl_(s) + H₂O_(L) (coefficients are one) therefore my Molarity would be 1 mol / (my volume in L) for my acid / base?? --Agester 23:38, 22 October 2006 (UTC)[reply]

You're right that it's a 1-1 mole ratio neutralisation, but if you don't know the molarity of either solution then I can't see how you can find the molarity of one from this experiment (Acid-base titration seems to agree with me). You'll need someone more knowledgeable to help you here... all I have is my 4-years-neglected recollections of A-level Chemistry, I'm afraid. --YFB ¿ 23:47, 22 October 2006 (UTC)[reply]

Yes. This lab is killing me because I know so easily Molarity_Acid X Volume_Acid = Molarity_Base X Volume_Base. I could find out the problem of this in 2 minutes if they gave me either moles, or molarity of either acid and / or base! --Agester 00:00, 23 October 2006 (UTC)[reply]

The required information may be "hidden". For example, your lab manual might say something like, "The solution you will use for the titration was made by dissolving x grams of solid y in z liters of water. --JWSchmidt 02:44, 23 October 2006 (UTC)[reply]

Hmm thanks for the input all! I got it. I found out the molarity of the acid (HCl) was given as .1000 Molars (constant for this experiment). It was confusing because the lab manual said the Normality of the base solution was 5 N. And we don't use normality in our lab we use molarity. In addition, i don't think we used a 5 N basic solution since we diluted our base so heavily. I doubted the manual which is why i came here for help. (if i did use that 5 N i would've been in big trouble.) Thanks for the input everyone! --Agester 01:51, 24 October 2006 (UTC)[reply]

October 23

The "Wholly Grail".....

Let`s talk about the Wholly Grail if we could, for a bit. More specifically, Virtual Particles. You know, those "pesky" electron-positron pairs that are alleged to exist at this level. My understanding of these 'alleged' particles is that they don`t exist long enough to be REAL...i.e. they exist for shorter than a Plank second, therefore, don`t REALLY exist at all. This could bring Hawking Radiation into question. I have, I think, a VERY interesting follow-up to this but I`d like to 'hear' everyones` ideas about these 'alleged' existances first, for fear of sounding a fool. Please, no jokes! TY Dave 205.188.116.74 00:27, 23 October 2006 (UTC)[reply]

Is your spelling of "Wholly" meant as some sort of pun? If so, I don't get it--64.12.116.74 01:17, 23 October 2006 (UTC)[reply]

Also, may I suggest, if you want to sign with a name, sign '''name'''~~~~~, that way the name stands out more than with name ~~~~--64.12.116.74 01:22, 23 October 2006 (UTC)[reply]

I didn`t want to include a spiritual or religious connotaion. You can call it "Holy", if you want, but I`m strictly aiming towards science. There was no pun intended, to answer your question. I have no clue what the person is talking about in regards to "signing", sorry. Please keep to the main subject. Dave 205.188.116.74 01:29, 23 October 2006 (UTC)[reply]

I too am utterly confused about the title. Since when is wholly an adjective? As for the question itself (which, funnily enough, has no question mark anywhere), who ever said virtual particles were real? I'm sure by Plank second you mean Planck time, but virtual particles can exist much longer than the Planck time (it depends on their rest energy: the more energy they have, the less time they can "get away with it", according to the uncertainty principle for energy and time). Anyway, how does any of this "bring Hawing Radiation into question"? —Keenan Pepper 04:42, 23 October 2006 (UTC)[reply]

First, will you folks please drop the "Wholly" already? It`s amazing how often you JUMP on someone over a single word, completely forgetting/ignoring the question/comment! Sheesh! Second, who said it was a question?..."Let`s talk..." Third, I DID say, ""MY" understanding of...". Perhaps I was wrong there...forgot about energy/time. Fourth, assuming I was correct, just for an instant, perhaps a Plank second even,(the tiniest piece of Plank Time, thought you`d get that!) if virtual particles didn`t "last" long enough to be 'seen' by Gravity, then that would CERTAINLY put Hawking Radiation into question! I guess I`ll save my VERY interesting follow-up for when I have some more-interested listeners/readers. Any thoughts about Vacuum Genesis? Thanks, Dave 152.163.100.74 05:49, 23 October 2006 (UTC)[reply]

The Reference Desk is not a soapbox, bulletin board or chat channel. It is meant for asking questions to which you could not find an answer. --Lambiam ^Talk 10:42, 23 October 2006 (UTC)[reply]

That question is crackpot physics. Hawking radiation is due to virtual particles. There's nothing 'holy grail'ish about virtual particles - they're a well-understood part of mainstream physics, unlike the question which doesn't seem to have understood any of the terms it's using: Planck time is an unrelated concept to virtual particles. You can have a virtual particle for as long or short time as you want. --83.145.46.141 14:06, 23 October 2006 (UTC)[reply]

How gyrocompasses work

After reading the gyrocompass article, I spent a long time trying to figure out how it works. The article describes 2 types of gyrocompasses: the first kind is filled with fluid, while the second kind has weights that pin down its gyroscope.

I pretty much understand how the first kind works. However, according to the article: "This friction force caused by the fluid results in a torque acting on the axis, causing the axis to turn in a direction orthogonal to the torque (that is, to precess) toward true North (to the North star)". Is the torque applied in the direction opposite to the direction the axis is travelling in? If so, how will this torque cause the axis to move toward the north star?

I have no idea how pinning down a gyroscope's axis will cause it to point north. Shouldn't it point east, because that's where the stars rotate around the celestial north pole as much as they do around the celestial south pole? Can someone explain what I'm missing here? --Bowlhover 01:52, 23 October 2006 (UTC)[reply]

The gyroscope's axis points north, the gyroscope spins around its axis. It points north because it is aligning it self with the earths rotation, the earth also rotates around an axis which points north. This might give you a broader insight: equatorial mount. Vespine 06:40, 23 October 2006 (UTC)[reply]

I know that, but how does it align itself with Earth's rotation? --Bowlhover 15:37, 23 October 2006 (UTC)[reply]

Because that is the only orientation where the gyroscope's axis will not experience a change to its angle due to the rotation of the earth. Hold a pencil vertically and then spin around with your arm out, if your friend looks at the pencil, it will look like it is pointing in the same direction while you are spinning. If you hold the pencil on an angle from vertical, the place where the pencil is pointing will change as you turn. You could do this with an actual gyroscope instead of a pencil and feel the actual force involved.Vespine 01:05, 24 October 2006 (UTC)[reply]

I don't think I've made myself clear enough. So, here is what I do and don't understand:

The gyroscope is spinning around its axis. Right now, its axis is pointing toward a certain part of the sky. Earth is changing its velocity by rotating, but the gyroscope tries to resist this change in velocity (due to Newton's first law of motion). As the result, the axis will point to the same piece of sky relative to the stars. This "piece of sky" will rotate around Polaris in a counter-clockwise direction, just like the stars do.

Since the gyroscope's axis is rotating around Polaris in a circle, it will rarely point directly north. As a result, torque has to be applied to decrease the size of this circle (i.e. force the axis toward Polaris). My question is: how will submerging the axis in a fluid force it toward Polaris? --Bowlhover 04:29, 24 October 2006 (UTC)[reply]

Mercury and Fillings

Can mercury from a 30 to 40-ish year old filling come out and be hazardous to you health? Deltacom1515 02:09, 23 October 2006 (UTC) 02:05, 23 October 2006 (UTC)[reply]

Current research and the accepted beliefs of dental health organizations hold that dental amalgam fillings do not harm your health. - Dozenist talk 02:10, 23 October 2006 (UTC)[reply]

Its not done me any haarm yet >Harr, Haa HAAA haarrr!! HHAAAAA!!! HAAAAAr haarrrr!. Where is that hatter? See mercury--Light current 02:13, 23 October 2006 (UTC)[reply]

It's believed to not be harmful. Although i'd say most people who've encountered mercury in a chem lab would have doubts about that statement. We actually have an article on this, take a look Dental amalgam controversy#Health effects generally --`/aksha 02:30, 23 October 2006 (UTC)[reply]

No one is disputing the effects of raw mercury. What is at question is the safety of amalgam in fillings. 8-)--Light current 02:39, 23 October 2006 (UTC)[reply]

.....amalgam's contain mercury. --`/aksha 02:56, 23 October 2006 (UTC)[reply]

Correct.... amalgamated with other metals and therefore not RAW! Read my original post again.--Light current 03:14, 23 October 2006 (UTC)[reply]

Yaksha, I certainly do not see mercury in a chem lab ever, but I work with amalgams frequently. To repeat, amalgam fillings are not considered harmful. - Dozenist talk 03:19, 23 October 2006 (UTC)[reply]

Salt is not harmful, even though pure chlorine and sodium are extremely harmful. Just because the original subsituents are harmful, does not mean the compound/alloy is. Generally, the more reactive (and therefore, usually the more dangerous) a chemical, the safer the compounds they create. If they release a lot of energy in a reaction, it is not easy to reverse the process. --liquidGhoul 12:09, 23 October 2006 (UTC)[reply]

The question was about "Can mercury from a 30 to 40-ish year old filling come out". Amalgam isn't meant to be harmful. It's believed the mercury doesn't do anything funny. But if the mercury came out? That's the point. And my comment about the chem lab was meant to be a joke. (Although i can't really imagine why a general chem lab wouldn't have some mercury somewhere, it's not that usual. Even my high school chem labs had it, never worked out why though, it's not like they ever let students near it...) The link to the Dental amalgam controversy was meant to be my answer. The article explains it all. --`/aksha 14:02, 23 October 2006 (UTC)[reply]

Archiving

This section has been moved to Wikipedia talk:Reference desk#Archiving. --hydnjo talk 00:20, 24 October 2006 (UTC)[reply]

Different roles for insects in a hive/nest

Hi - this might be a silly question, but can someone tell me the entomological term for the different types of members of, say, an ant nest - the soldiers, the workers, the queen - they're not separate species, of course; are they separate subspecies? - phenotypes? - breeds? - what's the correct word here? Thanks Adambrowne666 02:42, 23 October 2006 (UTC)[reply]

