Content deleted Content added

Inline

Revision as of 09:23, 20 May 2012

Welcome to the science section
of the Wikipedia reference desk.

skip to bottom

Select a section:

Shortcut

WP:RD/S

Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
Post your question to only one section, providing a short header that gives the topic of your question.
Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
Don't post personal contact information – it will be removed. Any answers will be provided here.
Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
Note:
- We don't answer (and may remove) questions that require medical diagnosis or legal advice.
- We don't answer requests for opinions, predictions or debate.
- We don't do your homework for you, though we'll help you past the stuck point.
- We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

May 16

Non-ethanol effects of alcoholic beverages

Can anyone find any information about why different alcoholic drinks have different effects independent of the volume of ethanol consumed? The effects of champagne, red wine, whisky, rum, gin and beer are all subtly different, yet I've never read anything to explain why. The volume of water has some effect with beer, but the hops have others; presumably the herbs used in gin make a difference too, as do the flavonoids in red wine. Are there are secondary metabolites that brewers yeast produces during fermentation that are bioactive in humans? SmartSE (talk) 01:00, 16 May 2012 (UTC)[reply]

Whilst we're at it, can anyone improve on the first sentence of Ethanol#Pharmacology? SmartSE (talk) 01:06, 16 May 2012 (UTC)[reply]

I'm guessing this is mostly observer bias. Someguy1221 (talk) 01:11, 16 May 2012 (UTC)[reply]

Someguy is probably close to the answer. There have been numerous studies which show that a large portion of drunkenness is psychological, and not directly related to the ethanol itself. I have seen countless studies and demonstrations which show the placebo effect on drunkenness; people who are given non-alcoholic beverages and told they are alcoholic show signs of intoxication. It is quite likely that different types of drinks make you feel certain ways, not for their chemical composition, but for the social implications of what they represent (champagne or fine wine feels sophisticated, big fruity drinks seem fun, etc. etc.) so your internal feelings likely represent something of that beyond the mere chemical effects of the drink. That's my guess, anyways. --Jayron32 01:47, 16 May 2012 (UTC)[reply]

There's a few other possible factors at play as well: Many people can have an adverse reaction to sulphites found in certain drinks, especially wine. Anecdotal evidence suggests that drinks high in sugar can result in a worse hangover, but I have yet to find a reliable source for this. Also, if you have had a bad experience with a certain liquor (as in, you end the evening worshiping the porcelain gods), it is more likely to make you feel ill in the future: see Taste aversion. -RunningOnBrains^(talk) 01:54, 16 May 2012 (UTC)[reply]

fusel oils are supposed to worsen hangovers. Staticd (talk) 07:15, 16 May 2012 (UTC)[reply]

I would personally expect drinks high in sugar to result in a worse hangover for the simple reason that eating and drinking a lot of sugar can give you many of the same symptoms as a hangover, in my experience. Presumably for similar reasons (dehydration). 86.161.213.137 (talk) 09:14, 17 May 2012 (UTC)[reply]

Ripping someone a new one

In among all the -ostomies and -plasties and whatnot, there's got to be a medical term describing the forcible creation of a new asshole. What would it be? --Carnildo (talk) 01:56, 16 May 2012 (UTC)[reply]

Giving someone a new asshole is actually a necessary medical procedure following certain types of rectal cancer. I see various articles calling this "rectal reconstruction", but I haven't seen anyone apply a fancy name to it. Someguy1221 (talk) 02:02, 16 May 2012 (UTC)[reply]

Ok, that actually made me laugh... I guess if we take the lead from tracheotomy, it would make it a rectumotomy, not sure if that sounds quite right, maybe rectotomy might need someone with some latin skills to figure out the correct spelling.. This reminds me of defenestration, the technical word for throwing someone (or something) out a window. Vespine (talk) 02:51, 16 May 2012 (UTC)[reply]

Maybe that medical procedure could help me care about things. DMacks (talk) 21:49, 19 May 2012 (UTC)[reply]

Newborns also can have an imperforate anus, which, as you can imagine, needs to be fixed fairly quickly. StuRat (talk) 03:18, 16 May 2012 (UTC)[reply]

What about colostomy which is quite a common procedure for people with advanced colonic cancer where removal of the colon is necessary. The exit is usually placed on the lower left or right front abdomen for the obvious reason that anastomosing the colon to the anus is difficult and likely to give poor control. Richard Avery (talk) 07:17, 16 May 2012 (UTC)[reply]

As proctology is the branch of medicine dealing with the rectum and anus, would it be proctostomy? As an aside, the proctologist at my local hospital is Mr Shatwell. Prime example of nominative determinism. --TammyMoet (talk) 08:44, 16 May 2012 (UTC)[reply]

Wouldn't a proctostomy be the removal of a proctologist impacted in your nether regions ? StuRat (talk) 16:15, 16 May 2012 (UTC) [reply]

The imperforate anus article mentions a "perineal anoplasty". --Sean 20:15, 16 May 2012 (UTC)[reply]

Some cats might or might not have had a Fistulotomy, which is the most common reason for the surgical procedure in question. Tevildo (talk) 21:06, 19 May 2012 (UTC)[reply]

List of Self Limiting diseases?

I'm trying to find this list. It appears that wikipedia does not have such a category. Many definitions I've found will list a few, but so far I've got gastroenteritis, hepatits, the common cold, dyptheria, tonsillitis, llaryngitis. I imagine an exhaustive list would be huge, but can anyone find (or has seen) a list of the most common 20 or so? Vespine (talk) 02:43, 16 May 2012 (UTC)[reply]

I don't think many of those qualify as "self-limiting" as they are limited by our immune response. A truly self-limiting organism would control it's own population, say with waste which is toxic to itself in sufficient quantities. StuRat (talk) 03:13, 16 May 2012 (UTC)[reply]

Self limiting in medicine has a slightly diffent meaning to that of general biology. From the article: the term may imply that the condition would run its course without the need of external influence, especially any medical treatment. Vespine (talk) 04:16, 16 May 2012 (UTC)[reply]

Don't all diseases run their course with or without medical treatment? 112.215.36.183 (talk) 10:16, 16 May 2012 (UTC)[reply]

If you count "death" as an allowable outcome, then yes. I presume, however, the OP is asking about diseases that don't normally result in death if left untreated. If that is the case, then I'd remove diphtheria from that list. --TammyMoet (talk) 11:06, 16 May 2012 (UTC)[reply]

If you search wikipedia for articles containing self limiting, you get 2464 results. The description usually has enough info to know if it is a self-limiting disease. Examples: Scheuermann's disease, Scleredema, Vernal keratoconjunctivitis, Transient synovitis, Epiploic appendagitis, idiopathic scoliosis, Mondor's disease, Acute posterior multifocal placoid pigment epitheliopathy, Pityriasis alba, Cricopharyngeal spasm, Necrotizing sialometaplasia... For a list of the most common ones, I guess the best way would be taking a list of the say 50 most common diseases and check which ones are self-limiting. Problem is finding such a list, I can't even find two sources that agree on the nr 1 (gum disease according to one tabloid), even found a list that had sociopathic personality disorder in the top five. Ssscienccce (talk) 16:55, 16 May 2012 (UTC)[reply]

I thought there might have been a strict medical meaning but it seems there might not be.. Diptheria says it has a fatality rate of 5%-10%, lol, ok, I just searched for the source of the claim that diptheria is self-limiting back to Herbert_M._Shelton. Not exactly a reliable source.. But Ssscienccce has given me a great start. Thanks. Vespine (talk) 22:51, 16 May 2012 (UTC)[reply]

red giant's maximum diameter

How big is red giant at the maximum diameter. Can some red giant be as big as Mars orbit. How we know how big will our red giant be? Does some red giant get as big as 2 AU or as big as 3 AU? What is the size range of red giant from the smallest to the biggest?--69.228.133.188 (talk) 02:44, 16 May 2012 (UTC)[reply]

Does red giant answer any of this ? Note that the "size" is usually given as mass, which tends to be fairly constant, versus volume or radius, which can change dramatically. They list 0.5 Suns to 10 Suns as the range, with any stars more massive than that called red supergiants. (The illustration in that last article answers your Mars question.) As for how they know what the Sun will do, that's based on it's mass. StuRat (talk) 03:02, 16 May 2012 (UTC)[reply]

If your question was specifically "What is the biggest star?", see VY Canis Majoris. Its radius is estimated to be between 8.4 and 9.8 astronomical units, which is almost out to the orbit of Saturn. You may also be interested in our List of largest known stars. As far as our sun, it is estimated its radius will only be about 1.2 AU at its maximum, so the Earth will likely escape destruction due to its increased orbital radius by then (not that any of us will be around to care). -RunningOnBrains^(talk) 03:26, 16 May 2012 (UTC)[reply]

1.2AU isn't the maximum, it continues to say that it will eventually grow to 2AU. Vespine (talk) 04:14, 16 May 2012 (UTC)[reply]

My question was the biggest red giant in size, not the red supergiant or biggest star. I am asking for largest red giant only.--69.228.133.188 (talk) 05:34, 16 May 2012 (UTC)[reply]

Well then the answer is about 200 solar radii (2AU); the cutoff between red giant and red supergiant is quite fuzzy, if not arbitrary. For instance, Epsilon Aurigae at 135 solar radii is described as a red supergiant, while Rho Persei at 164 solar radii is considered merely a Asymptotic giant branch star (i.e., red giant).RunningOnBrains^(talk) 05:50, 16 May 2012 (UTC)[reply]

An inquiry into the nature of solenoids

The equation $B_{z}={\frac {\mu _{0}}{4\pi }}{\frac {2\pi IN}{L}}\left[{\frac {d+L/2}{\sqrt {(d+L/2)^{2}+R^{2}}}}-{\frac {d-L/2}{\sqrt {(d-L/2)^{2}+R^{2}}}}\right]$ gives the axial magnetic field strength $B_{z}$ in a solenoid. This is true for the case where every turn of conductor in the solenoid has the same radius $R$ . Would you expect the magnetic field strength measured in a coil consisting of many layers of turns, each with a slightly different radius, to be higher or lower in magnitude than that predicted by this equation using the inner radius of the solenoid? --130.56.84.118 (talk) 02:53, 16 May 2012 (UTC)[reply]

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our policy here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Anonymous.translator (talk) 03:23, 16 May 2012 (UTC)[reply]

GRRROOOAAAANANNNN. My line of reasoning was that the answer is higher because the many layers reinforce each other. Is this right? --150.203.114.37 (talk) 03:42, 16 May 2012 (UTC)[reply]

We actually can't answer your question because you didn't tell us enough about the multi-layered solenoid. Are you taking a solenoid with the same number of turns and simply altering the radii of the turns? Are you wrapping an additional solenoid around the original? Are you scrunching the solenoid up so that it's shorter and fatter? What is the distribution of radii with respect to the original radius? If you make the total circuit length longer, are you changing the total voltage to keep the current constant, or are you dropping the current to keep the total power output constant? Someguy1221 (talk) 04:08, 16 May 2012 (UTC)[reply]

I don't suspect this is a homework question, as it is more a test of algebra skills than magnetic understanding, and a generic math teacher is not likely to set such an applied question. On the other hand, electronics technicians and radio hams often ask this - they tend to only find the single layer formula in books etc.

Assuming you want to know what happens if you have the same total number of turns in a multilayer solenoid as in a single layer solenoid, with the current, wire gauge, and everything else the same, for comparison, the asnwer is simple: Apply the formula to each layer separately. The total axial field strength is the sum of the fields contributed by each layer. As the layer radius is lower (denominator) terms in the calculation, the layers are progressively less effective moving out from the inner layer. Therefore the total axial field strength in a multi-layer solenoid is less than that for a single layer coil of the same turns, and less than that estimated by taking the inner layer radius. However, if wire gauge is reduced so that the radial distance occupied by the turns is the same, you'll get near enough the same field. This is because the improved contribution of the inner layers is balanced by the reduced contribution of the outer layers. Keit120.145.9.168 (talk) —Preceding undated comment added 04:40, 16 May 2012 (UTC).[reply]

There seems to be some laziness in presenting an equation which contains 2 pi in the numerator and 4 pi in the denominator. Please simplify and ask again. Is it correct to infer that mu nought is the permeability of free space, I is current in amperes, and L is inductance in ohms? Is R in meters or centimeters? (My education included physics textbooks using both units of distance). Edison (talk) 05:34, 16 May 2012 (UTC)[reply]

You made me look at his formula again. It doesn't look right - for a start the Ln (natural log) symbol is missing. However, L in this case is not inductance (& inductance is not measured in ohms), but the length of the coil. Keit124.178.152.203 (talk) 07:18, 16 May 2012 (UTC)[reply]

Looks right to me, L being the length, and d the distance from the center (between -L/2 and L/2). If I understand the question correctly, it's whether using this formula for a multilayer solenoid, with R equal to the inner radius, would underestimate or overestimate the field strength? Assuming we're still talking about a solenoid of length L with N turns, I'd say the measured field strength would be lower, since the real radius is larger. Ssscienccce (talk) 17:44, 16 May 2012 (UTC)[reply]

Orbital Chainsaw

The Orbital Chainsaw would consist of thousands of small "teeth" satellites in Low Earth orbit. Each such tooth would have a reasonable sized solid state laser that was powered by ultracaps that were then refilled by solar power during the 99% of the orbit when that tooth wasn't over a useful target.

Each tooth is only over a target area for a short period of time, so only needs enough power storage to cover lasing for that short period of time.

The Orbital Chainsaw would have a thousand and one uses. From laser propulsion for laser powered aircraft and satellite launches (to fill in more teeth satellites), to weapons uses such as shooting down hostile ICBMs or satellite launches and dealing with gatherings of terrorists or other dissidents.

So why hasn't this been implemented yet? (What exactly is the flaw in this brilliant (pun intended) scheme of global domination?) Hcobb (talk) 10:58, 16 May 2012 (UTC)[reply]

You would need literally millions of satellites. Not very practical. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 11:09, 16 May 2012 (UTC)[reply]

Expensive. Plasmic Physics (talk) 11:10, 16 May 2012 (UTC)[reply]

Very—it would require more money than the entire U.S. gross national product. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 11:12, 16 May 2012 (UTC)[reply]

I don't know about the number but I agree cost seems to be a big factor. Consider say the cost of Iridium satellite constellation (which only involves 66 active satellites) or Globalstar or Orbcomm involving communications satellites. For example if we take Iridium, scale up the cost for your fancy laser satellites, times it by 50 for your proposal involving thousands of satellites (with some discount for the large number) and you probably end up with something more then the entire Military budget of the United States. Also while probably not banned by the Outer Space Treaty, I think it's clear such an extreme Militarisation of space is unlikely to be popular. The Strategic Defense Initiative was rather unpopular, and considering the existing ability of the US to nuke the entire planet many times over, the advantages seem slim considering the cost and unpopularity of such a move, mutually assured destruction is generally considered to remain a powerful deterence to any large enemy. In terms of 'rogue states', the US already have their Missile Defense Agency allegedly for that purpose. And smart bombs and UAVs to deal with gatherings of people they don't like. I think many would question how achievable most of your stated goals are anyway (using them from satellite launches sounds like wishful thinking to me). Nil Einne (talk) 11:37, 16 May 2012 (UTC)[reply]

Just compare the cost of the system against what is spent on commercial jet airliner fuel every year and it pays for itself in no time.

As for the number needed, assuming a reach of 200 km, fewer than 5000 satellites are needed to cover the entire Earth. Hcobb (talk) 12:05, 16 May 2012 (UTC)[reply]

Unlikely. Assuming this implausible amount, you would require 6332.573979 satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:13, 16 May 2012 (UTC)[reply]

Exactly who is going to pay to get them up there? Plasmic Physics (talk) 12:19, 16 May 2012 (UTC)[reply]

Yeah. It is much easier to get citizens to pay for things to move them places than it is to get anyone to pay for orbital weapons already racked with other problems. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:46, 16 May 2012 (UTC)[reply]

And capacitors to store power collected in one orbit (1 hour) would have to store megawatts of energy. These would be big and heavy. Graeme Bartlett (talk) 12:39, 16 May 2012 (UTC)[reply]

And that would give complete 100% global coverage at all times (die penguin scum!). A few dozen would be sufficient for satellite launches. (At which point the launch costs go way down.) Hcobb (talk) 12:22, 16 May 2012 (UTC)[reply]

"And that would give complete 100% global coverage at all times." Wrong. Because circles do not tesselate, you would need at least twice that number. And practically, you would need twice that, to provide redundancy and allow for failures. Which comes out to over 25,000 satellites. And a number of satellites this large would be extremely difficult to keep from occulting each other, colliding with each other, dragging each other out of orbit etc. Dyson sphere#Dyson swarm lists the problems with such a large swarm of satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:33, 16 May 2012 (UTC)[reply]

They only have to worry about occulting each other if they block a significant fraction of the sunshine striking the Earth. Given a a few meter diameter satellite every hundred km, this isn't much of a problem. As for energy storage the satellite would only be in a position to fire for an average of 200 km (not the full diameter of 400 km, because passing directly overhead is rare) and since it is traveling at 7 km/s it only fires for half a minute. Given a solid state laser of 100 kw, this is only three megajoules of power storage or 30 kg of ultracaps. Hcobb (talk) 12:59, 16 May 2012 (UTC)[reply]

Were the satellites light enough to be significantly helped to orbit by each others' lasers, then whenever one of the satellites fired its laser, the recoil would be enough to throw it into a higher orbit, possibly causing it to tumble as well, and forcing EITHER the launch of a replacement satellite, in which case the satellites would be literally one-use-only, necessitating replacement whenever fired, which would VASTLY increase the costs, OR the expenditure of large amounts of fuel to bring it back down to its rightful orbit.

Also, for attacks on ground-based targets, atmospheric attenuation might be a problem for objects near the satellite's horizon.

Firing the laser against property of another nation would be an act of war. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:04, 16 May 2012 (UTC)[reply]

It is also doubtful whether another space-faring nation would allow the development of such a system without deploying its own laser-armed killer satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:11, 16 May 2012 (UTC)[reply]

Because of the energy conversion issues, the Solar sail effect of the solar cells would be much greater than the laser recoil and almost, but not quite, vanishing small. It still makes a tiny bit of sense to be able to tilt the solar cell wings to (very very slowly) bleed off excess rotational momentum from the stabilization reaction wheels. One imagines that any country able to achieve Prompt Global Strike against any target on or near the Earth within one second of locating that target wouldn't care very much about what other countries thought about it. Hcobb (talk) 13:21, 16 May 2012 (UTC)[reply]

See my post above about enemy killer satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:24, 16 May 2012 (UTC)[reply]

Also, the force exerted by the laser on another satellite would be exactly equal to the force exerted on the satellite firing its laser. Have you any knowledge of Newton's third law of motion or the law of conservation of momentum? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:30, 16 May 2012 (UTC)[reply]

See Space-based solar power. I'm afraid you're golfing pretty far into the woods this time. There are limits on how precisely you can focus a laser from orbit. And using light defeats much of the purpose of space-based power since it gets absorbed in the atmosphere anyway. And a geosynchronous array loses only 75 minutes of sunlight to the Earth's shadow twice a year. And the satellites can be useful for power generation without being military assets. Wnt (talk) 14:54, 16 May 2012 (UTC)[reply]

Whether it is practical or not, I can't see why it would be useful. You can cover the whole Earth's surface with just a handful of geosynchronous satellites, so why bother building thousands of LEO ones? --Tango (talk) 23:06, 16 May 2012 (UTC)[reply]

In LEO the laser beam would have a much smaller spot on the ground. So it could be more concentrated, causing less damage outside its target area. Also from your geostationary orbit fixed location you may get a building in the way of your target. Graeme Bartlett (talk) 21:40, 17 May 2012 (UTC)[reply]

I'm totally out of my depth on this one, so this is probably moronic, but ... is it conceivable to make a "virtual" satellite to focus the signal from the geosynchronous microwave satellite tightly, by etching some sort of perfect lens into the ionosphere with high-frequency radiation to establish more and less conductive regions? Wnt (talk) 19:03, 18 May 2012 (UTC)[reply]

Due to air currents, I don't think so. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:55, 19 May 2012 (UTC)[reply]

Compatibility (mechanics)

Can someone explain to me in ver simple terms what compatibility in mechanics is? I've read the article on Wikipedia & many books but they all confuse me. Thanks.Clover345 (talk) 13:50, 16 May 2012 (UTC)[reply]

It means that a certain kind of parts can be used with ("are compatible with") a certain device. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:54, 16 May 2012 (UTC)[reply]

Well, I've read the Wikipedia article http://en.wikipedia.org/wiki/Compatibility_(mechanics), and I'm none the wiser. But it has nothing whatever to do with what Whoop whoop said. The word "compatibility" is apparently used in the sense that certain conditions must be satisfied for the math theory of deforming/bending a solid by distributed force to be valid, such as not tearing. The article needs to be re-written, or at least a preamble added, in ordinary language, so us ordinary mortals can figure out what possible use it might be. The wiki article was written by some math nut who likes the sound of his own gibberish, and thinks that links to other math nut stuff means something. Wickwack120.145.133.138 (talk) 15:30, 16 May 2012 (UTC)[reply]

It's unusual to see an article entirely written by one editor, but no, it's not total gibberish. You might like to read our article on Finite strain theory first, but unless you are "into" tensors, that might not make much sense either. I agree that the lead needs to be written for non-specialist readers. Dbfirs 16:04, 16 May 2012 (UTC)[reply]

I think it has to do with: 1) the deformation must be continuous, ie a curve or surface must still be a curve or surface after the deformation, and 2) the formulas or matrix representing the distortion have more unknowns than there are independent variables, so you'll have additional constraints, which are the compatibility equations. Edit: just found the Continuum mechanics article. Ssscienccce (talk) 19:05, 16 May 2012 (UTC)[reply]

Whether I saw the moon.

I saw a thin white cresent, perhaps waxing, not far above the horizon, roughly in the east, from London at 9:00 this morning. Was it the moon? 82.31.133.165 (talk) 14:01, 16 May 2012 (UTC)[reply]

Very likely yes. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:06, 16 May 2012 (UTC)[reply]

The Moon is in its waning crescent phase right now, and would appear as a thin crescent near the eastern horizon (and not too far from the sun) in the mornings. Also, there just isn't anything else that looks like a crescent moon in the sky, really, ever. TenOfAllTrades(talk) 14:10, 16 May 2012 (UTC)[reply]

Actually, the crescent was indeed waning. I confused waxing and waning. Thanks. 82.31.133.165 (talk) 19:20, 16 May 2012 (UTC)[reply]

Definitely waning; it needs to be a new moon soon so we can have this weekend's eclipse. You can always tell waxing vs waning by if the moon is visible at sunset or sunrise: visible at sunset means it's waxing, visible at sunrise means it's waning. -RunningOnBrains^(talk) 20:37, 16 May 2012 (UTC)[reply]

Or remember the word DOC, D for waxing, O for obviously full moon, C for waning, note the curve of the letters. Richard Avery (talk) 22:06, 16 May 2012 (UTC)[reply]

Maybe I'm missing something, but that doesn't seem like a very helpful mnemonic. Won't it depend on what direction you're facing? Also how do you keep track of which is waxing and which is waning? -RunningOnBrains^(talk) 22:13, 16 May 2012 (UTC)[reply]

Won't it depend on what direction you're facing? - Technically yes, but only if you can get yourself to somewhere beyond the far side of the Moon to be able to observe it from that side. From the Earth, which is where most of us live most of the time, it appears the same everywhere, at least in terms of general shape and orientation. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:29, 16 May 2012 (UTC)[reply]

I agree DOC does not seem useful, DOC seems to refer to crescent or gibbous but the moon goes throug each phase both waxing and waning: Lunar phase. Waxing or waning indicates whether the moon is going towards full or away from full (towards new), which you can not easily tell from casual observation, except to note whether it is morning or evening, as Running on Brains pointed out. Vespine (talk) 22:41, 16 May 2012 (UTC)[reply]

Doesn't a moon with a crescent on the right side appear to have the crescent on the left side when the viewer is upside down ? And, being an Aussie, you must be aware that you are upside down from the civilized world. :-) StuRat (talk) 22:41, 16 May 2012 (UTC)[reply]

(edit conflict)If I'm facing south and the moon is curved like a "C", then when I turn and face north the moon will look like a "D". I know I'm not mistaken... -RunningOnBrains^(talk) 22:40, 16 May 2012 (UTC)[reply]

Thank you guys, I'm really glad I'm not crazy :) -RunningOnBrains^(talk) 23:03, 16 May 2012 (UTC)[reply]

If you're facing south and looking at the moon and then turn and face north, then you can't see the moon any more because it's behind you... --Tango (talk) 23:13, 16 May 2012 (UTC)[reply]

And, if what you say is true, the HOLLYWOOD sign would read DOOWYLLOH from the front, depending on which way you're facing the sign. Is that your actual experience? If so, I suggest a good optician or even a psychiatrist. -- ♬ Jack of Oz ♬ ^{[your turn]} 23:33, 16 May 2012 (UTC)[reply]

Okay, but does the Hollywood sign move across the sky and rotate over the course of the night? When the moon is near the horizon, it is neither a C or a D, it is a U (the spines of a crescent always point up from the horizon). When it is directly above, it could be any of these, depending on what direction your body is facing as you look up. I'm not trolling, I'm legitimately trying to point out that this method just doesn't work. -RunningOnBrains^(talk) 00:45, 17 May 2012 (UTC)[reply]

And when I say "always" points up, I'm talking seeing the moon at night. I assume you can't see a small crescent while the sun's up (although I've admittedly never tried to look for it). -RunningOnBrains^(talk) 00:51, 17 May 2012 (UTC)[reply]

You can but it's more like hide the Sun behind an object and look x degrees along the ecliptic for it. Sagittarian Milky Way (talk) 18:26, 18 May 2012 (UTC)[reply]

What is the connection between the letter D and the word waxing? Or the letter C and the word waning? Even if you are restricting yourself to one hemisphere (as StuRat says, the moon is the other way around in the opposite hemisphere), I don't see how the word DOC helps you remember anything. --Tango (talk) 23:13, 16 May 2012 (UTC)[reply]

I'm not sure I get what Jack is saying, the moon DOES look upside down when viewed from the other hemisphere. You'd have to flip Hollywood along the horizontal axis too to get the correct picture, but since a C and a D are symmetrical along that axis, you don't notice it as much with the moon. That's why Orion and most of the other constellations are upside down when viewed from Australia, because northerners made them up. Australian Aboriginal constellations like the emu will look upside down from the north. Vespine (talk) 23:52, 16 May 2012 (UTC)[reply]

Actually, I might be wrong about the emu because I think even us southerners look south at emu, you'd probably have to be on the south pole to look north at it. Vespine (talk) 23:58, 16 May 2012 (UTC)[reply]

Forget this D and C stuff. "Waxing" means "getting bigger" and "waning" means "getting smaller" - as in the size of the illuminated portion of the moon as seen from earth. Next time, class, I'll explain what "gibbous" means. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:59, 16 May 2012 (UTC)[reply]

Resembling a gibbon ? :-) But seriously, why do we use such an archaic term as "waxing" instead of "growing" ? What would it be like if we used "waxing" for anything else ? "Hey mom, can you measure me ? I want to know how much I waxed today !" :-) StuRat (talk) 00:06, 17 May 2012 (UTC)[reply]