I don't know what they are. But i do know they're not seperate subspecies or breeds. It's not to do with genotype/phenotype either. Insects, especially community insects like bees and ants, commonly have permanent gene loss during development. So your queen and your soldier ant could have almost the same genes, it's just that the soldier ant losses a part of its genome permanently during development. Where as the queen keeps all the genes, obviously...since the queen does the reproducing. I'd assume different 'classes' in a nest are due to different genes lost, since all ants within one nest/community have very similar genetically. --`/aksha 02:55, 23 October 2006 (UTC)[reply]

They're called castes. - Nunh-huh 04:02, 23 October 2006 (UTC)[reply]

Effectively, they're just different genders. In a honeybee colony for example, the queen is female, the workers are infertile females (they have under developed ovaries, and can't mate, being inhibited by the influence of an existing queen) and the drones are males. The same is true of the ant colony (in this case, queen, worker, soldier respectivally). M a rtinp23 23:03, 23 October 2006 (UTC)[reply]

In case you're interested, there is only one queen in a honeybee colony, and she lays eggs which hatch into workers and drones. The difference between the two eggs is that the worker one is fertilised with sperm from a drone, and the drone egg isn't. Therefore, interestingly, the males have no father in a honeybee colony. To produce a queen, an egg is laid fertilised, like that for a worker, and is fed on an enriched diet of Royal Jelly, allowing her ovaries to fully develop. The queen mates once in her life with around a dozen drones from different colonies (who die after impregnating here) and she stores the sperm for the rest of her life (up to 5 years) within her body. </beekeeping lesson> :) M a rtinp23 23:03, 23 October 2006 (UTC)[reply]

Castes! Of course! Thank you very much. Adambrowne666 23:07, 23 October 2006 (UTC)[reply]

how may calories are in... me?

To start.... this is NOT A MEDICAL QUESTION. I do not intend to be eaten or eat myself. I am purely curious about this matter.

I am very curious about how many calories there are in me. I am a 5'1" 140lbs woman with a 32% body fat percentage. (I know, I know, I need to lose weight). This means that my "lean body mass" is appx 96 pounds. It would be easy enough to add together the calories of 96 pounds of say, ham, and 44 pounds of pure fat, but there is the issue of bones, hair, brains, teeth, fingernails etc. I am assuming that my organs will not be eaten (although it would be interesting to find out how many calories are in a human kidney or liver). So is there any way to find out what is the weight of my muscle, or non-soft-tissue? All help is appreciated.

PS: Is it legal to eat myself, say, if I am wearing a mask? What about a red mask? ;) 74.117.47.205 03:20, 23 October 2006 (UTC)[reply]

We have a dog and four cats and they seem to be trying (to eat themselves) on a daily basis and without masks. I don't know about the calorie part as they seem to neither lose nor gain weight. ;-) --hydnjo talk 04:13, 23 October 2006 (UTC) addendum: If you want to do the math, one pound of fat equals about 3500 - 4000 calories (actually it's kilogram calories or kcals) and one pound of muscle (meat) equals about 1600 - 1900 calories (using the same colloquial term). Water makes up a significant part of your body weight (about 70 pounds in your case) so you'll need to subtract that out. Also, so as to confound your calculations even further, your 32% body fat percentage may include some of the water weight as does the muscle (meat) weight. All in all, your body properly preserved and so on. should provide enough sustenance for one person for several weeks however I would suggest including some vegetables and fruits for a balanced diet. ;-) --hydnjo talk 04:50, 23 October 2006 (UTC)[reply]

I'm not sure what you are trying to get at. If all you care about is the total amount of energy within your body then a calorimeter should be used. Ideally, we put you into the "large" calorimeter and burn you alive and measure the total amount of heat release. Short of that, we use a cadevar in a body similar to yours and burn it instead. 202.168.50.40 04:27, 23 October 2006 (UTC)[reply]

Do bones have calories? What about hair, cartilage, eyeballs etc? I am not talking about my body as a whole, but rather the parts of my body that a human would consume if he were to eat me. 74.117.47.205 04:34, 23 October 2006 (UTC)[reply]

Since you weight 140 lbs, your body mass has about $5.70735888\times 10^{18}$ joules of energy on it (E=mc²), which gives $1.36409151\times 10^{15}$ kcal. Good luck on losing that. :) ☢ Ҡi∊ff⌇↯ 04:31, 23 October 2006 (UTC)[reply]

Umm, don't forget to subtract out the water and bone (hair, nails etc). --hydnjo talk 04:55, 23 October 2006 (UTC)[reply]

That's if you made her into an atom bomb... --Wirbelwindヴィルヴェルヴィント (talk) 05:23, 23 October 2006 (UTC)[reply]

Wow! I sure would like to get the concession rights for that one. Step right up ladies and gentlemen ... --hydnjo talk 05:46, 23 October 2006 (UTC)[reply]

Atom bombs work by exploiting the mass defect between different atoms to create its energy. Atom bombs are far less potent than the energy amount quoted above. That amount of energy (e=mc²) could only be released if you annihilated every atom (e.g. with an anti-person).Richard B 23:13, 23 October 2006 (UTC)[reply]

Fat and muscle contain water, I dare say that the kcals figure quoted above includes that water, it isn't kcals of "dehydrated muscle or fat". So I don't think you subtract the water from the calculations.... As far as "edible" goes, most of the body would be and have some sort of nutricius value, as far as eyeballs, cartilage, kidneys, lungs, brains. About the only thing I think isn't digested is teeth and hair, you'd have trouble eating the bones, but you would digest them, if you were to grind them up for example.Vespine 00:42, 24 October 2006 (UTC)[reply]

Feral Horses of the World?

Hi, I've been trying to research feral horse populations, but haven't been able to find many, just the well-known ones. I.E. Sweden's Gotland Pony or North Carolina's Corolla "Wild" Horse/"Banker Horse". (The only true wild horse is the Przewalski's Horse.) Can anyone point me to a good site/book that lists feral horses of the world? If not, could I please put in a request for such an article? Thanks in advance:) —The preceding unsigned comment was added by StandardSmiley (talk • contribs) .

The horse section of feral suggests that there are hardly any real wild horses, so any horses in the wild are likely to be feral. Confusingly, the wild horse article uses the term 'equus ferus'. But I don't really know anything about the subject. I'm just doing your work for you, looking it up in Wikipedia. Note, btw, that feral horse simply redirects to 'horse'. As for requesting an article, you can do that at Wikipedia:Requested articles. I have never done that, so I don't know how successfull such a request might be. DirkvdM 08:21, 23 October 2006 (UTC)[reply]

The confusing thing is that a Latin word meaning "wild beast" has been given a rather more specialized meaning when applied to animals; specifically: domesticated animals having reverted to a "non-domestic lifestyle". Equus ferus is simply classical Latin for "wild horse". When your compatriot Pieter Boddaert introduced the scientific binomial name, the word in English just meant "wild", "savage". --Lambiam ^Talk 09:29, 23 October 2006 (UTC)[reply]

Chlorinated water and eye damage.

Would my eyes be damaged from exposure to chlorinated water for 20-30 minutes a day? I often swim with my eyes open, and haven't noticed any damage to my sight in the six months I've been swimming, but wanted to make sure. —The preceding unsigned comment was added by 202.10.86.63 (talk • contribs) .

I assume you've read the medical disclaimer, so I'll skip the lecture. Swimming pools are "chlorinated" with hypochlorite salts like sodium hypochlorite (a.k.a. chlorine bleach). A quick search on Google Scholar for cornea hypochlorite damage doesn't turn up anything worrying. It causes "irritation" (duh), but I don't see any mention of long-term damage. —Keenan Pepper 05:03, 23 October 2006 (UTC)[reply]

Fissile material

After obtaining enough fissile material, what would one need to do to create a crude nuclear bomb (assuming one also had trained nuclear technicians to do so)? Like, what kind of steps would ahve to be taken? For the record, I'm not a terrorist, I'm just writing a thriller for NaNoWriMo. —The preceding unsigned comment was added by 58.7.219.86 (talk • contribs) . 01:17, 23 October 2006 (UTC)[reply]

It's not just obtaining the fissile material - it's seperating it from non-fissile material. Fissile uranium and plutonium make up less than 1% of all teh uranium and plutonium in all of nature. So, you have to run it through very long pipes (3 miles or more); the lighter (fissile) material comes out first, and that's what you put in your bomb. The problem is that in order to make it run through the pipes, you have to bond it with flurine to make it Uranium hexafluoride gas. However, Uranium hexafloride is *extremely* corrosive. In early WWII test, it ate through a 3 inch think glass pipe in 10 minutes; and a 3-inch steel pipe in two hours. The next trick: coat the inside of the pipe with Telfon and your corrosion problem is solved (This is why Teflon was a major war secret). After that, it's relatevely easy to build a uranium bomb. A plutonium bomb requires a lot mroe work though. Raul654 06:04, 23 October 2006 (UTC)[reply]

That's only if you are trying to enrich the material yourself. No terrorist has the facilities to enrich their own material; they'd either try to buy it or steal it. Both would be somewhat difficult; I'm inclined to think stealing it would be slightly easier though not much (buying it already requires somebody to have access to it and be willing to stake their own life on it, and you also have a high chance of either ripped off by the seller or by accidentally trying to buy it from a CIA agent). --Fastfission 00:36, 24 October 2006 (UTC)[reply]

Read the Nuclear weapon article. I'm afraid that's all the help we can provide right now (and it's substantial). I usually don't speak for other users but in this case I'll make an exception; none of us are terrorists. Perhaps you could add your manuscript to Wikibooks as a token of thanks for our research. --hydnjo talk 06:04, 23 October 2006 (UTC)[reply]

Yep, and nuclear weapon design. If you've got enough highly enriched uranium, the task is conceptually simple - just slam two bits of it together at a reasonably high speed - the easy way to do this is essentially whack one bit of uranium in a custom-built cannon and have the other bit at the end of the barrel. With a bit of experimentation, I reckon I and the local blokes down at the local welding shop would be able to do that. By contrast, if you're trying to use plutonium, it's a much, much harder task. You need to create a set of explosives that can "pinch" a sphere of plutonium into a super-compressed state, and then just as the compression reaches its maximum release a supply of neutrons that starts the reaction going - if you get the timing of the explosives or the neutrons wrong, you may get no bang at all, or a much smaller one. From the information to hand, it seems like North Korea's nuclear test was just such a "fizzle". --Robert Merkel 06:30, 23 October 2006 (UTC)[reply]

In The Fourth Protocol (film), you can see Pierce Brosnan assemble a nuclear bomb. Not too much detail, but it's a good film, so if you might ever want to watch it, now would be a good time. Btw, I like your disclaimer. I wonder what answers you would have gotten if you would have said you were a terrorist, since that would obviously have meant you weren't. Or would it? Let the double guessing begin. :) DirkvdM 08:05, 23 October 2006 (UTC)[reply]

Uranium bombs are easier to build - you just fire two sub-criticle masses at eachother, when they are close enough to make a criticle mass they explode, assuming enough neutrons are floating around in the environment. The WWII bomb had a 10% chance (if memory serves) of not working due to a lack of available neutrons.

Plutonium bombs need an initiator to supply a burst of neutrons at the right time to start the reaction (during implosion). The WWII bomb used a Berillium based initiator that is still classified (as of twenty years ago). Modern plutonium bombs use a minuture linear accelerator and target material for the initial neutron burst. Dallas67 10:25, 23 October 2006 (UTC)[reply]

The Little Boy gun-device also had neutron initiators. They wouldn't have left anything to such a high chance as 10%. --Fastfission 00:36, 24 October 2006 (UTC)[reply]

If you obtain enough fissile material you've gone 90% of the way to making a very crude nuclear weapon — fissile material will be the biggest and most difficult step for any non-state entity. You could create a pretty nasty explosion just by throwing a critical mass together any which way. It might not be on par with a kiloton-range weapon but it would be pretty impressive on a human scale. It really depends what you, the hypothetical user, are going for. If it is just a scary and deadly explosion you can get that without it being elegant from an engineering standpoint. The fancy implosions and all of that are for when you really want to get a VERY big explosion out of it; if you are content with something which is less than a kiloton in yield (which is still very large from a human point of view; the Oklahoma City bombing explosion was only .002 kt in yield) and radiologically messy you can cut a lot of complicated engineering corners. --Fastfission 00:36, 24 October 2006 (UTC)[reply]

On a related note: I've heard whispers of the theoretical possibility of a fustion bomb without a fission trigger, relying on super accurate implosion techniques. Does anyone have any information on this? Dallas67 02:11, 24 October 2006 (UTC)[reply]

That possiblity is discussed in Induced gamma emission and Sam Cohen and the book Imaginary Weapons. (Hafnium bomb) Current consensus says it doesn't work. --GangofOne 05:16, 24 October 2006 (UTC)[reply]

Genetic tendancy towards AIDS amongst Black people?

It seems all statistics show that black people have a greater chance (up to 15 times!) more to get AIDS then a white person of European decent [15]. Is this a genetic tendancy or some other reason? Has there been any scientific studies? — Preceding unsigned comment added by 68.235.225.150 (talk • contribs)

The question is flawed. There is absolutely no genetic basis for AIDS.^{[citation needed]} Raul654 05:59, 23 October 2006 (UTC)[reply]

I believe Africans have a high rate of AIDs, did you mean that? See AIDS#Economic_impact and the graph there, for example. But no, AIDS isn't a genetic thing; it has to do with behavior, which might be impacted by upbringing, family, where someone grew up, etc. --Wirbelwindヴィルヴェルヴィント (talk) 06:09, 23 October 2006 (UTC)[reply]

I am seeing different responces from different people. It seems most black professors who have studied the subject claim that AIDS is in fact a genetic disease, while white scientists claim that it is not. I haven't been able to find any scientific studies. 68.235.225.150 06:25, 23 October 2006 (UTC)[reply]

Indeed. It is probably due to behavior. --Proficient 06:27, 23 October 2006 (UTC)[reply]

Okay... just like any disease, there is a genetic component to HIV infection, at least with regards to susceptibility, meaning that there is likely some degree of variation; I recall reading a paper about a number of African women appear to be entirely immune to the disease, probably because they carry a mutation in a gene that affects the expression of the glycoproteins that HIV attaches to. I cannot say whether any particular population is more or less susceptible to HIV infection, (and I have no doubt that the answer is complicated by the fact that the strains of HIV that are prevalent in Africa are not the strains more common to Europe and the Americas) but I would think that someone, somewhere, managed to get grant money for such a study. If you find an answer, let me know. – Clockwork Soul 06:38, 23 October 2006 (UTC)[reply]

For a start, i believe the question you're really asking is whether Black people are more suspectible to HIV, the virus which results in the syndrom called "AIDS".

The answer is, yes, black people more commonly get HIV and are more 'suspectible' to AIDS. But it has nothing to do with genes.

Let's start with HIV.

- HIV originated in poor parts of asia/africa area, and didn't enter western society for quite some years. Not surprisingly since people from those poorer parts don't often immigrate to the western world. In other words, HIV has had far longer to spread in africa, where it's all black people.
- no one knew what HIV was at first. When it hit the western world - the western world keeps medical records, has medical databases, has hospitals and doctors and the facilities. When people noticed a how bunch of gay guys suffering from immuno-deficient disorders, they had the resources to investigate. They worked out what it was, what caused it, and how it spread. I'm afraid africa doesn't have those luxuries. Since AIDS is a slow disease (you don't get sick and die immediately from contracting it), AIDS probably existed and had time to spread for ages in africa before it hit the western world and got identified. In africa, HIV has had the chance to integrate into the population so much that children are often born carrying the virus these days (because their mothers do), which further increases the rate of HIV in those countries.
- HIV is spread primarily as a sexually transmitted disease. Condoms are rare in african countries, people do not have a culture of using them. Rape is also common. It's also common for men to have many wives. All these things means VERY easy transmission of HIV. In the western world, people are on average more monogamous (sticking to one sexual partner), condoms are readily avaiable and used.
- HIV is not acute - you get it and have no symptons for ages. In africa, many people probably don't realize it when they contract HIV. In the western world, people can access a doctor very easily. If they just get a little sick, they go see a doctor. Blood tests are easy to do, and experienced doctors will notice the symptons and diagonose quicky. People can then take measures to not spread it. In Africa, people get sick a lot more often, and don't have access to doctors so easily.
- Even in western society, black people are more likely to get HIV. This is because black people tend to be of the poorer, low social-economic class. Poorer means less access to medical facilities. Low social economic class means less education about health, disease, spread of disease, and safe sex in terms of sexually transmitted diseases. Poorer also means living in areas of higher crime, basically meaning more cases of rape and so on. You'll find that black people iin general tend to have more health problems, because they're poor and of low social-economic groups. HIV is not an exception, it's pretty much the norm.

now moving on to aids. HIV can stay in the body for a period of time without causing any harm. Up to about 10 years without medical help. At some point, no one knows what the trigger is, HIV activates and the body starts showing the symptons. These symptons are what's called the "Acquired Immune Deficiency Syndrome" (acquired refering to how AIDS is aquired from the virus HIV). --`/aksha 03:42, 24 October 2006 (UTC)[reply]

One tiny point: AIDS is called "acquired" because it's acquired from other people (it's contagious), rather than being genetic. --Tardis 19:13, 23 October 2006 (UTC)[reply]

Uhh...isn't it "acquired" as oppossed to cognetial? ...uh...however you spell that word. The one which means you're born with something. --`/aksha 03:42, 24 October 2006 (UTC)[reply]

- Once aids starts to show, people are more or less doomed to die soon. There are drugs to slow this process down, so HIV can lie dormant in the body for a long time and the person will not start to suffer from AIDS. These drugs are expensive - not the kind of thing that's regularly found in Africa. The drugs are also most effective if the person starts taking them as soon as they get HIV. Once again, black people who are of the poor social-economic groups tend not to do this. Some don't realize they have HIV. Others can't afford the drugs. Others simply can't be bothered. Hence the HIV infected black people develope AIDS much more quickly.

"I recall reading a paper about a number of African women appear to be entirely immune to the disease" they're not actually immune. There were reports of people in Africa who seemed to be infected with HIV and just carried it with them all their lives without realizing it. In other words, they lived long (long as in normal age for healthy people in their society) and normal lives despite having the virus. So for whatever unknown reason, despite being infected with HIV, the HIV never really activates in their body. So not exactly immune...more like being sick but just not dying. Which i guess is 'immunity' in one sense of the world. --`/aksha 13:54, 23 October 2006 (UTC)[reply]

The term for that is a carrier, with Typhoid Mary being perhaps the most infamous example. Another problem in Africa is the persistent myth that having sex with a virginal woman will cure the man of AIDS, while, in reality, it only spreads AIDS to the woman. Also, women in Africa are so dependent on men for survival that they must do whatever they are told, including having sex without a condom, even knowing that their man sleeps around. StuRat 19:20, 23 October 2006 (UTC)[reply]

Um, no there is a genetic component to susceptibility to AIDS. The HIV virus requires several proteins in order to attach to and infect a cell. One of these is the chemokine protein CCR-5. Patients who lack CCR-5 are relatively immune to infection by HIV!

why do i have a feeling that the only way someone can lack CCR-5 is if they don't have a functioning gene for it? --`/aksha 03:42, 24 October 2006 (UTC)[reply]

Does the CCR-5 gene do anything useful ? If not, we could wipe out AIDS by creating a virus that destroys that gene, then let the virus go free. StuRat 16:57, 24 October 2006 (UTC)[reply]

Gamma rays

Why gamma ray is not a particle but a ray only? It is told that waves also behave as a particle, then why is it not true with gamma ray?