My interest in this topic is waning. ;) Vespine (talk) 00:25, 17 May 2012 (UTC)[reply]

Except for this bit "it appears the same everywhere, at least in terms of general shape and orientation." That's definitely wrong, even the Moon phases article points out that the moon looks different from the northern and southern hemispheres. Vespine (talk) 00:27, 17 May 2012 (UTC)[reply]

OK, I overstated my argument. I'm trying to get to the bottom of what Runningonbrains said: If I'm facing south and the moon is curved like a "C", then when I turn and face north the moon will look like a "D". If there are 30 people standing in a circle out in a moon-lit field, and they all look up at the Moon, ALL of them will report seeing a C-shaped object, or ALL of them will report seeing a reverse C-shaped object, or ALL of them will report seeing a (roughly) circular object. Is this not true? I cannot get my head around the claim that it will appear C-shaped to some but the reverse orientation to others in exactly the same place at exactly the same time, depending on which why they happen to be standing on the ground. That would be true of an object that is MUCH, MUCH closer, such as a photo of the crescent Moon stuck on your bedroom ceiling. But the actual, real Moon is WAY further away than that and it simply does not change its orientation in a matter of seconds as the observer changes their attitude to it. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:06, 17 May 2012 (UTC)[reply]

So if the dean is waxing wroth, he's getting bigger, Bugs? That will certainly make it harder for Roth to wax the dean when it's his turn. Deor (talk) 00:37, 17 May 2012 (UTC)[reply]

EO is way ahead of you on that one.[1] :) "Waxing wroth" is an obscure way of saying "growing angry". As to the verb form of "wax", which appears to have nothing to do with the noun form, my old Webster's indicates the word is cognate with Latin and Greek terms that are also the basis for the term "augment". So instead of a "waxing crescent", we could have an "augmenting crescent". But that doesn't alliterate with "waning" very well. :( As regards the moon looking "upside down", the "tilt" of the crescent is going to vary depending on your latitude, just like the sun's angle will vary - however, the "midpoint" of the crescent moon is always going to be "aimed" at the rising or setting sun. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:35, 17 May 2012 (UTC)[reply]

If you google [waxing] you'll see other relatively common uses of that old-fashioned term. "Waxing poetic" seems the most obvious, along with alliterative pair, "waxing and waning", which can refer to most anything, like "ebb and flow". ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:41, 17 May 2012 (UTC)[reply]

One other thing: On the night of May 5th we had that "super" full moon, so it stands to reason that it would be in waning crescent phase 11 days later. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:48, 17 May 2012 (UTC)[reply]

There's actually heaps of sites that explain this if you google "does the moon look upside down", but I can't find any that have a clear illustration showing why. I might try to make one later. Vespine (talk) 00:40, 17 May 2012 (UTC)[reply]

Indeed, just thinking outer-spatially about it, the moon is going to appear 180 degrees different from the north pole to the south pole. -RunningOnBrains^(talk) 00:49, 17 May 2012 (UTC)[reply]

Controversy on Gravitational Singularities

If the universe had begun from a single point of zero volume and infinite density and mass, then why is the present universe so imperfect? — Preceding unsigned comment added by 205.178.233.168 (talk) 14:04, 16 May 2012 (UTC)[reply]

"Imperfect" in what way? If you mean the uneven (yet still highly homogeneous and isotropic) distribution of matter, our article on cosmological inflation gives credit to quantum fluctuations magnified to large scales during that period, and further references the galaxy formation and evolution and structure formation articles. — Lomn 14:15, 16 May 2012 (UTC)[reply]

So far as I know - correct me if I'm wrong - this "point" never really existed, not even in theory. To quote our article, "This singularity signals the breakdown of general relativity. How closely we can extrapolate towards the singularity is debated—certainly no closer than the end of the Planck epoch." Given how much stuff is said to have happened in the first second after the Big Bang - see Chronology of the universe - it is hard to think of any part of the process as "actually" being "instantaneous" in any meaningful sense. It is simply absurdly fast and absurdly small by our yardsticks - by the vibrations of cesium atoms, for example. But in the philosophical sense - the sense abused by some as described in Religious interpretations of the Big Bang theory - I think it would make more sense to view our universe as infinitely "old" (in the sense of a logarithmic time), but "receding to the horizon" as we look backward to ages where different properties of matter predominated. So there is no need to explain this contradiction - indeed, if we extrapolate from any time we can actually observe, the hypothetical first point, if it existed, would be an imperfect point. Wnt (talk) 14:44, 16 May 2012 (UTC)[reply]

Yes, since things tend to happen much more quickly, and in a much smaller area, at higher energies, it's reasonable to take this logarithmic view of the early universe. Still, the laws of physics presumably bottom out somewhere, so there would still be a beginning in traditional big bang cosmology, though it wouldn't be a singularity.

The other issue is that inflationary cosmology explicitly replaces the early universe, including the supposed singularity, with something quite different, which has no obvious beginning. So in a sense we're back where we used to be, with a universe that has existed for an unknown time, perhaps forever, with the details seemingly beyond the reach of present-day experiment. -- BenRG (talk) 20:17, 16 May 2012 (UTC)[reply]

Spontaneous symmetry breaking Hcobb (talk) 16:52, 16 May 2012 (UTC)[reply]

Do not complain of an imperfect universe. It is written that the actual world is the Best of all possible worlds. 84.209.89.214 (talk) 14:51, 18 May 2012 (UTC)[reply]

Prove it. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:57, 19 May 2012 (UTC)[reply]

Beryllium barrier

What would be happened if beryllium-8 were a stable isotope? --84.61.181.19 (talk) 17:09, 16 May 2012 (UTC)[reply]

The entire universe would scream in horror. Looie496 (talk) 17:32, 16 May 2012 (UTC)[reply]

Big bang nucleosynthesis#Heavy elements briefly discusses one consequence. I'm not sure how much sense it makes to discuss this happening in isolation though - would it be possible to change physics in such a way that this isotope becomes stable, but nothing else is affected? 81.98.43.107 (talk) 18:00, 16 May 2012 (UTC)[reply]

If beryllium-8 were 10 eV lighter, it would be stable. --84.61.181.19 (talk) 19:20, 16 May 2012 (UTC)[reply]

Yes, but in order to achieve this you would have to change fundamental laws of physics, which would likely have dire consequences for life as we know it. -RunningOnBrains^(talk) 19:59, 16 May 2012 (UTC)[reply]

10 eV or 100 keV? --84.61.181.19 (talk) 20:28, 16 May 2012 (UTC)[reply]

Irish Great whites

If global warming is going to make the sea levels rise and warm up a bit does this mean that sharks like the great white might end up appearing in irish coastal waters. — Preceding unsigned comment added by 86.41.81.98 (talk) 21:20, 16 May 2012 (UTC)[reply]

Actually, quite possibly the opposite effect would happen in Ireland. The Gulf Stream, which brings warm tropical water to the coast of Ireland, is driven by a complex circulation in the North Atlantic. Melting northern polar ice is, in many models, predicted to disrupt this process, actually interrupting the flow of the Gulf Stream, causing the water (and likely the climate) of Ireland and other parts of Western Europe, to actually get colder. That's because, while the term is "GLOBAL" warming (that is the average temperatures of the entire world, averaged together) are getting warmer, the effects of those changes are complex, and can result in some local areas getting colder. That's why the preferred term is "climate change" as it captures the more complicated nature of what melting ice and overall warming will be. See Shutdown of thermohaline circulation which explains in some more detail how global warming can have a cooling effect on Ireland, which is named specifically. --Jayron32 21:30, 16 May 2012 (UTC)[reply]

I prefer the term "global weirding". StuRat (talk) 21:50, 16 May 2012 (UTC)[reply]

I have always had a problem with the "Gulf Stream Shutdown" theory, mainly because it's not purely a thermohaline circulation; it is also a dynamically-driven western boundary current. Seems like an incredibly unlikely scenario to me. Computer models seem to agree with me too, from the lede of the "shutdown" article: "In coupled Atmosphere-Ocean General Circulation Models the THC tends to weaken somewhat rather than stop, and the warming effects outweigh the cooling, even over Europe" -RunningOnBrains^(talk) 23:08, 16 May 2012 (UTC)[reply]

Indeed, the Gulf Stream itself will continue as long as the Earth continues to rotate at roughly its present rate and the Atlantic Ocean is not ice covered. Short Brigade Harvester Boris (talk) 00:26, 17 May 2012 (UTC)[reply]

Probably. The question becomes whether the nature of the Gulf Stream, such as the specific direction it travels, the temperature and salinity of the water it caries, and the way it affects the climate of Europe, which is in question. It is a complex situation, and does not have a simple answer, or even necessarily a known one. That's part of the major issue with climate change: If its effects were well understood, they could be prepared for. --Jayron32 00:29, 17 May 2012 (UTC)[reply]

There's no "probably" about the continuation of the Gulf Stream, or its direction, just basic fluid dynamics. See western boundary current. You're right about questions of thermal transport, salinity, and thus effects on climate of Europe (though in strict terms the latter are attributable to extensions of the Gulf Stream such as the North Atlantic Drift, rather than the Gulf Stream proper). Short Brigade Harvester Boris (talk) 01:45, 17 May 2012 (UTC)[reply]

I wouldn't call it "basic" fluid dynamics! That'd imply that it's the sort of thing we could use a simple equation for. Or, that it's the sort of topic that gets covered in an introductory textbook on fluid motion! The hydraulic equation is basic fluid mechanics. Bernoulli's law is basic fluid dynamics. The statistical dynamics of an ideal gas flowing in a perfectly thermally isolated closed box is fairly intermediate or advanced fluid dynamics. But climate and oceanic currents? That's some serious world's-largest-supercomputer kind of cutting-edge theoretical physics research. We have fairly sparse coverage in Wikipedia. One day I intend to work on improving our articles on global circulation models, but it's really dense stuff. Nimur (talk) 03:42, 17 May 2012 (UTC)[reply]

The equations which lead to a western boundary current are simple... or as simple as you get on the scale of fluid dynamics, anyway. They are laid out here quite nicely. Western boundary currents are unavoidable even in extremely simplified models of large bodies of water on a rotating sphere. -RunningOnBrains^(talk) 03:51, 17 May 2012 (UTC)[reply]

Urgent Antihistamine Question

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

If you have a question about the effects, uses, or dosage of medication, consult a medical professional. --Jayron32 23:14, 16 May 2012 (UTC)[reply]

Fair does. However, since "the deed is done" and I am genuinely not asking about my situation any more I will repost what I believe to be a compliant version just in case it interests people:

Antihistamine and general drug dosage question

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

If you have a question about the effects, uses, or dosage of medication, consult a medical professional. --Jayron32 23:14, 16 May 2012 (UTC)[reply]

Since you made your intent known in the previous version, people here cannot unremember the past. It is best to just drop this line of questioning, and ask a doctor if it is of concern to you. Or use the Wikipedia or Google or WebMD search function where no one will try to stop you. --Jayron32 23:35, 16 May 2012 (UTC)[reply]

I appreciate that but I'm not at all asking about the medicine as it relates to me, or tbh that medicine at all. I'm genuinely interested in the theoretical question. Although as the actual medicine is starting to kick in now, I will just ask tomorrow at some time when it is obvious it's just for interest, then we can all rest easy

Egg Centri c 23:40, 16 May 2012 (UTC)[reply]

Talk to someone who's qualified to answer, such as your doctor or maybe your pharmacist. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:50, 16 May 2012 (UTC)[reply]

One suspects that the complainants here must be associated with the medical profession, as once they have branded the defendant as "sick" and permanently denied him the rights other people would have on this basis, their work is done! Wnt (talk) 04:35, 17 May 2012 (UTC)[reply]

We can't answer any questions on specific medical matters, it doesn't figure into that how hypothetical it might be. It's a moral and a legal issue, if someone reads the archives or the thread here and makes an assumption they could become seriously hurt. It's not just because of the original question-asker, but anyone who might read the advice and be tempted to follow it rather than the guidance of a certified medical professional. HominidMachinae (talk) 04:39, 17 May 2012 (UTC)[reply]

No, I think you've misunderstood. We can't offer medical advice. We can indeed answer questions about medicine. --Trovatore (talk) 05:29, 17 May 2012 (UTC)[reply]

Physicians can bluster about the importance of following medical advice, but unless the person can pay them their shakedown, even the most medically necessary medicine will be banned from the patient by law, even though it is otherwise affordable - just as their advice on this would be limited to paying customers. I am wearied of their ethics. Wnt (talk) 04:53, 17 May 2012 (UTC)[reply]

Some of us are fortunate enough to live in a country where medical advice is free at the point of use. Here in the UK, if I need medical advice of the sort that's often asked for here on WP, I can go and ask my GP, or a pharmacist, and pay nothing for the privilege. (I've already paid for it in my taxes, but I'll complain about tax spending on aircraft carriers or unimplemented national databases before I complain about tax spending on keeping me and everyone else healthy.) AlexTiefling (talk) 08:15, 17 May 2012 (UTC)[reply]

Wnt apparently would stake his very life on the free (and worth the price) advice from strangers on the internet, rather than pay a trained professional. That's up to him. And if he suddenly stops editing one day, we'll know what happened. But wikipedia has rules against giving medical advice, and if Wnt doesn't like that rule? Too bad, too sad. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:27, 17 May 2012 (UTC)[reply]

Yeah, well, here in the U.S. they just closed down the local hospital, one which had stood for decades, in a sensible business decision reflecting that there is just not that much demand for medical care in the Third World. The prisons for people who sell allopurinol without a prescription, of course, well, those received a perky plus with all the money the state saved cutting the education budget. Wnt (talk) 21:32, 17 May 2012 (UTC)[reply]

What has that got to do with wikipedia's rules against giving medical advice? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:04, 17 May 2012 (UTC)[reply]

Nobody here ever pretended to give medical advice; then someone called it a "moral issue". Wnt (talk) 03:10, 18 May 2012 (UTC)[reply]

Advising someone NOT to see a doctor qualifies as medical advice. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:43, 18 May 2012 (UTC)[reply]

So make an appointment with your doctor so you can ask him if you should make an appointment with your doctor. :-) StuRat (talk) 05:52, 18 May 2012 (UTC) [reply]

May 17

What is the formal definition of a chromosome, or is there one?