It's not... Although the name is a bit confusing, gamma rays are simply the highest-energy example of electromagnetic radiation. Their particle is the photon. – Clockwork Soul 07:21, 23 October 2006 (UTC)[reply]

Make that the highest-energy range of EM. Just like visible light is an EM range, ranging from red to violet. Gamma rays, however, are not defined to have an upper limit. Something tells me that in reality there is a limit, but the article doesn't seem to say. DirkvdM 07:57, 23 October 2006 (UTC)[reply]

Cosmic Rays are supposed to have an even shorter wavelength.

But Cosmic rays are particles. ☢ Ҡi∊ff⌇↯ 15:14, 23 October 2006 (UTC)[reply]

At shorter and shorter wavelengths, dont these things appear more like high energy particles. All particles can be considered as waves and vice versa.--Light current 16:08, 23 October 2006 (UTC)[reply]

As Feynman said, they're particles, because you can hear the individual clicks they make in a Geiger counter. There's no such thing as half a gamma ray. —Keenan Pepper 19:36, 23 October 2006 (UTC)[reply]

But again, there's no such thing as a "particle" when one looks at the proper scale - gamma rays are just very high frequency/short wavelength light (with the attendent wave/particle duality involved). Virogtheconq 04:10, 24 October 2006 (UTC)[reply]

Application of Nature Cure for treatment psoriasis

Can you kindly explain whether Naturopathy can be successfully applied in the treatment of psoriasis of skin, where treatment by allopathy, siddhavaidya etc., have not yielded satisfactory results. If so. what are the sources (hospitals etc.,) available in Hyderabad? Regards Chandru — Preceding unsigned comment added by 59.144.66.3 (talk • contribs)

A very natural treatment is by doctor fish. I'm afraid this may not yet be available in Hyderabad. --Lambiam ^Talk 08:50, 23 October 2006 (UTC)[reply]

Some people have suggested that a change in diet can be helpful. Also, some preliminary evidence exists for positive results from topical application of some botanicals. --JWSchmidt 13:56, 23 October 2006 (UTC)[reply]

alluvial sedimentary rock

what is an alluvial sedimentary rock? — Preceding unsigned comment added by Ali khoobani (talk • contribs)

Here "rock" is a mass noun. Alluvial sedimentary rock is an alluvial deposit consisting of rock. --Lambiam ^Talk 08:40, 23 October 2006 (UTC)[reply]

And I'll say it's a rock composed originally of alluvial deposits. :) --Zeizmic 12:43, 23 October 2006 (UTC)[reply]

heat

what generates the heat to melt the rock in the earth core? — Preceding unsigned comment added by Ali khoobani (talk • contribs)

According to the prevailing theory of planetary formation, planets are formed by dust and gas clustering together in increasingly larger clumps, finally collapsing together to form a planet-sized body. The heat generated by the kinetic energy lost in the impacts made the Earth melt. Since then it has only been cooling down. The generally accepted explanation is that the heat is simply left over from the Earth's initial formation; see Structure of the Earth. --Lambiam ^Talk 08:33, 23 October 2006 (UTC)[reply]

I sure wish I had cared more about global geophysics. Our article seems wrong about the source of heat for the earth. This [16] is more in line with what I understand. --Zeizmic 12:40, 23 October 2006 (UTC)[reply]

I agree, the primary source of heat in the earth is radioactive decay. If the earth had no internal heat source it would have cooled down a lot more already. Wikipedia does have some articles that mention this - see Geothermal power, Age of the Earth and Uranium. JMiall 12:57, 23 October 2006 (UTC)[reply]

This played an interesting role in the history of science. Before radioactivity was understood, people such as Kelvin claimed that their physical science theories for rates of cooling of the Earth and rate of chemical fuel depletion in the Sun showed there had not been enough time for biological evolution to take place on Earth. A good example of why scientific claims should always begin with, "If my assumptions are correct...." --JWSchmidt 13:39, 23 October 2006 (UTC)[reply]

What about all the pressure on the earth's core? That's what I always assumed made the nickle and iron in the core liquid. - AMP'd 14:09, 23 October 2006 (UTC)[reply]

No, high pressure forces most materials to become solids. This is why the very core of the Earth is thought to be solid, even though the mantle is molten, and the core is even hotter. StuRat 19:01, 23 October 2006 (UTC)[reply]

So I had it backwards? - AMP'd 02:20, 24 October 2006 (UTC)[reply]

In general, yes, although there are exceptions, as some materials, like water, are actually less dense as a solid. StuRat 16:53, 24 October 2006 (UTC)[reply]

Radioactive decay is what generates the heat. since nothing can burn in the conventional sense, this is the only possible answer (I think).--Light current 14:53, 23 October 2006 (UTC)[reply]

Also, tidal forces from the Moon and Sun generate some heat in the interior, but likely far less than radioactivity. StuRat 18:57, 23 October 2006 (UTC)[reply]

Current researches about earth core neutrinos try to tell the percentage of heat caused by radioactivity. It could be only 30-40%. -- DLL ^{.. T} 18:59, 23 October 2006 (UTC)[reply]

Unidentified Mushrooms

What mushrooms are in this image? It's a very good image, and I would like to put it in the appropriate area, but I don't know what kind of mushroom it is. Chanterelle, maybe? NauticaShades 07:49, 23 October 2006 (UTC)[reply]

They certainly look like it. I wouldn't eat them, though, unless they are printed on edible paper. Did you think of asking the photographer and uploader André Karwath aka Aka? --Lambiam ^Talk 08:16, 23 October 2006 (UTC)[reply]

Oh man, so many witticisms in one reply, I think I'm going into repartory arrest! Anchoress 09:47, 23 October 2006 (UTC)[reply]

Google has 280,000 hits for "repertory arrest", and it is my sad duty to report that a lot of them seem to be talking about cessation of breathing, eg. respiratory arrest. JackofOz 09:56, 23 October 2006 (UTC)[reply]

Dood, you need to bone up on your ironical vocab. I meant repArtory as a mangling of 'repartee'. I assure you that clever word play is not sufficient to induce respiratory arrest, in me anyway. ;-) Anchoress 10:19, 23 October 2006 (UTC)[reply]

Oh, I detected the irony and the humour alright, although I admit I thought it was a mangling of "repErtory" rather than of "repartee". What's it called when you laugh at a joke for completely the wrong reason? Dorkness, maybe? But I don't think it matters as long as you're having jolly fun with your chums. Anyway, we've discovered a further reason to be cautious when a doctor uses big words. :-) JackofOz 19:58, 23 October 2006 (UTC)[reply]

I love collecting edible mushrooms, but that looks like a pile of several types. --Zeizmic 12:25, 23 October 2006 (UTC)[reply]

Looks like chanterelles to me. I wouldn't worry too much about eating them myself. I've heard more than one mycologist say that chanterelles don't resemble any poisonous fungi. (And are therefore recommended to inexperienced mushroom-pickers). --83.145.46.141 13:53, 23 October 2006 (UTC)[reply]

Um, ah, well, it all depends where you live. If you happen to live somewhere that the Jack o'lantern mushroom grows (and glows), you can make yourself sick as hell if you confuse the two pretty orange mushrooms. We really oughta give the same disclaimer for mycology here that we do for medical and legal advice... --jpgordon^∇∆∇∆ 17:17, 23 October 2006 (UTC)[reply]

Chemistry/ Heat curves

How much energy (kj) is required to convert a 15.5 g ice cube at -5.0 to water vapor at 180 degrees Celsius?

Yeah, I had those questions in school too. You'll want to divide the problem into a few steps:

Heating the ice to the temperature where it can melt
Melting the ice
Heating the water to the temperature where it can evaporate
Vaporising the water
Heating the vapour to the final temperature

—Bromskloss 13:15, 23 October 2006 (UTC)[reply]

Sodium Dodecyl Sulfate

I did a Chemical Engineering lab last week to investigate the surface tensions of various concentrations of solutions of sodium dodecyl sulfate, also known as sodium lauryl sulfate. One of the aims was to discover the maximum surface concentration at the liquid/gas interface, and the minimal area of a molecule of said compound. This was done using the maximum slope of a plot of surface tension against log of concentration. Does anybody know the correct value for either of these, so I can analyse the accuracy of my results in my lab report? I only need one correct value because there is a simple relationship between the two. I already tried to google it but no luck, and I can't access the library at the moment.

Thanks! 7Munkys 13:07, 23 October 2006 (UTC)[reply]

Bad horror-movie bacteria!

How the hell do you kill these things??? Just reading the article is personally scaring the crap out of me! I mean, it can take heat, cold, vacuum, radiation, dehydration, even acid! Do you lock them up in jar and watch them starve them over three years? Do you try to poke it with a sharp stick? What? I really want to know! I need to be able to sleep at night! 83.250.208.83 14:57, 23 October 2006 (UTC)[reply]

Just put it in the oven and bake it for a few hours at 300 degrees Celsius. I'm sure it can't take that much heat. --Bowlhover 15:34, 23 October 2006 (UTC)[reply]

I'd prefer a plasma arc, if in fact, those things actually exist! --Zeizmic 15:46, 23 October 2006 (UTC)[reply]

One should consider whether it is, in fact, pathogenic before getting worried about them. Also, just because it is an extremophile doesn't mean it will flourish in normal environmental conditions, given that it is suddenly confronted with both competition and predators. —The preceding unsigned comment was added by 192.91.171.42 (talk • contribs) .

Exactly. They're probably harmless to humans. And Strain 121 can kick their asses anyway. =P —Keenan Pepper 19:45, 23 October 2006 (UTC)[reply]

Don't waste time; break out the phasers and photon torpedoes. B00P 20:38, 23 October 2006 (UTC)[reply]

Being extremophilic, I doubt they've had much experience of antibiotics, so a good dose of those could probably kill it. La ï ka 21:37, 23 October 2006 (UTC)[reply]

If you're afraid of foreign creatures which can stand high temperatures, does that make you an xenophobic in regards to extremophilics? --Fastfission 00:23, 24 October 2006 (UTC)[reply]

Very cool - are there other bacteria like these two? Aaadddaaammm 03:38, 24 October 2006 (UTC)[reply]

Reishi and Birth Control Pills

Dear Wikipedia:

Good Day! My wife is taking some birth control pills but she's also taking some reishi mushroom for her health. Does reishi can hinder the effect of pills?

Thanks> Gerry

Please read the rules. "If requesting medical or legal advice, please consider asking a doctor or lawyer instead.--Russoc4 16:21, 23 October 2006 (UTC)[reply]

You could also ask your pharmacist. Make sure you can specify the specific pills used. --Lambiam ^Talk 16:25, 23 October 2006 (UTC)[reply]

The pill can be made ineffective if you experience vomiting or diarrhoea and this mushroom seems to have symptoms of gut clearing, so I would definitely check with a professional. Vespine 23:07, 23 October 2006 (UTC)[reply]

watershed management

how can watershed management be done? can you give an example of a place where this is being implemented?61.246.87.148 16:51, 23 October 2006 (UTC)abcd[reply]

I'm not sure what you mean. Do you mean management to prevent floods, to limit soil run-off, prevent water pollution, reserviors for irrigation and human consumption, sewage treatment, hydroelectric generation, etc. ? :-) StuRat 18:41, 23 October 2006 (UTC)[reply]

...or trying to get your kids to bed by 9pm?--Shantavira 08:48, 24 October 2006 (UTC)[reply]

Make sure they pee before turning in, or the bed could become a water shed. :-) StuRat 20:00, 24 October 2006 (UTC)[reply]

Alpha Helix

I am trying to do a quesstion for an university assignement and after looking through 3 text books, I am still puzzled. The question is as follows:

"The peptide chain below folds into an alpha helix. /an alpha helix is 0.54 nm long (5.4 Angstrom units) for every complete turn, and there are 3.6 amino acid residues per turn of the helix.

Lys pro val leu gly ala Ile ala asp leu val val gly leu glu ile leu ala val ala gly tyr ser"

I know this is kidna a homework question but I am officially stumped.

I am also to describe the nature of the protein (i.e. polar, non-polar, predominantly charged)

and finally,

If the hydrocarbon core of biological membranes is approximately 30 Angstroms, suggest a plausable function for the peptide.

Was there supposed to be a question already before you came to "I am also to"? If so, I couldn't find it! —Bromskloss 18:33, 23 October 2006 (UTC)[reply]

I apologise, it says I am to calculate the length of the helix. I got 3.45 nm.

You might find it useful to read this article. It has more detail than do Wikipedia articles such as this one about how to make guesses about the functions of alpha helical domains.--JWSchmidt 22:14, 23 October 2006 (UTC)[reply]

You might want to visit the amino acid article and take a look at each individual amino acid in the peptide. Make a list of the charge and polar nature of each residue. For instance, the first residue is charged and polar. Look at the second residue very closely, as it will do something weird to the helix. Also, you basically answered one of your questions yourself: put the membrane thickness and helix length into the same units, and you might see a similarity.Tuckerekcut 23:07, 23 October 2006 (UTC)[reply]

Black Engineers

Hello. I was wondering, where could I find an article or a list of African American Mechanical Engineers? I searched this topic 3-4 times before, but couldn't find anything specific. Thank you for your time.

Thank you,
A Troubled User

See Category:African American inventors. StuRat 21:12, 23 October 2006 (UTC)[reply]

Crucell

Hie I am a wondering if you could possibly put up the Crucell's profile on wikipedia. Crucell is a leading Dutch Biotech company based in the city of Leiden. I am a press officer at Crucell so should you need some help compiling the profile, please feel to contact me on email removed

Thanks for the good work

Best regards

Bruce

We want money! --Zeizmic 21:20, 23 October 2006 (UTC) :)[reply]

Of course, Wikipedia does not change editorial standards for pay. A good spot to list this would be Wikipedia:Requested articles. Standards for the inclusion of a corporation are noted at WP:CORP. Cheers, --TeaDrinker 23:12, 23 October 2006 (UTC)[reply]

a) That sounds like advertising, b) This isn't the request section, and c) I removed your email as per policy at the help desk. --Wirbelwindヴィルヴェルヴィント (talk) 23:14, 23 October 2006 (UTC)[reply]

Algorithms

Just as articles need to be factual and accurate and subject to correction so do algorithms. So as not to break with the tradition of making encyclopedic articles as esoteric as possible is there an alternate place to post scientific and mathematical algorithms within the Wikipedia? Adaptron 21:38, 23 October 2006 (UTC)[reply]

Try Wikibooks. --JWSchmidt 22:19, 23 October 2006 (UTC)[reply]

Some can be posted here, like common sort algorithms. What does your do ? StuRat 23:13, 23 October 2006 (UTC)[reply]

I have many algorithms I am interested in posting but the current one involves the computation of sunset and sunrise from latitude and longitude. Except for one freely available from an online magazine I have yet to find any that are complete and error free. With a place set aside in the Wikipedia for publishing such algorithms I would hope such incompleteness or error would soon be eliminated. Adaptron 23:18, 23 October 2006 (UTC)[reply]

You've got a good idea for a new wiki there (WikiCode ?). I could post a few myself. What does everyone else think ? StuRat 23:39, 23 October 2006 (UTC)[reply]

If it doesn't belong in either WikiBooks or WikiSource, I would definitely be behind such an idea. It's a pity the Wikimedia Wiki-creation process is in a state of flux, or I'd suggest putting a proposal up right away. Confusing Manifestation 02:10, 24 October 2006 (UTC)[reply]

Here is a sunup/sundown computational algorithm in need of new wheels and a fresh coat of paint which it might acquire if left in a wiki instead of parked on the street.

Starting with the number of days since Nov. 17, 1858 or a modified Julian date of

JD = 2454032.24616, a computed event (sun up or down) set to UT = 0 or noon and an

estimated event set to UTO = 6 an iteneration is begun to compute the actual value

of UT.

(Note: addition of functions to convert degrees to radians may be necessary)

'initial values

Lat = latitude

lon = longitude

PI = 4 * Atn(1)

HORIZON = -50 / 60

while (abs(UT-UTO) > .0008

'compute the number of centuries since the epoch of Jan. 0, 2000.

T = (JD + UT / 24 - 2451545#) / 36525#

'compute the solar mean longitude

l = 280.46 + 36000.77 * T

'Compute the solar mean anomaly in degrees.

G = 357.528 + 35999.05 * T

'Compute the ecliptic longitude in degrees.

M = l + 1.915 * Sin(G) + 0.02 * Sin(2 * G)

'Compute the obliquity of the ecliptic in degrees.

E = 23.4393 - 0.013 * T

'Compute the equation of time.

ET = -1.915 * Sin(G) - 0.02 * Sin(2 * G) + 2.455 * Sin(2 * M) - 0.053 * Sin(4 * M)

'Compute the Greenwich Hour Angle in degrees.

GHA = 15 * UT - 180 + ET

'Compute the sun's declination in degrees.

DEC = arcsin(-Sin(E) * Sin(M))

'Compute the hour-angle at UT (the current estimate)

HA = arccos((Sin(HORIZON) - (Sin(Lat) * Sin(DEC)) / (Cos(Lat) * Cos(DEC))))

'Compute the time of sunrise, sunset, or twilight.

'RISESET is +1.0 for morning (rising) and -1.0 for evening (setting).

RISESET = 1#

UT = UT0 - (GHA + lon + (RISESET * HA)) / 15#

'The result is in hours.

'Continue until the UT and UT0 estimates are close together.

Wend

'print the computed event time.

print UT

Adaptron 08:42, 24 October 2006 (UTC)[reply]

Rate of lunar transit

What is the calculation or rate at which the moon 'crosses' the night sky? In other words, if you were watching a shadow on the ground cast purely by the moon's reflection, is there a calculable rate to ascertain how long it would take to travel an inch, or a foot?

The moon or a shadow cast by it is easier to imagine travelling in degrees, not feet or inches. Imagine a big tree and a little fence post, in one night the tree's shadow may travel dozens of feet while the fence post's only looks like it moves a couple of feet, the commonality is that both shadows travel the same number of degrees. The shadows turn, like the hands of a clock.

Obviously the 'turning' is caused by the earth spinning on its axis. The earth spins fully around once every 24 hours, meaning the moon looks like it is travelling 360 degrees every 24 hours, or 15 degrees an hour, (ignoring the much longer monthly period of the moon). If you do include the monthly revolution of the moon around the earth, it actually adds up, the moon travels forward by an extra 28th of a full revolution every day, or about 13 degrees, which is almost one hour, so every day the moon is almost an hour earlier.Vespine 22:47, 23 October 2006 (UTC)[reply]

Dear Vespine: Dynamite answer, in every fashion -- I'm very grateful for your help. Wolfgangus 01:17, 24 October 2006 (UTC)[reply]

I think the Moon lags behind the Sun; while it apparently travels from East to West like the Sun, it does so more slowly. See the picture at Lunar phase. This also explains why the path of the Moon's shadow during a solar eclipse travels West to East; see e.g. the animation at Solar eclipse of 2006 March 29. So "Moon rise" is a bit later each day. Using the value of 29.5 days for the synodic month, so that we have on the average 29.5−1 = 28.5 Moon rises in that period, the time from Moon rise to Moon rise is (29.5/28.5) × 24 hrs = 24.8 hrs = 24 hrs 50 mins. The overhead Moon traces out 360° in that period, which amounts to about 14°/hr = 0.25 rad/hr. --Lambiam ^Talk 06:49, 24 October 2006 (UTC)[reply]

Dear Lambiam: Thank you so much for the valued additional input. Now, if I may ask a question that perhaps begs the obvious: with all this in mind, does the Moon have a verified annual apsis and periapsis with the Earth? If so, how is it calculated?Wolfgangus 09:18, 24 October 2006 (UTC)[reply]