I cannot seem to find a definition anywhere, Wikipedia contains conflicting information across different pages, as do Google results. Hoping someone here with a genetics background can enlighten me. Chromosome says that it is DNA complexed with protein, but I've read elsewhere that it is DNA complexed specifically with chromatin. The latter definition would exclude prokaryote circular DNA from being classed as chromosomes then.

Also, in regards to karyotype, I've heard at various times both a chromatid and a pair of homologous chromosomes being referred to as a chromosome in the singular. Humans have 23 pairs of chromosomes, right? So 46 different (heterozygous?) chromosomes in a normal interphase cell? Sorry if I have overcomplicated this, I just can't get my head round it at the moment. Thanks for any clarification in advance -Zynwyx (talk) 09:30, 17 May 2012 (UTC)[reply]

The first part is is a very good question, which urges me to do some remedial reading on prokaryote biology - thanks for asking it. As explained in chromosome, the term is used "loosely" to refer to the bacterial chromosome, which is technically a genophore. But when the literature is full of people speaking of the bacterial chromosome, that attempted redefinition seems like a failure.

So what is the criterion to distinguish a natural chromosome from a mere plasmid?

First, let's consider what it's not. A plasmid has an origin of replication, or it couldn't replicate; and the vast majority of the time it contains a partitioning locus (see segrosome), or otherwise strains would be spontaneously cured of it with fairly high frequency. (See [2] for some pretty data) The plasmid apparently is associated with the nucleoid at least the majority of the time (see previous and [3], though I'm a bit tentative about this). The distinction I find in one book is "Probably the best distinguishing characteristic of a plasmid is that it has a more typically plasmid origin of replication with an adjacent gene for a Rep protein rather than a typical chromosome origin with an oriC, along with a dnaA gene and other genes typical of the chromosomal origin of replication".[4]

Problem is, bacterial artificial chromosomes and phage artificial chromosomes don't meet this definition! As explained here, they're based on the F episome (F plasmid), but as our article explains, they do have the Rep gene. Also, lest we have any confusion, these are closed circular constructs which often undergo homologous recombination after injection to a mammalian host to form linear transgene arrays [5] - they are not eukaryotic chromosomes with telomeres and such like yeast artificial chromosomes. So they're called "artificial chromosomes" in the bacterial sense.

So what distinguishes a plasmid from an artificial chromosome? I'm going to guess that much of the distinction is practical: how much stuff you can cram into it. A high copy number plasmid is going to recombine with itself, sometimes lose pieces of the gene, and the ones with the least stuff in them will replicate best. So when you want to shove a lot of stuff into bacteria, you want one of these more well behaved low copy number vectors with more natural-ish performance. And so, in the absence of a very clear definition, we have this sort of continuum from natural chromosome to artificial chromosome to plasmid to the odd random piece of lost circular DNA, without as clear a defining line as we might like. But I might still be just not thinking about something important... Wnt (talk) 15:24, 18 May 2012 (UTC)[reply]

-----Thanks for the detailed response! I think genetics has become a victim of the speed of its own success in this respect, as there aren't concrete definitions for loosely thrown about terms. So at least there are genetic differences between a plasmid and a bacterial genophore (in terms of origin of replication, which is good to hear. A chromosome in eukaryotes though, as I have now learnt, can refer to:

a single strand of DNA, what might commonly be referred to as a chromosome when you look at a karyotype. E.g one member of the homologous pair of chromosomes we inherit from each parent.

in mitosis, two strands of identical DNA joined at the centromere (the X-shaped things, to put it in very non-technical terms). Each strand is in itself an individual 'chromosome' as by the first definition, and indeed one of the strands is a replica of the other, but in this situation each strand is called a chromatid and the two identical 'chromatids' joined together are known as sister chromatids. When the sister chromatids separate, each one then becomes known as a (daughter) chromosome again.

So this was confusing me when reading up on mitosis/meiosis as I found it hard to keep track of what chromosome/chromatid was going where and so on. I don't personally think it is a good use of a word, much like bringing two identical twins together and calling them one person, but I guess that is personal opinion and its something that is ingrained and unlikely to change. By the wider definition though of simply a large 'section' of genomic DNA (chromatin or not), both situations above are fine (albeit still slightly muddling to me). Another semantic muddle-up I was having was homozygous and homologous. Homozygous refers to when the two 'sister chromatids' are identical in gene sequence and allelic content (barring random mutation). Homologous refers to when two separate chromosomes or two 'chromatids' are the same in terms of what genes they encode for and the sequence of the gene loci, but differing in alleles. So in chromosomal crossover, two 'chromosomes' (which are composed each of two homologous and homozygous 'sister chromatids') come together to form what is known either as a tetrad or bivalent (more confusion...) and then homologous but heterozygous 'chromatids' (ie one chromatid from each X-shaped 'chromosome') recombine to form in total four homologous (but heterozygous to one another) chromatids, which then separate to become four homologous chromosomes. Phew! I'll definitely be thinking more carefully in future in using these terms!! -Zynwyx (talk) 13:25, 19 May 2012 (UTC)[reply]

Or maybe not! According to this page the two sister chromatids together can also be called a dyad. Is it correct to call the two sister chromatids (the "X") a chromosome then? -Zynwyx (talk) 13:42, 19 May 2012 (UTC)[reply]

I'm not following your reasoning with the last step. If you have 46 pairs of chromosomes, they first replicate their DNA to form 92 chromatids. They're still 46 chromosomes because each has 1 centromere. (That isn't as odd as it sounds because, remember, there's a period when the DNA isn't fully replicated, and you only definitively know that's over after they are starting to be pulled apart) Now the chromosomes pair up to form 23 bivalents, with exchanges between sister chromatids. I'm unaware of people calling them anything but 46 chromosomes and 92 chromatids at this point. After the exchanges, which involve crossing over and thus are incomplete, the bivalents get pulled apart, so you have 23 chromosomes and 46 chromatids in each. Then the chromatids are pulled apart and you have 23 chromosomes and 23 chromatids in each gamete (1n). So 1 chromosome = 1 to 2 chromatids, 1 bivalent = 2 chromosomes, 1 tetrad = 4 chromatids, 1 dyad = 2 chromatids. Unless I fouled up ;) Wnt (talk) 14:38, 19 May 2012 (UTC)[reply]

Thanks, that makes a lot more sense. I have been thinking this whole time of a chromosome as being a helix of DNA which holds one allele of a gene, and thus couldn't get round in my mind how the X things are called chromosomes as well. Obviously each X derives from one chromosome that is being replicated, and the replicas are joined at the centre, so it makes a bit more sense in my mind now. So 23 pairs of homologous chromosomes > 46 individual chromosomes > 92 chromatids during mitosis (but still 46 chromosomes). That clears it up in my mind a bit, thanks! -Zynwyx (talk) 08:30, 20 May 2012 (UTC)[reply]

Metal loses its magnetic properties when in a liquid state?

I was just reading this, http://en.wikipedia.org/wiki/Electromagnetic_projectile_devices_%28fiction%29#Literature

And it stated that "Later it was shown that molten metal cannot be accelerated by a magnetic field as metal loses its magnetic properties in a molten state, and Clarke admitted his error gracefully." Is that true? In a railgun however, a piece of metal need only be conductive, not magnetic. Would molten lead or an eutectic of lead and bismuth still remain electrically conductive? ScienceApe (talk) 15:28, 17 May 2012 (UTC)[reply]

Mercury (element) is a metal in it's liquid state at room temperature, and, I believe, retains electrical conductivity. Perhaps what they mean is that, if significantly perturbed, the liquid will then break into droplets, making it more difficult for an electric charge to pass between them. StuRat (talk) 16:25, 17 May 2012 (UTC)[reply]

Mercury is absolutely conductive in its liquid state, see mercury switch for a common electrical use of mercury. --Jayron32 17:23, 17 May 2012 (UTC)[reply]

It must be possible because you can buy one. This link is to the Permanent Magnetic Pump (PMP), but the data sheet also mentions an Electromagnetic Pump (EMP). --Heron (talk) 18:35, 17 May 2012 (UTC)[reply]

A railgun uses an electric current to generate a magnetic field in the projectile, accelerating it through simple magnetic repulsion, and can use any projectile material that remains conductive during acceleration. A coilgun is essentially a linear motor that accelerates a magnetized projectile, and so requires that the projectile be able to remain magnetic during acceleration. You can (theoretically) fire a liquid metal from a railgun, but not from a coilgun. --Carnildo (talk) 23:04, 17 May 2012 (UTC)[reply]

If a nail is heated red hot, as with a propane torch, it reaches its Curie temperature and is no longer attracted by a magnet. Other ferromagnetic metals have varying Curie temperatures. Iron would still be electrically conductive when heated red hot, though its resistance would change.Molten iron and steel would still be conductive, since an electric arc furnace relies of current flow from an electrode to molten metal in the crucible. Metals typically have an increase of resistivity of 1.5 to 2.5 in the liquid form compared to the solid form at the melting point per [6]. Edison (talk) 23:54, 17 May 2012 (UTC)[reply]

I would agree with Heron, induced eddy currents would make it possible. An eddy current separator accelerates nonmagnetic metals in the same way. Ssscienccce (talk) 08:14, 18 May 2012 (UTC)[reply]

Could the molten metal imagined by Clarke have been above its Curie temperature but contain a suspension of a powdered metal with a higher Curie temperature? 84.209.89.214 (talk) 14:40, 18 May 2012 (UTC)[reply]

Yes! A ferrofluid! You can even make one yourself fairly easily. That said, it's useful to understand why something is no longer magnetic in liquid phase. It's essentially the definition of a liquid - the material no longer has any defined ordering of its atoms (or constituent molecules), and ferromagnetism in particular depends on electrons being able to align in some fashion and remain that way. It's a useless suggestion if the atoms are turning and tumbling over each other as happens constantly in a liquid. That said, one could have something putty or gel-like which flows like a liquid but has defined crystal structure at a small scale. This is kinda a cheat, but it might work for some sci fi - a very-much-zoomed-in example would be like a giant tank of those mini-magnetic spheres. They will flow. SamuelRiv (talk) 02:11, 21 May 2012 (UTC)[reply]

Space radiation

Lets say you are in outerspace and you are protected from everything (pressure, you have oxygen, temperature, etc) except radiation. About how long would it take before you die from radiation? ScienceApe (talk) 15:46, 17 May 2012 (UTC)[reply]

- which part of outer space? Some parts are full of radiation, and you'd die immediately. Some parts have very little, and you'd die of old age. 91.125.207.125 (talk) —Preceding undated comment added 16:03, 17 May 2012 (UTC).[reply]

I don't think there's enough radiation, near the Earth, to kill from radiation sickness. There is enough, however, to cause genetic damage and cancer. But these things aren't always fatal, so there would be a decreased life expectancy, not certain death. If you were closer to the Sun, the radiation damage might be more severe.

Note that I'm assuming that you are excluding UV light. If that is included, I'd expect bare skin in space (if somehow protected from the cold and vacuum) to quickly burn, crack, and bleed. Death might occur within hours or days, from dehydration and infection. StuRat (talk) 16:04, 17 May 2012 (UTC)[reply]

The article Health threat from cosmic rays may be of interest to you. LukeSurl^{t c} 18:46, 17 May 2012 (UTC)[reply]

Define "radiation". What portions of the electromagnetic spectrum are you including, and what portions are you excluding? --Carnildo (talk) 23:05, 17 May 2012 (UTC)[reply]

Quoting from a NASA report (Shielding Strategies for Human Space Exploration dec 1997):

In prior manned space missions, the GCR have been considered negligible since the mission times were relatively short and the main radiation concern was the very intense SEP events which can rise unexpectedly to high levels, delivering a potentially lethal dose in a few to several hours which could cause death or serious radiation illness over the following few days to few weeks if precautions are not taken [4]. The most intense such event known occurred on August 4–5, 1972 between the Apollo 16 and Apollo 17 missions [5].

GCR = galactic cosmic rays, SEP = Solar Energetic Particles Ssscienccce (talk) 15:21, 18 May 2012 (UTC)[reply]

Mario Rabinowitz

I was reading through the Ball lightning article, and it has some stuff about black holes by someone named Mario Rabinowitz. At a first look the Mario Rabinowitz article makes him out to be someone very impressive, but I can only see links to ArXive papers, and I can't see any affiliation with physicists I've actually heard of, or a position at a university or laboratory. Looking at the article's history, it looks like almost all of it was added to wikipedia by people who haven't done other things. So I'm a bit suspicious. I can't find anything worthwhile about this person by searching Google (I find facebook and patents and whitepages and stuff, and copies of the papers). So I'm concerned that a] this person doesn't really exist at all (that the article is a hoax) or b] that this person does exist, but isn't a physicist anyone has heard about (and so maybe should't be on wikipedia). Or is he really a famous physics guy who I've just failed to hear anything about? 91.125.207.125 (talk) —Preceding undated comment added 15:59, 17 May 2012 (UTC).[reply]

He's a real person,[7] but it sure looks to me like he's a non-notable person who wrote an article about himself as an autobiography, and the article should be deleted. But WP:AFD would be the place to bring that up, not here. Red Act (talk) 19:29, 17 May 2012 (UTC)[reply]

Cooperative lightning? Intelligent grass?