It has a perigee and apogee. I imagine this will be calculated assuming the Moon's orbit is a perturbed ellipse. Typical values for perigee is ~363000 km, and for apogee ~406000 km. Richard B 11:31, 24 October 2006 (UTC)[reply]

After edit conflict:

Stuff like this can even affect politics. After the Venezuelan coup attempt of 2002, there was controversy over when certain footage was shot. The Revolution Will Not Be Televised (documentary) had the shadows cast differently from another shot, so the argument was used that they were shot during different parts of the day. But how far apart those images were in time depends on the size of the building that cast the shadow. If cast by a tall building, the shadow would move very fast. DirkvdM 09:22, 24 October 2006 (UTC)[reply]

Abnormal waves and tides

Is tidal [and waves] data from ocean buoys available for download online? Adaptron 22:02, 23 October 2006 (UTC)[reply]

I don't think marine buoys are an effective means of measuring tide (which is an interaction with the shore). Other bouy data (NOAA data) can be obtained from the National Buoy Data Center. For tide information, you might try tides online (another NOAA service). --TeaDrinker 23:07, 23 October 2006 (UTC)[reply]

Sorry I meant ocean height (to include wave action) which includes changes caused by tides whether at the shore line or offshore. What I am ultimately after are the Fourier coefficients for all locations of measurement, either offshore or at the shoreline. Adaptron 23:28, 23 October 2006 (UTC)[reply]

Found it at harmonic_cons_defs Adaptron 23:42, 23 October 2006 (UTC)[reply]

October 24

Ivan Pavlov's Insomnia cure

Hello to all, I wanted to know about Ivan Pavlov's Insomnia condition, so far by other researches I found he developed some recipies or prescrpitions for his own condition, this help him to develope a new category of insomnia, is there anyone able to provide me with info about this prescriptios or recipies for insomnia treatment????

Sergio Mexico City

I'm drooling in anticipation of the answer. :-) StuRat 03:25, 24 October 2006 (UTC)[reply]

Beer, wine, or something else?

Would an alcoholic beverage fermented from coffee be considered a beer or a wine? I would intuitively say "beer", but I can't give a good reason for it. Does anyone know if such a drink exists? Bhumiya (said/done) 01:25, 24 October 2006 (UTC)[reply]

Coffee does not have (or has negligible amounts) of carbohydrates in it, so there's nothing to ferment into alcohol. --Pyroclastic 01:44, 24 October 2006 (UTC)[reply]

You can, however, get coffee stout :) GeeJo ^(t)⁄_(c) • 01:46, 24 October 2006 (UTC)[reply]

That stuff is God-awful. StuRat 03:22, 24 October 2006 (UTC)[reply]

Heresy! Respect the Gods X [Mac Davis] (SUPERDESK|Help me improve) 07:38, 24 October 2006 (UTC)[reply]

It is not uncommon to add flavours to alcoholic beverages, so why not coffee? (Apart from the fact that it might taste horrible.) Whether it is called beer or wine depends on what the basic ingredient is - grain or fruit, although the distinction is a bit blurred (and grain is actually a fruit). Note that if you add fruit to beer it's still beer, like with kriek. DirkvdM 09:33, 24 October 2006 (UTC)[reply]

I believe it's called Kahlúa, but that's not fermented from the coffee, just flavo[u]red with it.

Atlant 17:00, 24 October 2006 (UTC)[reply]

And there are drinks like Irish coffee which have alcohol and taste like coffee. But no, I don't think you can ferment coffee. Maybe if it wasn't dried and roasted and sat after harvesting in a vat for awhile? Then that might not taste the same as coffee at all. --Wirbelwindヴィルヴェルヴィント (talk) 17:51, 24 October 2006 (UTC)[reply]

Wet coffee grounds definitely mold, but the fluid that runs off is sure nothing I would ever choose to drink. StuRat 19:45, 24 October 2006 (UTC)[reply]

Bugs Bunny and shotguns

Having learned virtually everything I've ever needed to know from Warner Bros. cartoons, I was wondering about an effect ol' Bugs pulls off occasionally. If you were to jam your finger(s) into the barrel(s) of a firearm, would they cause the gun to be destroyed? Would the force be directed backwards or would you still get killed? Not that I'm planning on trying this at home or anything (I assume you'd lose your fingers and I need to count for a living), but you never know what places life will take you and I'd rather lose two fingers than, say, my head. There's some neat stuff at Physics of firearms about guns exploding by being fired underwater, but that's not quite the same thing. Matt Deres 02:45, 24 October 2006 (UTC)[reply]

I would think this might work on a cheap gun, like the proverbial Saturday night special, which has a weak, thin barrel incapable of handling the additional back pressure. The fingers and hand of the person plugging the gun would definitely be destroyed in any event, and the shards of the exploding gun barrel might hit him, as well. StuRat 03:09, 24 October 2006 (UTC)[reply]

MythBusters covered this in Season 3, see "Finger in a Barrel". --Canley 03:49, 24 October 2006 (UTC)[reply]

Basically, in that episode of Mythbusters, the dummy was killed (or would have been killed if it was a real person) when the gun fired. There was absolutely no damage whatsoever to the gun. The mythbusters even tried sealing off the barrel by soldering a large, iron cylinder into it. This time the barrel peeled back, but the cylinder was still fired out at a lethal speed. --Bowlhover 04:05, 24 October 2006 (UTC)[reply]

Right, but they probably used a decent gun. If you start with such a piece of junk that it's almost ready to explode on it's own, it doesn't take much to make it happen. StuRat 16:40, 24 October 2006 (UTC)[reply]

If you need your fingers to count, I would suggest abstaining from experiments with firearms. Or using Warner Bros. cartoons as a physics reference. And yes, your head is more important. - Samsara (talk • contribs) 09:21, 24 October 2006 (UTC)[reply]

One could just switch to octal, at least for the first two experiments.

Atlant 17:04, 24 October 2006 (UTC)[reply]

About the practical side of this. Just in case I'd ever come eye to eye with someone pointing a gun at me (however unlikely), if I could get my hand come close enough to the gun to put my finger in the barrel, that is not what I would do. I'd push the gun aside, possibly even take it from the gunowner. Faster, safer and more effective. Sorry to spoil this academic discussion with practical considerations. :) DirkvdM 09:42, 24 October 2006 (UTC)[reply]

I would stay away from the bunny, and go with the coyote! --Zeizmic 16:54, 24 October 2006 (UTC)[reply]

Relativity Paradox

I don't have much time to type this in but here is a quick and dirty relativity paradox.

Imagine a small black hole (in 2D space) with a circular event horizon with a radius of 1km. Therefore the circumference of the event horizon is 3.1416 * 2 * 1km = 6.28km

Now imagine a 10km (rest length) train travelling at near the speed of light entering the event horizon and going round and round in the event horizon. Because of the lorentz contraction, the 10km train is contracted to 5km so it would fit in the event horizon.

This is ridiculous that a 10km train can fit in a circular 6.28km event horizon "train track". What happens when the train slows down? The head of the train will crash into the tail of the train. Ridiculous!!! 202.168.50.40 02:53, 24 October 2006 (UTC)[reply]

I think I'm missing something, or the above is a little poorly explained. I don't quite see what the black hole is for in this example, why not just say a train track with a radius of 1km? And is the train crashing into its own tail really that ridiculous? Vespine 03:47, 24 October 2006 (UTC)[reply]

Why is it ridiculous for a train's head to collide with its tail? It's possible, isn't it?

Also, when the head of the train stops, the end of the train won't know about it until at least 33 microseconds later (10 km/299 792.458). (The real value is much longer, because the information actually travels at the speed of sound.) So for the first 33 micro-seconds, the end of the train would travel in exactly the same way as it used to. This will compress the train, squashing its unfortunate occupants to death... --Bowlhover 04:00, 24 October 2006 (UTC)[reply]

The train does not need to stop instantaneously. It can very very slowly slow down. Also, I put in the black hole to prevent the need for the train to exert a constant force, to cause a change in the velocity vector needed to get the train to go around an actual train track. 202.168.50.40 05:16, 24 October 2006 (UTC)[reply]

Doesn't the black hole prevent the train from slowing down? - Mgm|^(talk) 08:01, 24 October 2006 (UTC)[reply]

Also, it'd be unable to support itself in orbit at the event horizon - it'd end up in the sigularity - and if you were looking from far outside the event horizon, you wouldn't observe the train crossing the event horizon - you'd probably just see it increasingly redshifted. Richard B 12:13, 24 October 2006 (UTC)[reply]

I'm not sure this problem is well-posed, or is a paradox, but if we're going to bring black holes into the matter remember that in GR geometry isn't what you're used to. For instance, I believe (although I might have it backwards) that the circumference as measured by a (stationary) measuring tape at the event horizon is in fact greater than

2\pi r

, where r is the radius of the black hole as measured by a distant observer. (This means that the distant observer describes the tape as being length-contracted even though it's not moving!) It only gets weirder if you allow the tape (or the train) to fall into the event horizon... --Tardis 15:45, 24 October 2006 (UTC)[reply]

You cant go round ad round in the event horizon. You are either inside it or outside it. If inside you wlii eventually rech the singularity. Isuppose you could orbit around safely if outside it--Light current 15:57, 24 October 2006 (UTC)[reply]

This must be some new meaning of "safely" with which I'm unfamiliar. (Apologies to D. Adams.) --Trovatore 19:42, 24 October 2006 (UTC)[reply]

Smoking Question

Disclaimer: I am writing a novel! Be assured I have no intention of forcing this upon a real human being or doing it to myself. The situation is that, due to a horrible trauma in one of my main character's pasts, he's decided he deserves to be punished. He's mentally unstable, and has OCD. He's afraid to die, because he'll have to face the person he harmed. So his punishment of choice is to smoke a certain number of cigarettes(thank the OCD...I don't know the number yet, but let's say 5) in a 30-minute(or so) period, then go without for 7.5 hours(somewhere around there), then repeat over and over. He's been doing this(or somethign similar) for 7-8 years. My question is, what would this feel like? I've gotten some descriptions of withdrawal from smoker friends, but none of them have any idea if this would feel the same way or not. Thanks for helping! Tigger89 03:18, 24 October 2006 (UTC)[reply]

If your main character doesn't get lung cancer, it'll likely feel pretty good to be addicted. Quitting smoking permanently is hard. But quitting for 7,5 hours--that should be pretty easy. --Bowlhover 03:44, 24 October 2006 (UTC)[reply]

I can share with you my experience, but you're really not specific enough so I'm not sure if I can answer it correctly. I smoked for about 6 years, when I used to drink or played pool, I sometimes smoked a lot, not quite chain smoking 5 in 30 minutes, but maybe half a pack in an hour or two. However, keep in mind that I also more or less quit cold turkey. And I didn't smoke hourly etc, so it fits your character fairly well. From my personal experience, no, the withdrawal isn't stronger in between the binges, so to speak, even after 5-6 years. However, I also smoke slower, 7-9 minutes for a cigarette, I think, so I definately couldn't smoke 5 in 30 minutes. So it shouldn't be any different than the withdrawal from a normal smoker, but that's just from my personal experiences.