I'm sure this is a very dumb question, but assuming that what this website says is true: When lightning begins to travel downward from a cloud, many objects that have built up a charge emit streamers. This could come from anything such as a blade of grass or a power pole. The first streamer to make contact with the bolt defines the final path the lightning will take. - then how do the ground objects "know" when it's time to start emitting streamers? A simple answer please, for this admittedly ignorant non-scientist. Textorus (talk) 16:03, 17 May 2012 (UTC)[reply]

See our article on lightning. Fundamentally, though, positive streamers from the ground emerge for the same reason that negative streamers from the cloud (i.e. the formative lightning strike) emerge -- there's a large electric charge differential present. — Lomn 16:11, 17 May 2012 (UTC)[reply]

You might also want to know how gravity works. That is, how does the Earth "know" there is a star 93 million miles away which it should orbit ? Some rather non-intuitive explanations emerge, such as space being curved, or the even weirder gauge boson theory. StuRat (talk) 16:20, 17 May 2012 (UTC)[reply]

Now that you mention it, that is a fascinating question, but I'll save that one for another time. (The aether gets knotted up into a rope, maybe, like a yo-yo string?) Textorus (talk) 18:01, 17 May 2012 (UTC)[reply]

Lightning is a very complicated electrodynamic phenomenon. In addition to the visible incandescent stream of hot gas that you see, there are also wide-band electromagnetic waves (radio waves), preceding the lightning strike, occurring in tandem with the lightning strike, and coupling with the complicated streams of moving ions and electrons. The radio signals from a lightning strike are often called "sferics." Like all other radio waves, they travel approximately at the speed of light. The actual event of a lightning "striking" may be preceded by a very quick burst of radio-energy; and then as the streamer forms, all sorts of electromagnetic effects start happening and interacting with each other chaotically. The gas gets hot and incandesces, releasing visible light (incandescent light); but the gas is also ionizing and forming an electrically conductive stream, providing a current path, releasing more radio-wave emissions; and of course, the radio-waves emitted will affect air surrounding the lightning streamer. Here's a fairly advanced science web-site: Lightning Modeling, that reviews some of the physics necessary to accurately describe what's happening during a lightning strike. Nimur (talk) 17:10, 17 May 2012 (UTC)[reply]

The website is above my pay grade, but it makes sense that there must be some connecting force linking earth and sky. So when conditions are right for a storm, you're saying there's already a lot of ions and electrons moving between the two, invisible to our eyes? (I'm sure I must have learned that in Physics 101 but that was a l-o-n-g time ago.) Textorus (talk) 18:01, 17 May 2012 (UTC)[reply]

I think this is actually a very interesting question. The electricity clearly finds a quite specific path. My suspicion is that even in the absence of thunderstorms we are surrounded by an amazing display of static electricity which is simply, most often, too weak for us to see. Certainly I know that during a thunderstorm I can feel frequent little shocks from a mattress if it contains metal springs - sort of the sensation of being first bitten by a mosquito, but of course without the mosquito or subsequent irritation. Sometimes I've ever observed sparks from a window screen though lightning was not nearby. Has anyone ever sought to visualize the wider web of static electricity, or is it simply impossible, or indeed, am I deluded? Wnt (talk) 18:04, 17 May 2012 (UTC)[reply]

HAIL Project sought to visualize the wider web of electrodynamics in the atmosphere on massive geographic scales. By monitoring perturbations in the continuous background of electromagnetic signals (specifically, several LORAN transmitters), data was collected to drive a complete realtime model of the ionization and the electromagnetic environment for the continental US. Nimur (talk) 20:58, 17 May 2012 (UTC)[reply]

A former colleague of mine attempted to measuring electric fields near clouds by photographing polarization changes through an optical telescope. This is called the Kerr effect, and refers to the change in optical properties of certain materials (like atmospheric air) when exposed to very strong electrostatic fields. I recall thinking the idea was crazy (the signal should be well buried in the noise); but that's why it's research... Nimur (talk) 21:02, 17 May 2012 (UTC)[reply]

I can't search videos online at work, but I encourage you to find the ultra slow motion videos of lightning, it's frikken awesome! The "streamers" they talk about are a LOT slower then the speed of light and there are videos of them propegating though the sky, once they "contact" eachother, the lightning bolt actually fires like a flash. It's one of the most incredible natural phenomena on earth I think. Vespine (talk) 22:56, 17 May 2012 (UTC)[reply]

And of course, the fine folks in Gainesville, shoot off rockets trailing metal wires. Nimur (talk) 00:33, 18 May 2012 (UTC)[reply]

The streamers start due to electrostatic forces: attraction between positive and negative charges. To use an analogy: Imagine a cardboard box with a bunch of nails, screws, iron filings and stuff. If you hold a magnet above it, the objects start moving, pointing in the direction of the magnet. Get the magnet closer and eventually one of the objects will jump up and attach to the magnet, usually pulling others along. Electric charges act in a similar way. Ssscienccce (talk) 08:24, 18 May 2012 (UTC)[reply]

Thank you for that plain-English answer. I thought it must be something like that. Textorus (talk) 18:50, 18 May 2012 (UTC)[reply]

Of course, there's still the question of exactly why forces can act at a distance, whether electromagnetism, gravity, etc. StuRat (talk) 18:54, 18 May 2012 (UTC)[reply]

Where can drugs get in your body? What barriers are there?

I'm aware of the blood-brain barrier but presumably there are plenty of other barriers. Where can orally or intra-venously taken drugs get in your body? Presumably anywhere blood can get, but where is that? Can drugs get inside cells? What about bones? What about your eye lenses? What effects whether they DO get there?

I think that's enough questions for now... although I've got a lot more.

What would be a good place to start looking for the basics of this stuff?

Egg Centri c 19:50, 17 May 2012 (UTC)[reply]

Hair analysis shows that detectable levels of many drugs can be found in your hair. Long-haired drug users effectively carry a timeline of their drug consumption imprinted in every hair of their head. --Jayron32 20:31, 17 May 2012 (UTC)[reply]

Plus, if a man has long hair, can't we just assume he's a drug addict ? :-) StuRat (talk) 18:56, 18 May 2012 (UTC) [reply]

An oral drug should get into a cell at least once, in the intestinal epithelium, in order to enter the body. These and many other drugs usually have their effects inside a cell. Sort of an exception are antibiotics, which act inside a cell, but not one of yours! ;) There are exceptions, though - any drug which blocks a cellular receptor, for example. So far as I know any injected drug (well, "biologic") with a name ending in "-mab" (monoclonal antibody) will not get into a cell for meaningful purposes (it might get endocytosed with a receptor and have a trip to the lysosome, but that hardly counts). Vaccines don't get into cells, at least not the old fashioned kind available on the market. There's nothing quite like the blood-brain barrier and even that allows some things to pass. Certainly bones are visited by Fosamax and its ilk. The lens of the eye is a curious case, as it receives sustenance from the aqueous humor from the ciliary body; thus drugs must go by this indirect route; nonetheless they can arrive. For example, acetaminophen overdose can form cataracts in experimental animals receiving the drug systemically;[8] however, this occurs after it is first processed by the liver to form a more toxic metabolite. [9] Wnt (talk) 21:28, 17 May 2012 (UTC)[reply]

The place to look for basic information is any introductory pharmacology textbook. Generally speaking in order to get into cells a chemical needs to be lipophilic, meaning capable of dissolving in fats or oils. That's basically the same thing required for a drug to cross the blood-brain barrier. The exception is substances that are transported by active uptake mechanisms. Looie496 (talk) 23:47, 17 May 2012 (UTC)[reply]

Thanks all

Egg Centri c 17:57, 18 May 2012 (UTC)[reply]

Note that the skin acts as a barrier to some, but not all, drugs. Hence transdermal patches. StuRat (talk) 18:58, 18 May 2012 (UTC)[reply]

May 18

sun become a red giant 5 or 7.5 billion years?

Wait, I am a little confused here. said sun enter red giant about 5 billion years from now. Dr. Schroeder and Smith's website said tip of RGB is 7.59 billion years from now. Does it take 2.69 billion years for the sun to become a red giant, or sun last of red giant for 2.69 billion years. So when sun leaves main sequence in 5 billion years, it becomes a yellow subgiant first, and it slowly work the way to red giant by gradual increase of size/luminosity, or once it branches off main sequence it goes directly to red giant by large increase of size/luminosity. Is the end of red giant alot larger in diameter and luminosity then the beginning of red giant?--69.233.254.22 (talk) 01:32, 18 May 2012 (UTC)[reply]

According to Formation_and_evolution_of_the_Solar_System#The_Sun_and_planetary_environments, the process that leads to the formation of a Red Giant starts at around 5.4 billion years, and the process of growing into that phase will last until 7.5 billion years, so yes, it seems to take roughly about 2-3 billion years from starting on the red giant path and becoming a full-fledged red giant. --Jayron32 01:39, 18 May 2012 (UTC)[reply]

(edit conflict)Unfortunately I can't find the original website (I'd appreciate if someone could link to it) but 7.59 billion years seems like way too much precision to tell when the Earth will be consumed by the sun, especially since it is up for debate whether that will even happen.

The timeline you have sketched is about right though. See the text and charts at Sun#Life_cycle. The sun will start fusing helium at its core in around 5 billion years, which would technically be the start of its red giant phase. From there its luminosity and size will continue to increase over the course of 2-3 billion years, while the surface temperature will decrease. At some point in this expansion phase the earth may be destroyed, likely towards the end if it does happen. -RunningOnBrains^(talk) 01:49, 18 May 2012 (UTC)[reply]

The diagram sketches [10].--69.233.254.22 (talk) 02:35, 18 May 2012 (UTC)[reply]

Eclipse question

I've heard that there will be a partial solar eclipse in my area on Sunday, and I plan on watching it. What type of eye protection do I need? Thanks! 24.23.196.85 (talk) 05:59, 18 May 2012 (UTC)[reply]

Regular sunglasses aren't enough. I believe the usual precaution is not to look at the Sun at all, but rather look at a projection of the Sun. If you have no equipment for this, you can create a makeshift pinhole lens by poking a pin through a sheet of paper (preferably dark colored) and letting it shine on a white sheet of paper. StuRat (talk) 06:06, 18 May 2012 (UTC)[reply]

Welding goggles are sufficient protection if you can get your hands on them. There are a good number of eclipse-viewing goggles available online (I tried to link to amazon, but apparently even admins are blocked by the spam filter. How rude!), though I'd be sure I bought from a reputable vendor if I was going to risk my vision. Alternatively, it is safe to look at the sun for brief periods close to sunset ([11]); depending on where you are it may still be visible then. Just don't stare, and I'd still recommend sunglasses.-RunningOnBrains^(talk) 06:50, 18 May 2012 (UTC)[reply]

And it should be number 14 welding glass, not any other number. Sagittarian Milky Way (talk) 17:06, 18 May 2012 (UTC)[reply]

This construction or variants of it should give good results compared to other designs as it will (a) prevent stray light from hitting the screen, (b) reduce the loss of light from transmission through the screen. An adjustable back board(made by using one box inside another) will help you make the image-size brightness tradeoff. Use a binocular instead of the pinhole for better imagesStaticd (talk) 09:11, 18 May 2012 (UTC)[reply]

You can also get good projections simply by walking under some trees - pinholes in the leaves will project images of the sun (of varying focus) for a unique effect. Wnt (talk) 14:09, 18 May 2012 (UTC)[reply]

I was at Watford Junction train station at the time of the 1999 total eclipse, and was fortunate enough to see multiple images of the eclipse reflected off one of the dark glass windows of the waiting room! --TammyMoet (talk) 15:36, 18 May 2012 (UTC)[reply]

And in 2004 I didn't have a real filter, I saturated a piece of paper with butter, started looking through it for the shortest instant needed to form an image. Constantly blinked or waited 1 or 2 seconds depending on brightness. Was this bad? I stopped doing it when it when it seemed too bright (how high I don't remember but no more than about 12°). Another time I was holding eclipse glasses over 7x35 mm binoculars (the front, never the back), (I made a shield cause the glasses frames were too small), the sun was ~26° high, and either I put them up too late or probably forgot to cap the other lens despite planning to do that but holy crap, I saw sun! I shut my eyes as quick as I can. Thank goodness at least an eye doctor later didn't say anything (and I didn't ask). So don't do anything stupid/risky like this. They got so hot from the method above that a piece of glass broke off inside. Sagittarian Milky Way (talk) 17:57, 18 May 2012 (UTC)[reply]

What if you pointed a video camera at the eclipsed sun for 15 seconds or so (without looking through the view scope of course) and then watched it later on the TV? Obviously, the harmful rays would not be recorded and conveyed through the TV screen when you watched it, but the question is if 15 seconds pointed at the sun would seriously damage a video camera's CCD. 20.137.18.53 (talk) 15:43, 18 May 2012 (UTC)[reply]

Yes, I would expect that to damage the camera. Specifically, I'd expect the area of sensors which had the Sun focused on them to lose sensitivity. StuRat (talk) 18:13, 18 May 2012 (UTC)[reply]

That sounds a lot like electronics advice that should be given by a trained professional. ;) It seems worth saccing a videocamera for a total eclipse, but not an annular. ;) Wnt (talk) 16:54, 18 May 2012 (UTC)[reply]

Or get one of these, only $100,000 for the 95mm x 95mm CCD alone :) 20.137.18.53 (talk) 17:15, 18 May 2012 (UTC)[reply]

I suppose you could design a system where a camera has something like a "finder scope" that measures the max light level and puts the appropriate filter in front of the main lens. However, sudden changes in brightness, like the Sun coming out from behind the Moon, might still damage it. StuRat (talk) 19:23, 18 May 2012 (UTC)[reply]

Maybe a suitable photochromic lens material could be developed, like the one in those eyeglasses that become sunglasses outdoors? Which works too slow and too mild for this purpose. Even photographic solar filters transmit 0.01% of light, 10 times that of filters for eyes. Sagittarian Milky Way (talk) 21:49, 18 May 2012 (UTC)[reply]

An important thing to remember when picking a filter is that direct sunlight is a mix of infrared, visible, and ultraviolet light, any of which can cause eye damage. A lot of filters (eg. filters designed for cameras) only block visible light, leaving the infrared and ultraviolet components to blind you. --Carnildo (talk) 23:20, 18 May 2012 (UTC)[reply]

Phd stipend life sciences

Please don't post the same question to multiple places. Your question will be answered at Wikipedia:Reference desk/Miscellaneous, section "phd salary". Nyttend (talk) 12:44, 18 May 2012 (UTC)[reply]

What's the most likely problem with my tire inflator/spotlight?

I have a tire inflator with spotlight from the same manufacturer but not the exact same model as this one. I can charge it with the AC adapter all night until the battery charge status indicators (as seen on the fourth page of the PDF) show full charge with two red LEDs and one green one. But after I unplug the AC adapter, within a few minutes, if I push the battery charge status pushbutton, only two red LEDs come on, indicating a supposed medium battery charge level. I opened the thing up after having fully charged it and seen the indication of only medium charge (when the adapter was unplugged). The thing has two 6V sealed lead acid batteries. I tested both with my multimeter every couple of hours for the past two days. One has maintained a level of 6.57V, and the other has maintained a level of 6.43V. What is the most likely reason I get only two red LEDs and not the green one when the adapter's not plugged in? 1) Something's wrong with the charge indicator circuitry, 2) Something's wrong with the batteries (are sealed lead acid batteries labeled 6V supposed to be getting up to 6.4-6.5V? do bad things happen when they are overcharged by this much?), or 3)something else (ideas appreciated)? Thanks. 20.137.18.53 (talk) 12:59, 18 May 2012 (UTC)[reply]

The article Lead-acid battery indicates that 2.1V per cell i.e. 6.3V from your battery is a normal open-circuit voltage at full charge. The slightly higher voltage that you measure is not likely any fault and a gelled electrolyte battery would happly accept continuous 6.7V float charging. There could be a calibration error in the green charge indicator or you may be using the unit at an unexpected temperature. 84.209.89.214 (talk) 14:23, 18 May 2012 (UTC)[reply]

about sodium silicate

Hi, I want to remove or de-active sodium silicate property from west water of textile unit. — Preceding unsigned comment added by 115.248.240.58 (talk) 14:37, 18 May 2012 (UTC)[reply]

What is a textile unit? Sodium silicate is soluble in a highly alkaline solution. Plasmic Physics (talk) 14:40, 18 May 2012 (UTC)[reply]

Sodium silicate is often shipped with clothes, to absorb humidity and prevent mildew. It's normally grains in a small packet. Are you saying the packet ripped open and the grains are on the clothes ? If so, just brush them off, they aren't toxic to the touch (but wash your hands after, as they can be alkaline, and you wouldn't want them to get in your eye). If they leave a residue, run it through the washing machine. StuRat (talk) 18:19, 18 May 2012 (UTC)[reply]

I think the OP is not a native English speaker, possibly third world, and meant to ask: "I want to remove or de-activate sodium silicate properly from waste water produced by a textile (manufacturing) unit. How can I do this?" Wickwack60.230.203.253 (talk) 04:03, 20 May 2012 (UTC)[reply]

Gibbs free energy

Can someone explain what Gibbs energy is simply and give a biological example of Gibbs free energy. This article [12] is far too complicated. 176.250.232.230 (talk) 15:25, 18 May 2012 (UTC)[reply]

Simple explanations will invariably gloss over subtle details: this is tricky, because Gibbs energy is distinct from enthalpy and Helmholtz free energy, only by a small variation in definition. Roughly, Gibbs energy refers to the available energy from a chemical reaction, accounting for the pressure-volume work that must accompany that reaction. For example, if fermentation will release gaseous CO2, we use the Gibbs energy to quantify the work done, minus the "useless" work in expanding the CO2 as a gas. A worked example: ethanol metabolism thermodynamics. Nimur (talk) 15:47, 18 May 2012 (UTC)[reply]

thanks but people refer to Gibbs energy with things like linking amino acids to build proteins, converting ATP to ADP and the Krebs cycle and I just don't see the link with Gibbs which seems to be a thermodynamics concept rather than anything biological. — Preceding unsigned comment added by 176.250.232.230 (talk) 15:54, 18 May 2012 (UTC)[reply]

Lets try this really simply, by explaining "free energy" first. Free energy is energy availible to do work. Period. It just means that it is energy which could do something useful. There are forms of energy which are not free, that is there is energy which will cost you more energy to get to use. Roughly speaking, this is what entropy is. Basically, all of the energy in the universe is constant (First Law of Thermodynamics) but the amount of free energy is decreasing as the amount of entropy is increasing (Second Law of Thermodynamics). The different types of "free energy" and related measurements (like Gibbs Free Energy, Helmholtz Free Energy, Enthalpy) are just different variations on that theme; they are mathematical ways of expressing free energy in terms of highly constrained experimental set ups. One way to look at energy is to divide it into thermal energy and mechanical energy; that is energy which changed the temperature of a system, and energy which moves something around. So, free energy has a thermal component (changes in temperature) and a mechanical component (moving stuff). When you deal with gases, the mechanical aspect of their free energy deals with changes in volume (think, heating a balloon) and pressure (think, heating a steel tank). The difference between the various types of free energy is in how they treat that mechanical component of gases. In Gibbs free energy, your calculations assume that the system is at a constant pressure. For any system which is exposed to the earth's atmosphere, this works well, because any production or consumption of a gas will have a negligible effect on the entire earth's atmosphere, so any open container is a good Gibbs system. In Helmholtz free energy, your calculations assume a constant volume, which happens when you have a closed system, say a sealed tank, where the pressure will tend to vary a lot, but the volume remains constant. Gibbs free energy is also very important, because it is (via the Second Law of Thermodynamics) a mathematical way to calculate spontanaity. That is, any process which itself has a decrease in free energy associated with it will be spontaneous, because the universe spontaneously loses free energy, so any process that does that is likewise spontaneous. If you have a process which has an increase in free energy, TANSTAAFL, so there has to be a connected process which lowers the free energy by a greater amount, so the net change in free energy is always decreasing. So, to put it in simplest terms:

Free energy is energy which the universe has availible that you can tap into to do something useful
Gibbs free energy is a specific way of measuring that energy which works well in systems that are "open" to the environment
Gibbs free energy is important in calculating how "spontaneous" a process is; processes which cause a decrease in free energy occur spontaneously
(to your last question) Thermodynamics is inescapable. It's not like the laws of thermodynamics stop working in biological systems. That's why they are "laws" of the universe. They always work and continue to work, so when looking at, say, the assembly of a protein, if it has a negative Gibbs value, we know that it occurs spontaneously. This is kinda important info to know as a biochemist or molecular biologist who is concerned with how biological processes can occur.

Does this all help? --Jayron32 16:53, 18 May 2012 (UTC)[reply]

Thanks alot. That's a really good explanation. Most books or articles I look at just get too mathematical. — Preceding unsigned comment added by 176.250.232.230 (talk) 17:56, 18 May 2012 (UTC)[reply]

Zeta potential

I have a basic understanding of zeta potential (a property of colloidal systems). My question is, do hydrocolloids such as protein have a zeta potential? What about dissolved ions; do they have zeta potentials? ike9898 (talk) 18:48, 18 May 2012 (UTC)[reply]

Well, just trying the first example that came to mind, I searched "sickle cell" "zeta potential" and got actual numbers for red blood cells.[13] (also mentioned in Erythrocyte sedimentation rate). Trying the same for amyloid got what looked like a weaker set of results but nonetheless [14] [15]. My thought is that the measurement for proteins should be sort of weird because their aggregation depends so much on interactions that vary widely across the surface, but I really don't know. Wnt (talk) 19:00, 18 May 2012 (UTC)[reply]

Well, blood cells really don't fit into the category of things I am asking about. I'm taking about aqueous dispersions of hydrophilic polymers such as protein, starch, or polyacrylamide. Amyloid doesn't really fit into this category well, either. ike9898 (talk) 19:18, 18 May 2012 (UTC)[reply]

Apparently starch granules have a zeta potential. (Google dumps lots of results; apparently it's important for paper making) Polyacrylamide delivers more random results ... if it's a true gel, I don't know how you define a zeta potential, but that doesn't mean it can't be done. ;) Here's zeta potential for BSA [16]. Wnt (talk) 23:21, 18 May 2012 (UTC)[reply]

Thanks. The last sentence, especially, is a good lead. ike9898 (talk) 01:00, 19 May 2012 (UTC)[reply]

Flat universe have zero total energy?

I read from some dude on the internet (not exactly reliable) that the WMAP and Boomerang experiments (no idea what those are) indicate that the universe is flat and there is no net warpage of space time. Then he claimed that a flat universe can have zero total energy and thus come from nothing. What on earth is he talking about? Is this complete bs? ScienceApe (talk) 19:11, 18 May 2012 (UTC)[reply]

Those experiments (WMAP and BOOMERanG experiment) don't indicate that space is flat, they are simply consistent with it being flat. There is a margin of error in the results of any experiment, so all we can say is that zero curvature is within the range the experiments give. To conclude that it is absolutely flat, you need to use theoretical arguments, rather than experimental ones. --Tango (talk) 19:50, 18 May 2012 (UTC)[reply]

Check out Lawrence M. Krauss's 2009 lecture A Universe from Nothing. SkyMachine ⁽⁺⁺⁾ 20:52, 18 May 2012 (UTC)[reply]

Even theory doesn't imply that it's exactly flat, though I think the theoretical constraint is a lot stronger than the experimental constraint. -- BenRG (talk) 21:25, 18 May 2012 (UTC)[reply]

See Zero-energy universe. It's an idea that doesn't make much sense to me (because you have to break general covariance in order to talk about the energy of the universe), but plenty of legitimate physicists believe in it.

(Incidentally, the universe is spatially flat (approximately). Spacetime isn't flat.) -- BenRG (talk) 21:25, 18 May 2012 (UTC)[reply]

Can you explain what that means exactly? That the universe is spatially flat. ScienceApe (talk) 22:00, 18 May 2012 (UTC)[reply]

It means that if you look at the distances to all of the galaxies at a particular cosmological time (a particular era in their evolution), the number of galaxies within a distance R of you is proportional to R³. The number of galaxies would increase more slowly as R increased if space was positively curved, or more quickly if it was negatively curved. This is like measuring the amount of Earth's surface that's within a given distance of you—for small distances it grows like R², but the increase slows down for distances in the thousands of kilometers. The difference between spacetime and space in this context is like the difference between Earth and the surface of Earth. Earth is flat (it's a ball in 3D Euclidean space, approximately) but Earth's surface is positively curved on average. -- BenRG (talk) 23:08, 18 May 2012 (UTC)[reply]

Why isn't spacetime flat? If space is approximately Euclidean, isn't spacetime approximately Minkowski? I would call Minkowski spacetime flat - would I be wrong? --Tango (talk) 23:09, 19 May 2012 (UTC)[reply]

Mootractor

For lack of a better name. I made this image in the hopes that I can make a collage of images for different views of the moon.

Can I assume that a view from the north pole would be 180 deg different from that of the south pole and 90 deg different from the equator? I am hoping that people from around the world can hold their monitors up to the moon and let me know which angle shows on the top of the moon, or is there a way to do this with math once I get one from the equator?--Canoe1967 (talk) 22:26, 18 May 2012 (UTC)[reply]

You might find Commons:Category:Moonrises and Commons:Category:Moonset to be useful (hmmm, the first two photos I looked at from each, both from Germany, had about the same angle, though one was rising and one setting. Thinking about the moon is a good way to strain your spatial comprehension. ;) File:Gaisberg_and_rising_full_moon.jpg File:Monduntergang_2011-04-17_002.JPG) Wnt (talk) 23:13, 18 May 2012 (UTC)[reply]

http://wms.lroc.asu.edu/lroc_browse/view/WAC_GL000 may be the same as a view from the equator. I may have to query further on an astronomy site.Canoe1967 (talk) 00:11, 19 May 2012 (UTC)[reply]

Resolved

My original plan won't work. The wise people on the astronomy site tell me that there are variations in rotation of up to 180 deg of view from any one point on earth from moonrise to moonset. Our view in the northern hemisphere would be close to 180 degrees sideways from that of the southern hemisphere. Now I just need a practical use for the mootractor image I made.--Canoe1967 (talk) 02:51, 19 May 2012 (UTC)[reply]

May 19

Eye movement limit

Can you move your eye so that your iris and pupil are not visible to reasonably sizeably open eyelids? The reason I ask is that you quite often see in tv/movies etc folk with just the white of their eyes visible, but I just tried to do this both up and down in my webcam and it didn't work, and then I asked a girl friend of mine with a very different build to try it on webcam and it didn't work, so it isn't a male/female thing and probably isn't a build thing. Doesn't work with left/rights either with me at least, I just realised that when typing this. Indeed, I can't even move my eyes to the extent that I can't see anything (although she says she can). If it isn't possible and we're not freaks, is this a physical or a mental limitation? Egg Centri c 02:49, 19 May 2012 (UTC)[reply]

It would be limited by the flexibility and elasticity of the optic nerve and of the muscles moving the eyeball. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:52, 19 May 2012 (UTC)[reply]