However, I just have one question. Why is a person who's afraid to die smoking 3/4 of a pack a day? --Wirbelwindヴィルヴェルヴィント (talk) 03:47, 24 October 2006 (UTC)[reply]

The 3/4 pack a day thing...I really don't know, but I just know that's what he does. He started it when he was 15~, so maybe being a messed up teenager had something to do with it? Tigger89 04:27, 24 October 2006 (UTC)[reply]

Talk about unstable. If he's afraid of dying, he shouldn't be smoking that much. - Mgm|^(talk) 07:59, 24 October 2006 (UTC)[reply]

If you use 5 cigarettes in 30 minutes, as Wirbelwind hinted at, for 30 minutes this character will be a furious smoker, almost cartoonishly so. The 8 or so hours between these tar-manias will be immaterial vis-a-vis withdrawal symptoms. From personal experience, I would say physical withdrawal symptoms don't begin for about 24 hours. Cravings in those 24 hours, yes; but true physical withdrawals beyond mental cravings, no. Additionally, chain-smoking in that way is one way to really punish your lungs, meaning when you tie together a string of cigarettes, you feel a unique, rather nasty congestion in your chest, as well as in your mouth. This character's hands will certainly stink (perhaps only an OCD hand-washer could rinse the stench off), and he should have brown, distinct stains where he holds the cigarette -- an often telltale sign in World War Two that a man was a spy; only they had access to enough cigarettes to put stains on the fingers.Wolfgangus 09:34, 24 October 2006 (UTC)[reply]

Defibrillator voltaje, intensity and energy?

I have read so far the defibrillator article at Wikipedia but i am still unable to understand

why doctors use the term Joule (as energy) in the sense to give to the patient an electrical discharge instead of using more appropiate electrical terms such as volts or intensity?
What is the relationship of energy (joule) to volts and amperes given to the patient?.
What are the maximum and minimum volts and amperes (intensity) reached by a defibrillator?
Can a defibrillator be used to charge (electrically speaking) an car's uncharged battery?--HappyApple 03:38, 24 October 2006 (UTC)[reply]

Joule as you correctly identified is a measure of energy, neither electrical term such as volts or current. The relationship is V x I = P: Volts times Amps equals Watts. Watts are a measure of power, while energy is power applied over time. i.e. Joules = (Watts) x (seconds).

The maximum volts and amperes reached by a defibrillator seems not to be common knowledge, it is mentioned in the article that electrical burns are possible from insufficient contact with the electrodes so that suggest that the power is significant, since this power is achieved through capacitors it also suggest that the discharge is rather quick, which would also suggest that a defibrillator would not make a good car battery charger. A car battery requires a long slow low current charge, the defibrillator does the exact opposite..Vespine 04:44, 24 October 2006 (UTC)[reply]

Our Defibrillator page doesn't seem to have this info:( For some detail, check the inforesearchtizement for one manufacturer's equipment. DMacks 05:34, 24 October 2006 (UTC)[reply]

Ooo, if poor contact causes burns, what happens when there is good contact? Doesn't the heat just get produced in some other part of the body? —Bromskloss 07:42, 24 October 2006 (UTC)[reply]

Yes the heat gets produced in some other part of the body, but when the contact is good, the same amount heat is spread out over a much larger area and isn't enough to cause burns.Vespine 14:33, 24 October 2006 (UTC)[reply]

Also, there should less resistence if there is good contact. Therefore, the total heat produced should be reduced dramatically. StuRat 16:36, 24 October 2006 (UTC)[reply]

If the defibrillator delivers its full payload of energy under all conditions, then under all conditions n Joules (watt-seconds) of energy are deposited in the recipient's body. But with good contact, it isn't all dumped at the very surface of the skin, heating the skin to the burning point. Instead, it's much more-uniformly dissipated throughout the entire conduction path where the overall heating effect amounts to practically nothing. (4.185 joules of energy will heat one gram of water one degree C, so you can see that the overall heating effect of the impulse is pretty small, even at, say, 200 W-S.)

Atlant 17:11, 24 October 2006 (UTC)[reply]

If I have time tomorrow and no one else has answered definately, I'll look at it. But I found the 2005 American Heart Association Guidelines. From skimming, the human body is about 80 ohms, studies show that 30-40 A is probably optimal. The defibrillator resistance is around 50 ohms, and that's all I saw. I'll look at it more tomorrow. --Wirbelwindヴィルヴェルヴィント (talk) 08:22, 24 October 2006 (UTC)[reply]

Volts are the same as the unit Joules per coulomb. Amperes are the same as coulombs per second Richard B 11:18, 24 October 2006 (UTC)[reply]

enzymes

How many enzymes are needed for a human to stay alive. If memory serves me right, it is 500. I don't know for sure, though. --JDitto 04:06, 24 October 2006 (UTC)[reply]

I've never heard an estimate, but it would depend on how you classify enzymes. For example, do you count PKA as a single enzyme or a family of them? How about the PKCs? CaMKs? Hemoglobins? Acetylcholine receptors? You get the idea. --David Iberri (talk) 04:54, 24 October 2006 (UTC)[reply]

One list of enzymes is available at Wikipedia:MeSH_D08. --Arcadian 05:58, 24 October 2006 (UTC)[reply]

When you take into account the amount of enzymes you need for even a single step in some cell transport processes, I find it hard to believe 500 is enough. I think it's far more, but I can't give a definitive number. - Mgm|^(talk) 07:58, 24 October 2006 (UTC)[reply]

Why Does My Wife Kill Battery-Powered Wristwatches and Computer Hard Drives?

My wife kills battery-powered wristwatches in one to six weeks. We tried 8 different brands. She kills computers in three weeks to four months of daily use. She has done this for the last thirty years. We experimented with a wristwatch that had died and which began running again when she took it off. It quit working immediately upon being placed on her skin. The watch quit running a few seconds after being placed against her clothing. The watch began running again when it was placed more than eighteen inches from her body. My wife's cousin knew another lady who killed battery-powered wristwatches in a few weeks. In a medical test for metals retained in the body performed about 10 years ago, my wife tested perfectly metal-free. I had aluminum and copper and mercury and magnesium. Might this datum be related to her watch-killing phenomenon? Is watch-killing a common phenomenon? Is this related to a person's EM field? Can this watch-killing emanation be neutralized by some electronic device, similar to how sound waves can be neutralized with an electronic device? Does this watch-killing emanation have other aspects? What other methods might be used to neutralize this emanation? Would this emanation seem to be harmful to other people? Please comment. Marlin

That sounds really really odd. Though humans do generate an EM field (as do all living things), it's not very strong - certainly not strong enough to cause electrical interference (the body will couple with antennas, but that's an unrelated phenomenon). This seems more like a cause of really really bad luck, or something. I've actually never heard of such a phenomenon before, other than apocryphal bad-luck stories. Virogtheconq 06:00, 24 October 2006 (UTC)[reply]

At some point, you have to ask which is more painful to replace :).--Tbeatty 06:07, 24 October 2006 (UTC)[reply]

I used to wear my father's presumed dead pocket watch during carnaval. To my surprise it started running again, and accurately at that. After carnaval it died again, to be revived again next carnaval. Of course, that's a matter of moving the watch about, and actually the opposite of what your wife experiences, so this story is probably not relevant here. Sorry about that. :) DirkvdM 09:51, 24 October 2006 (UTC)[reply]

And now for something more relevant. Does the computer's hardware die (and which bit then?), or the OS? Does she install all sorts of stuff on it? If it's a hardware thing it might possibly be static discharge. Does she wear clothes that 'charge her up'? Do you get a discharge when you touch her? (No, I don't mean that.) DirkvdM 09:53, 24 October 2006 (UTC)[reply]

Sounds like my aunt, who is the only person I know who has this effect on anything electronic, mostly computers. Once we tested the phenomenon, she tried to dial-up with a modem and her daughter tried the same thing, and every time her daughter tried it worked. Weird. I would love to have an explanation for this. Her 'problem' is not as consistent or powerful as your wife's though; finally my aunt has found a laptop that seems to be behaving itself for the last few years. Could be due to a specific EM frequency that interferes with certain devices, as we are all beings of energy emitting a frequency ourselves. I wonder if it is linked to intelligence, as she is the cleverest person I know. BTW I've read somewhere that our EM should be in sync with the solar EM, or it leads to health problems - could anyone verify this? Sandman30s 12:10, 24 October 2006 (UTC)[reply]

Perhaps it's a static electricity thing. Maybe your wife wears insulating shoes/slippers and drags here feet when she walks, building up a static charge that's hard on sensitive electronic components. You might want to buy "ruggedized" computers and watches, which may have some degree of EM shielding, so aren't quite as sensitive to that type of thing. StuRat 16:27, 24 October 2006 (UTC)[reply]

Could be just odds: Someone in the world *has* to win the zillion dollar lottery, and someone in the world has to break everything they touch. --Zeizmic 16:52, 24 October 2006 (UTC)[reply]

Perhaps she's a Stepford Wife and it's a result of the EMI that she produces?