That's the physical limit. Is there a mental one? (i.e. brain preventing demands to the muscles that "don't make sense") Certainly when I move my eyes it isn't really a conscious process of trying to move the muscles as such, if you see what I mean. I just move the eyes to what I want to look at. Egg Centri c 03:00, 19 May 2012 (UTC)[reply]

I would guess not, seeing as it is possible to roll back one's eyes until it hurts, implying to me that that is where the physical limit kicks in. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:03, 19 May 2012 (UTC)[reply]

Sure, but for me at least that hurt is very similar to the sensation of looking at the sun. How do you perceive it out of interest? (And where does looking at the sun 'pain' come from now I think to add it, I was just assuming it was psychological) Egg Centri c 03:19, 19 May 2012 (UTC)[reply]

You know this is the thread that is actually going to get WMF sued... ;) Wnt (talk) 14:19, 19 May 2012 (UTC)[reply]

Are you saying I should not suggest trying to do eye exercises to gain stretchiness? Unique Ubiquitous (talk) 16:58, 19 May 2012 (UTC)[reply]

Clomethiazole

Just watching trainspotting at the moment (which is actually where my eye question inspriation came from!), and Renton is going through all the drugs he and his fellow junkies enjoy aquiring illicitly. As an opioid addict myself, which I've mentioned before (although never one who has ever stolen things or manipulated people or in any way whatsoever behaved like a twat to get drugs, and in fact I resent that stereotype for a million reasons - but primarily cause it means that a lot of so called medical "professionals" treat me like shit - oops rant detected... in my view it comes about because twats are far more likely to use drugs, rather than drugs make you a twat, I've even met highly functional crack and meth addicts, I know a head of a desk at a bulge bracket bank who's one... but I digress...) I've never heard of this drug, unlike the others, and looking it up... ok, it's a sedative and a hypnotic. That usually means it may have some recreational potential. But is that the only reason they would want it, and does anyone know of it actually being abused? Is it a potentiater (sp?) for something? One suspects there was another reason for including it in the list. Egg Centri c 03:19, 19 May 2012 (UTC)[reply]

Clomethiazole ... hmmm... I see it affects GABA signalling; but GABA-A rather than the GABA-B of Gamma-Hydroxybutyric acid ... still, apparently both have been attempted as withdrawal-medications for alcohol according to the worst translation of a medical article I've seen in some time. It appears on a long list of abusable substances, but this seems to be mostly a weird practice of alcohol addicts. [17] [18] My suspicion is that all this GABA crap, including GHB, is a poor substitute for ethanol which is a straight up GABA agonist and much better-known to partiers. Wnt (talk) 03:56, 19 May 2012 (UTC)[reply]

I wouldn't say GHB is a poor substitute for ethanol, speaking from personal experience (admittedly with GBL but AIUI that's for all intents and purposes the same thing) - indeed I'd say it's probably better due to the lack of calories! They are very similar admittedly. Lots of thingies effect GABA signalling though, right? Egg Centri c 04:36, 19 May 2012 (UTC)[reply]

Well, some Russians I've known said they used to get high on isopropanol in Gorbachev's time. All alcohols seem to have certain things in common - they taste bad, they're toxic, and they get you drunk - some are just worse than others. The nice thing about beer is you can usually figure out the right dosage even after you're drunk. ;) Wnt (talk) 14:42, 19 May 2012 (UTC)[reply]

Proton-Proton Chain Reaction

In the artile (http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction), the third paragraph reads:

“In the Sun, deuterium-producing events are so rare (diprotons being the much more common result of nuclear reactions within the star) that a complete conversion of the star's hydrogen would take more than [ten billion] years at the prevailing conditions of its core.”

I am confused by this, as well as the next sentence, which states

“The fact that the Sun is still shining is due to the slow nature of this reaction; if it went more quickly, the Sun would have exhausted its hydrogen long ago.”

Regarding the first sentence; I find several problems for me to conclude that the article is accurate. First of all it claims that ‘deuterium-producing events are so rare in our sun, that it would take 10 billion years to convert all the sun’s hydrogen into helium’ - yet this IS what the sun’s lifetime is predicted to be.

Secondly; the article goes on to explain this very deuterium process (one proton, one neutron) in the subtitle: The proton-proton chain reaction - opposed to the “much more common” nuclear reactions the author proposes is taking place in our sun.

Thirdly; the sentence, in parenthesis, infers that the sun’s main fusions product is “diproton” - or, helium-2. The article for helium-2 (http://en.wikipedia.org/wiki/Diproton#Helium-2_.28diproton.29) explains that diproton is only a “HYPOTHETICAL” helium isotope. It appears to me that helium-2 not only defies the Pauli exclusion principle, if even possible, it would convert the sun’s hydrogen into helium exponentially faster than 10 billion years!

As for the second sentence; it could be accurate IF it unambiguously defined WHICH “reaction” (deuterium or diproton) is responsible for the “slow nature” of the sun’s actual proton-proton fusion process.

To me, it seems the contributor may have cut-and-pasted information from another stellar nucleosynthesis web-page, without making the necessary changes to the wording to fit the context of the actual article. — Preceding unsigned comment added by 64.180.243.107 (talk) 03:50, 19 May 2012 (UTC)[reply]

The diproton, once formed, immediately decays back into two protons. Therefore, the only reactions that produce anything are those extremely rare ones that produce deuterium. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:09, 19 May 2012 (UTC)[reply]

See the present state of the article. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:51, 19 May 2012 (UTC)[reply]

Equilibrium in analytical mechanics

My textbook asked me to prove that a system with scleronomic constraints is in equilibrium if and only if ${\frac {\partial V}{\partial q_{j}}}=0$ for all generalized coordinates q_j, assuming that all non-constraint forces are conservative. (V is the system's potential energy).

I was able to prove this, but as far as I can tell this proof holds for all holonomic systems, not just scleronomic ones. But because it was stressed that the constraints are scleronomic, I suspect that I'm making a mistake somewhere.

Proof of above statement: System is in equilibrium iff ${\vec {F}}_{i}=0$ , where ${\vec {F}}_{i}$ is the total force on the ith particle. $Q_{j}=\sum _{i}{\vec {F}}_{i}\cdot {\frac {\partial {\vec {r}}_{i}}{\partial q_{j}}}$ , where Q_j is the generalized force associated with the jth generalized coordinate. So, if ${\vec {F}}_{i}=0$ , then Q_j = 0. But $Q_{j}=-{\frac {\partial V}{\partial q_{j}}}$ , so ${\frac {\partial V}{\partial q_{j}}}=0$ is a necessary condition for equilibrium.

Now we prove that it is a sufficient condition. To do this, we find the ${\vec {F}}_{i}$ 's as a function of the Q_j's by making virtual displacements $\delta q_{j}$ to the generalized coordinates. The the virtual work is $\delta W=\sum _{j}Q_{j}\delta q_{j}=\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}$ . Writing $\delta q_{j}=\sum _{i}\nabla _{i}q_{j}\cdot \delta {\vec {r}}_{i}$ (we've tacitly expressed the generalized coordinates as functions of the r_i's; $\nabla _{i}q_{j}$ stands for ${\hat {x}}_{i}{\frac {\partial q_{j}}{\partial x_{i}}}+{\hat {y}}_{i}{\frac {\partial q_{j}}{\partial y_{i}}}+{\hat {z}}_{i}{\frac {\partial q_{j}}{\partial z_{i}}}$ ).

From this, it follows that $\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}=\sum _{i}(\sum _{j}Q_{j}\nabla _{i}q_{j})\cdot \delta {\vec {r}}_{i}$ , implying that ${\vec {F}}_{i}=\sum _{j}Q_{j}\nabla _{i}q_{j}$ . Therefore, if Q_j = 0, system is in equilibrium. QED.

Now, as far as I can tell I haven't used the assumption that the constraints are scleronomic, but maybe the assumption sneaked in there somewhere.

This is a pretty pedantic question, but it's been bugging me for a little while now, and I'd appreciate anyone's help in alleviating this confusion. 65.92.6.118 (talk) 06:26, 19 May 2012 (UTC)[reply]

Plastic cities

http://rt.com/usa/news/donna-summer-cancer-singer-627/

Donna Summer believed that it was the September 11 deadly smokes that gave her cancer.

During the WW2, the U.S. and Nazi Germany destroyed many cities by carpet bombing: Dresden, Coventry just to name a few. These WW2 cities certainly did not have much polymers and advanced man-made materials. People could live in the ruins. The cities were rebuilt within years after the end of the wars.

What will happen if a today's city is destroyed by a major earthquake or a meteor impact? Will they become too toxic for anyone after the fire? -- Toytoy (talk) 07:02, 19 May 2012 (UTC)[reply]

Despite Donna Summer's beliefs, most people don't worry too much about a comparatively low probability of harm from pollutants in city air (except possibly in Tokyo). A major earthquake or meteor impact might temporarily increase these pollutants, but it is likely that any single city so afflicted would be rebuilt within a few years, by which time nearly all air-borne pollutants would have been blown away (possibly to harm life on the rest of the planet, but the risk to any one individual is usually considered to be really small). Dbfirs 07:37, 19 May 2012 (UTC)[reply]

Plain old wood ashes and fumes are plenty toxic, when inhaled. People just don't think of it in those terms, when they say people died from "smoke inhalation". StuRat (talk) 22:11, 19 May 2012 (UTC)[reply]

While I know that Donna Summer didn't smoke, and died of lung cancer, she did work most of her life in an industry historically riddled with smokers. For her to blame the September 11 attacks is pretty non-scientific, but of course she wasn't a scientist. However, we haven't conducted many experiments on the longer term health impact of destroying several large, modern city buildings, and hopefully won't, so some speculation is valid. HiLo48 (talk) 22:40, 19 May 2012 (UTC)[reply]

I saw a clip a couple of days ago, of her in a recording studio, smoking a cigarette. So she definitely smoked. How much, I couldn't say. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:27, 20 May 2012 (UTC)[reply]

TMZ confirms she was a smoker.[19] This pic[20] looks like what I saw on TV. My guess is she was in denial. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:29, 20 May 2012 (UTC)[reply]

Hmmmmm. Our article on her says "...despite being a non-smoker, and the cancer was unrelated to smoking". Is this bullshit? HiLo48 (talk) 00:38, 20 May 2012 (UTC)[reply]

She was certainly a smoker at some point. I didn't record the coverage the other day, but I had the impression she had quit relatively recently. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:22, 20 May 2012 (UTC)[reply]

Species identification

Could I have help identifying this flower species (1, 2) and this monkey species? — Crisco 1492 (talk) 09:13, 19 May 2012 (UTC)[reply]

No. 2 looks like Bouganvillea spectabilis which illustrates our article. Richard Avery (talk) 19:45, 19 May 2012 (UTC)[reply]

I believe the monkey is a macaque, possibly a crab-eating macaque but there are lots of macaque species that all look generally similar. Looie496 (talk) 22:02, 19 May 2012 (UTC)[reply]

And number 1 is clearly a lily, and looks a lot like Lilium candidum, better known as the madonna lily, although that isn't native to Indonesia. Looie496 (talk) 22:09, 19 May 2012 (UTC)[reply]

Unidentified houseplant

Could anyone help identify this houseplant please? I suspect it of being some sort of geranium. DuncanHill (talk) 09:21, 19 May 2012 (UTC)[reply]

No it's a Begonia tiger. --TammyMoet (talk) 10:38, 19 May 2012 (UTC)[reply]

Of course it is, I should have known it's a Begonia! Many thanks :) DuncanHill (talk) 10:50, 19 May 2012 (UTC)[reply]

Geiger tube

Find the potential difference between the tube and the wire in a Geiger tube. Variables are $r,L,R,Q$ ; you can guess what they stand for. Doing these integrals is prone to mistakes all the time. This is what I have tried so far. We must find the potential difference between a ring segment of the tube and the corresponding segment of the wire, each of width $dL$ . This makes the ring segment carry charge $-{\frac {QdL}{L}}$ and the little wire segment carry charge ${\frac {QdL}{L}}$ . To find this, we must find the potential difference between a point on this ring segment and the little wire segment. By Gauss's law, the electric field due to the little wire segment a displacement ${\vec {x}}$ away from the wire is $k{\frac {QdL}{x^{2}L}}{\hat {x}}$ if $x>r$ and zero otherwise. This potential difference is therefore $-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}{\hat {x}}\cdot d{\vec {x}}=-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}dx=k\left({\frac {Q}{RL}}-{\frac {Q}{rL}}\right)dL$ . I am wondering if you have to do lots more integrals; is there an easier way perhaps? The version of Gauss's law I have been given is: the electric field in the tube is solely contributed by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. --150.203.114.37 (talk) 10:59, 19 May 2012 (UTC)[reply]

Do microwave ovens use less energy than stoves?

For comparable tasks, I mean. Suppose a 1400kw microwave versus an ordinary GE electric stove (I have no idea what it's kw rating is, but it plugs into a standard 220v outlet). I wonder if anyone can make a reasonable calculation of energy required in two cases:

1. Raising a 16 oz. mug of instant coffee + tapwater to drinking temperature, which takes 1 minute in my microwave, versus perhaps 5 minutes on the stovetop. (Actually I haven't heated coffee on the stovetop in the last twenty years, so I'm guessing at the 5-minute idea.)
2. Heating a small frozen dinner, 5 minutes in the microwave versus 30 minutes in a 350-degree oven, which takes 5 minutes to preheat.

Am I truly saving energy by using the microwave for these tasks, or not? A plain-English answer, please. Textorus (talk) 15:05, 19 May 2012 (UTC)[reply]

In general, yes, the microwave oven is more efficient, because you heat less air. However:

A) Excess heat isn't a bad thing in winter. That means that much less heat your home heating system needs to generate. If you have electrical home heating, this should cost about the same (unless you pay a different rate for that electricity). Of course, the reverse is true in winter, if you find yourself running the A/C more because of the excess heat generated by the oven and stove.