Atlant 17:18, 24 October 2006 (UTC)[reply]

There is this Wal-mart that I go to that always charges me and whenever I reach for anything, I get shocked. I don't know why it happens and why it doesn't happen to people I go with either. So it's probably some sort of body chemestry, the way you walk, the clothing you wear, something to that nature. Of course, I don't break electronics often, as I'm usually the one who gets asked to fix computers and it works when I use it and they're like, "what did you do?" and I'm like, "Nothing." Also, when my ex-suitemate and I walked around anywhere, lights would go out very very often. Anywhere from garages to malls to you name it. I'm not saying it's supernatural powers, but I think there's something that amplifies or react to what's already there, which might be something as simple as more iron in the blood or how conductive you are etc. Maybe if you're not very conductive, you build up a charge and that affects things. I have no idea how it happens scientifically, or it's just very whacky coincidence. --Wirbelwindヴィルヴェルヴィント (talk) 17:45, 24 October 2006 (UTC)[reply]

application

can provide me the uses of molecular markers in relation to fruits and vegetables?

It would help if you gave a little more information about what you're trying to do. Have you googled for molecular markers or read our article about them? - Mgm|^(talk) 07:55, 24 October 2006 (UTC)[reply]

Touch lamps

How do those touch lamps that you can touch and they light on work? Body capacitance claims it is the body's natural EM field, but that sounds hardly right. X [Mac Davis] (SUPERDESK|Help me improve) 07:54, 24 October 2006 (UTC)[reply]

how about this[17] Xcomradex 08:17, 24 October 2006 (UTC)[reply]

A friend of mine built one based on capacitance for his computron. —Bromskloss 08:50, 24 October 2006 (UTC)[reply]

This question came up here before. The body of the lamp is charged and discharged repeatedly. The time it takes to charge and discharge is easy to measure and stays pretty much the same. When you touch it, your body changes the time it takes for the lamp to charge/discharge, allowing the lamp to know it is being touched. --Kainaw ^(talk) 15:21, 24 October 2006 (UTC)[reply]

How much energy does it waste with this constant charging/discharging?

One can charge and discharge an ideal capacitor (a capacitor with no leakage or series resistance) endlessly while wasting no power. But practically speaking, the complete controller probably wastes less than a watt; the gizmos are small and can be built entirely into lamp sockets, so they certainly can't be dissipating much power or they'd get hot.

Atlant 17:24, 24 October 2006 (UTC)[reply]

science

give information about nutriton

Welcome to Wikipedia. You can easily look up this topic yourself. Please see nutrition. For future questions, try using the search box at the top left of the screen. It's much quicker, and you will probably find a clearer answer. If you still don't understand, add a further question below by clicking the "edit" button to the right of your question title. --Shantavira 08:57, 24 October 2006 (UTC)[reply]

Great answer, hi-five! X [Mac Davis] (SUPERDESK|Help me improve) 18:41, 24 October 2006 (UTC)[reply]

Sitting up while asleep

I have never sleepwalked, but ever since I was a child I have often found myself sitting upright in my sleep, sometimes several times in a night. Sometimes I simply open my eyes to find myself sitting up, and other times I seem to be vaguely aware of sitting up as though "half-asleep". I'm not sure if it prevents me from getting the rest I need, but it might be bad for my back since I don't have good posture when I'm asleep...I'm not really worried enough to consult an expert though. I just want to know if anyone has heard of this before. I guess I am also curious about the causes of weird sleep behaviour - why me, what causes me to do this? --Grace 11:38, 24 October 2006 (UTC)[reply]

This sounds as though it is a harmless precursor to sleepwalking. Check out the last external link on that page. It would also be worth mentioning to your doctor.--Shantavira 12:26, 24 October 2006 (UTC)[reply]

I do stuff like that, although not as often as you. I never got it checked out though, I never cared that much. I will say, though, that I am far more likely to exhibit those types of behaviours when I'm very busy/stressed. Anchoress 12:41, 24 October 2006 (UTC)[reply]

Premature rigor mortis ? :-) StuRat 16:02, 24 October 2006 (UTC)[reply]

See parasomnia for more information. X [Mac Davis] (SUPERDESK|Help me improve) 17:56, 24 October 2006 (UTC)[reply]

audio timescaling

Can you tell me if pitch synchronous granular synthesis is equivalent to pitch synchronous overlap and add (psola) method? THNX! --Ulisse0 14:59, 24 October 2006 (UTC)[reply]

Unexpected Item In The Bagging Area

When my mum and I went to pick up a few groceries recently at our local ASDA supercentre, we decided to use the automatic checkout system where you scan your own items. We had to get the assistant to scan her ID card and pin at least four times! What is wrong with these machines? If we were going to steal something, we wouldn't put it in the bagging area, and these machines are so flaky that even if we did, the ASDA lady would just swipe her card and we'd get away with it anyway. At one point the machine told us to remove the unexpected item and as soon as we did, it told us that we'd removed an item and to put it back! --Username132 (talk) 17:08, 24 October 2006 (UTC)[reply]

The computer "knows" how much each item should weigh (based on its SKU number). After an object is scanned, the computer expects the total weight in the bagging area to increase by that much. But very light objects can fool it, especially if the scale, err, "bagging area" is already burdened by heavy objects.

Atlant 17:28, 24 October 2006 (UTC)[reply]

Yes, there's just no way the same scale can handle a 50 lb bag of dog food and yet still know when a greeting card is placed on it, and still have a reasonable price tag. They just need to trust their customers better. Perhaps they could have a "preferred customer card" (for those who have never tried to rip the store off) and only allow them to use the automatic check-out. StuRat 19:06, 24 October 2006 (UTC)[reply]

Chemistry

File:FG08 03.jpg

Write the usual group configuration notation for each d-block group. How do the group numbers of those groups relate to the number of outer s and d electrons?

Aha! I think someone is trying to get away with not doing their homework themselves! :-) If you ask about the principle, rather than request the solution of your particular problem, you are much more likely to get an answer. As can be seen somewhere among the vast amounts of text in the beginning of this page, people here won't do your homework, but can perhaps help you learn how to do it yourself. (Also, it's a good thing to sign your comments with "~~~~".) —Bromskloss 16:14, 24 October 2006 (UTC)[reply]

This isn't homework. dudewhoaskedthequestion 11:43, 24 October 2006 (UTC)

Sure as hell sounds exactly like it. See electron orbital and the table to the right. X [Mac Davis] (SUPERDESK|Help me improve) 17:55, 24 October 2006 (UTC)[reply]

you owned me

Low testosterone and high hemoglobin/iron test

I have been on testosterone for 6 months because it was very low (way lower than norm). My dr. suggested I take a topical cream version which worked great. My problem is my iron was to high and getting higher each time I took a test. I gave blood as often as they would allow. My hemoglobin stayed high and got steadily higher which put me in a higher risk of heart attack or stroke. So the dr. as of this moment, is taking me off testosterone cream and wants to send me to another dr. that deals with this kind of thing. Can anyone explain in easy lay terms what is going on? How can a 46 yr. old man have second to the lowest testosterone level a dr. has ever seen and a high iron/hemoglobin count in the blood? I am 30 lbs. overweight and 6'5" tall. I am not sedentary, but work in an office long hours and travel a lot.

Any information would be great, but please only reply to this question if you know by experience or knowledge.

thank you, richard

first sex

on 17 april 2007 i m going to married and till today i did not seen any nude girl so i am very much worried how i do a safe sex and how i find that she had the hymen tell me detail about the sexual intercourse

Although I doubt this is a serious question, read the articles on safe sex, hymen, and sexual intercourse. Click on any links you are not familiar with. Be warned though, there's some scary stuff out there. :yikes!: If you want to see a nude girl than go to nude. Or type it in google. X [Mac Davis] (SUPERDESK|Help me improve) 17:52, 24 October 2006 (UTC)[reply]

Enough internet access to ask a question on Wikipedia, but never seen a naked woman? That's like half the content of the internets... -- Scientizzle 19:36, 24 October 2006 (UTC)[reply]

Where Art Though Contact Lens

If I rub my eye, sometimes my contact lens dissapears off somewhere under my eyelid and after much trauma comes back out, often folded. Where does it go and what happens to it on its journey? --Username132 (talk) 17:00, 24 October 2006 (UTC)[reply]

Mine also like to go for a "morning swim", somwhere under the eyelids, when I first put them in. StuRat 18:56, 24 October 2006 (UTC)[reply]

vacuolated glandular endometrial cells

i have apparently got vacuolated glandular endometrial cells in my pap smear results - does anyone understand the pathology of these? Are they inevitably cancerous, or precancerous, or just a bit odd? I have to wait at least a month for a colposcopy, hence the need to ask here.

Thanks,

Scared 81.153.246.135 17:58, 24 October 2006 (UTC)[reply]

@@ Line 708: / Line 708: @@
 [[dudewhoaskedthequestion]] 11:43, 24 October 2006 (UTC)
 :Sure as hell sounds exactly like it. See [[electron orbital]] and the table to the right. [http://en.wikipedia.org/wiki/User_talk:Mac_Davis <font color="#6644FF">X</font>] '''[<i></i>[[User:Mac Davis|<font color="#006600" face="Times"><i>Mac Davis</i></font>]]<i></i>]''' <small>([[User:Mac_Davis/Desk|<font color="#6688AA">SUPERDESK</font>]]|[[User_talk:Mac_Davis/Improvement|<font color="#666666">Help me improve</font>]])</small> 17:55, 24 October 2006 (UTC)
+you owned me
 == Low testosterone and high hemoglobin/iron test ==