B) Old microwave ovens tend to lose efficiency.

C) You compared with an electric stove and oven. The calculations are different with gas, since it's usually about 1/3 the cost.

D) Certain things (large frozen items) are not well suited to microwaving, and you may need to nuke them for a very long time on the lowest setting, to allow the heat to conduct into the center. This lowers the microwave's efficiency, as the heat generated near the surface is also radiating out.

E) The efficiency of a stove is highly variable. If you have the heating element extending beyond the edges of the pot, and the pot is uncovered and boiling away, you are wasting lots of energy.

To find the answer in your specific case, you might want to buy an outlet monitor, which will show exactly how much electricity is used by an outlet. Of course, in your case, you'd need one that can handle 110V/220V, so you can compare the stove and microwave. And the oven/stove may be hardwired in or have an inaccessible outlet.

If so, you can try watching your house electricity meter. Note the rate of electricity usage before you turn each device on, and after. Then multiply this by the time needed. This isn't perfect, though, as a fridge or something might just happen to kick on or off at the same time. So, you might want to do this a few times.

Incidentally, I bet the cost of the electricity to heat the coffee will be trivial in either case, when compared to the cost of the coffee. This is because you are heating such a small amount of water. Compare this, say, with the amount of water heated for a bath or shower. StuRat (talk) 15:11, 19 May 2012 (UTC)[reply]

Thanks for these observations, Stu, but I'm not concerned with the cost (convenience is paramount in these situations), and I'm not going to buy an outlet monitor that I would never use again for anything. It's an academic question - I was thinking that someone who knows what must surely be a very simple equation or two could quickly calculate the amount of energy used in each case (e.g., 1 min. @ 1400kw/hr. = 23.33 kw, no? But how much for five minutes on a stove burner?). Textorus (talk) 15:39, 19 May 2012 (UTC)[reply]

Unfortunately, you won't have the inputs needed for such equations, like the insulation factor on the oven, amount of heat that escapes around the edge of your pot, etc. So, the best we can do, short of measuring those, or measuring electricity use by the devices directly, is to make generalizations. StuRat (talk) 15:45, 19 May 2012 (UTC)[reply]

Stu, if we were building an ultra-precise set-up for a rocket launch or a nuclear pile, we might want to consider those other variables. But it should be a very simple, straightforward calculation to determine how much energy is drawn from the household electrical system during the time the stove is turned on. That's what I'm asking, and I've given specific values for those times. Textorus (talk) 15:53, 19 May 2012 (UTC)[reply]

The answer is, obviously, to check the power consumption of your stove. It's probably printed on the stove's UL sticker. A typical GE electric range or cooktop, like this one, is rated around five kilowatts, or about 1500 watts per burner at full power. Total energy is almost perfectly approximated by "power times duration" because the coil is almost a perfect resistive load. Nimur (talk) 16:04, 19 May 2012 (UTC)[reply]

1500W×4 = 6kW. I take it the smaller burners are more like 1000W ? StuRat (talk) 16:29, 19 May 2012 (UTC)[reply]

I'm in the habit of avoiding false precision; as the exact stove model is unknown, the wattage is unknown, the number of burners is unknown, the altitude above sea-level is unknown, and the total time required to boil a "cup" of unknown volume (between 8 and 16 fluid ounces?) is also unknown, I'd avoid even using as much precision as that, StuRat. One significant figure and an approximate order of magnitude should suffice; I trust that our OP knows how to multiply, as long as they can find the relevant parameters. In the words of a pedigreed physicist, "approximately four burners times approximately 1500 watts each is approximately 5 kilowatts, which is approximately the safe level for a typical household kitchen circuit breaker in the USA." Incidentally, this fast-and-loose definition of "equality" is the difference between analysis and arithmetic, which is why they pay physicists more than accountants. (Or less. I can't recall; there may be a lost minus-sign in my figuring; that sort of arithmetic nuance is left as an exercise for the reader). Nimur (talk) 16:40, 19 May 2012 (UTC)[reply]

How much energy the oven and stove uses, is, of course, variable. So, your best bet is to look at the house electricity meter, as I suggested, to see how much is actually used when heating the given items. StuRat (talk) 16:00, 19 May 2012 (UTC)[reply]

There is a label of some kind inside the oven door, but it's long since become illegible from heat and cooking stains. I don't need a precise, down-to-the-last ion answer, just a general comparison between the two heating methods. Using standard ratings, such as in the link Nimur provided, would be sufficient. And no, I'm not going to go stand in the yard and guestimate how fast the little wheel in the electric meter is turning; that would be highly inexact for anyone but the meter reader, perhaps, and not worth the trouble. Textorus (talk) 17:03, 19 May 2012 (UTC)[reply]

OK, then, without having any idea how much power your stove or oven uses, the best we can do is go with the general trend and say that it uses more than the microwave. We could even guesstimate it at twice as much. Good enough ? StuRat (talk) 17:17, 19 May 2012 (UTC)[reply]

Yes, possibly rather more than twice. "1 min. @ 1400kw/hr. = 23.33 kw" should read "1 min. @ 1400watts. ≈ 23 watt-hours" (or 84000 joules) and a 1.5Kw stove ring for 5 minutes would use 5 min. @ 1500 watts ≈ 125 watt-hours (or 450000 joules) so perhaps up to six times as much (though with the benefit of room heating mentioned above). Dbfirs 07:09, 20 May 2012 (UTC)[reply]

Specific entropy

Is it possible to express specific entropy in terms of the gas constants cv, cp and pressures p1 and p2 only? I can't derive anything close to this without using v1 and v2. The closest I've got is s2-s1=cvln(p2/p1)+cpln(v2/v1) but i don't want the vs in there. This is an is isentropic process so were saying PV^n is constant. 94.116.0.41 (talk) 15:17, 19 May 2012 (UTC)[reply]

I haven't done chemistry in years. I remember PV/nT(1) = PV/nT(2). Pressure, Volume, molecule count (not moles) and temperature. I think you can remove any variable from each side, so P/nt(1) = P/nT(2). I don't know if this helps.--Canoe1967 (talk) 19:07, 19 May 2012 (UTC)[reply]

Why isn't dinitrogen toxic?

Since dinitrogen, the acetylide dianion, and the nitrosonium cation are isoelectronic with carbon monoxide and the cyanide ion, why aren't they as deadly poisonous as carbon monoxide and the cyanide anion? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:24, 19 May 2012 (UTC)[reply]

Different different elements have different electronegativity, formal charge, HSAB interactions, bond strength, intrinsic stability/transportability to the biochemical target, etc. I think even your own hypothesis flawed, assuming that CO and CN- are "deadly poisons" by the same biochemical pathway. DMacks (talk) 17:18, 19 May 2012 (UTC)[reply]

They are deadly poisonous in the same way—they bind to iron in hemes to the exclusion of oxygen. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:03, 20 May 2012 (UTC)[reply]

Cyanide poisoning is not caused by reaction on bloodborn hemoglobin or exclusion of oxygen transfer. Carbon monoxide is not instantly deadly down to the cellular level even at low concentrations. DMacks (talk) 02:49, 20 May 2012 (UTC)[reply]

Is this similar to iron, cobalt and nickel being the only three normally magnetic elements but stainless steel made from iron, cobalt and nickel is not normally magnetic?--Canoe1967 (talk) 19:12, 19 May 2012 (UTC)[reply]

One other thing to think about: Why would life have evolved to be incompatable with dinitrogen, which has always been the predominant gas in the atmosphere? If you reach a possible conclusion that is incompatable with easily observable facts, then your conclusion is obviously wrong, in other words, since all life has been bathed in dinitrogen for billions of years, there's no need to assume it would be toxic in any way. Any line of thinking that leads you down that road isn't very productive, so your assumptions must be horribly flawed. --Jayron32 20:28, 19 May 2012 (UTC)[reply]

Because the lone pair on the carbon of CO is not as tightly bound as those on dinitrogen, CO is the stronger Lewis base. The same goes for the cyanide anion. Nitrosonium's lone pairs are even more tightly bound than dinitrogen's. Therefore, CO and cyanide coordinate much more easily to the iron than dinitrogen or nitrosonium. This is my best explanation of the toxicity of these isoelectronic species.--Jasper Deng (talk) 02:58, 20 May 2012 (UTC)[reply]

But then why isn't the acetylide dianion even more toxic than either CO or cyanide? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:09, 20 May 2012 (UTC)[reply]

As I noted before, how were you planning to get that dianion to the heme–Fe target? DMacks (talk) 03:12, 20 May 2012 (UTC)[reply]

How do you think? It's obvious, isn't it? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:16, 20 May 2012 (UTC)[reply]

It's obvious that you did not actually think about that. DMacks (talk) 03:28, 20 May 2012 (UTC)[reply]

Two reasons:

Acetylide cannot exist in solution. It would immediately hydrolize to acytelene and hydroxide ion: C₂^2-(aq) + 2H₂O(l)→ 2 OH^-(aq) + C₂H₂(g).
Even if it could exist in solution, the lone pair is more tightly bound on acetylide than CO because in CO, the bond is polarized toward O, lessening C's grip on its electrons, which is not the case with acetylide.--Jasper Deng (talk) 03:31, 20 May 2012 (UTC)[reply]

That depends on whether the iron atom is a stronger Lewis acid than the proton or vice versa.
Non sequitur from what you said earlier about that. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:39, 20 May 2012 (UTC)[reply]

The proton is stronger by a lot. I know this for a fact for that particular anion; it would hydrolyze before coming anywhere near the iron.--Jasper Deng (talk) 03:42, 20 May 2012 (UTC)[reply]

Dasyuromorphia and carnivores

Since Dasyuromorphia are carnivores, why there are ranked as separate order and not suborder of Carnivora? Also, what hinders thylacine from being placed in Canidae or Caniformia?--176.241.247.17 (talk) 22:27, 19 May 2012 (UTC)[reply]

For the first part, Carnivora are placental carnivores, and the Dasyuromorphia are marsupials. Now, as to why they don't have an order of marsupial carnivores, I do not know. StuRat (talk) 22:31, 19 May 2012 (UTC)[reply]

The simple answer is that the order Carnivora is not inclusive of all carnivores. Being carnivorous does not put one, taxonomically, in the order Carnivora. As for your second question, again, marsupials are quite separate from the other carnivorous orders and quite different, no matter what type of taxonomical scheme you are using. Their apparent visual and behavioral similarity is an example of parallel evolution and does not indicate that they are closely related. --Mr.98 (talk) 23:05, 19 May 2012 (UTC)[reply]

Problems with xenon tetroxide xenate formulas

Didn't the xenic acid article say that there is no such thing as a completely de-protonated xenate salt?--Jasper Deng (talk) 23:22, 19 May 2012 (UTC)[reply]

Xenate. Not perxenate. Completely deprotonated perxenates are known—and xenon tetroxide is perxenic anhydride, not xenic anhydride. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:06, 20 May 2012 (UTC)[reply]

Yeah, but the xenon tetroxide article makes mention of XeO₄^{2-, which is a completely deprotonated xenate. See the synthesis section.--Jasper Deng (talk) 01:16, 20 May 2012 (UTC)[reply]
That has been fixed. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:33, 20 May 2012 (UTC)[reply]
My only concern is that I'm not too certain of the products of that reaction.--Jasper Deng (talk) 01:36, 20 May 2012 (UTC)[reply]}

May 20

Locomotives

There are 0-4-0s, 2-2-0s, and 0-2-2s. Why aren't there 0-2-0s? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:57, 20 May 2012 (UTC)[reply]

In France an 0-4-0 would be an 020. Sticking to American notation, an 0-2-0 would have a single axle, which seems unsatisfactory, rather like a car with one axle. Acroterion (talk) 02:03, 20 May 2012 (UTC)[reply]

Although, strangely, there have been 0-3-0s. Gandalf61 (talk) 08:51, 20 May 2012 (UTC)[reply]

Geiger tube continuation

I calculated the potential difference between the wire and the tube of a Geiger tube to be ${\frac {1}{2\pi \epsilon _{0}}}{\frac {Q}{L}}\ln {\frac {R}{r}}$ approximately. The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. This means that the air breakdown potential has been reached (and they provide a value for this and for all the other variables). Calculate this power-supply voltage. How do you do this calculation? Do you just substitute the numbers into the above equation or is it more involved than that? --150.203.114.37 (talk) 04:54, 20 May 2012 (UTC

It would help if you told us what L, Q, R, & r signify. Operating a geiger tube at a voltage high enough to cause a gas discharge (as indicated by a glow) is harmful to it. Normally, you just set the voltage to the value stipulated by the gieger tube manufacturer. This is at a "plateau" of maximum sensitivity that occures just below the glow point. Ratbone60.230.203.253 (talk) 07:23, 20 May 2012 (UTC)[reply]

Maybe; I didn't actually observe this; it's just a theoretical problem. Q = charge on the wire, R = radius of the tube, r = radius of the wire, L = length of the wire/tube. --150.203.114.37 (talk) 07:30, 20 May 2012 (UTC)[reply]

Then the formula given is the relation between the potential difference and charge as for any coaxial pair of conductors separated by a perfect insulator of permitivity e. It is a concept useful in calculating the capacitance http://en.wikipedia.org/wiki/Capacitance and has no relavence whatsowver to the operating voltage of a gieger tube. Under operating conditions,gieger tubes are operated with DC voltage, and no current flows in the tube capacitance, except for the very brief recharging current after each detected particle. Note that the term e_o, strictly speaking, being the symbol for the permitivity of free space, is in any case incorrect, it should be the permitivity of the gas used (e₀.k, which will however be sensibly close to e₀). Keit120.145.31.247 (talk) 09:22, 20 May 2012 (UTC)[reply]

Domestic Heat Pump connections

I am planning to install an air to water domestic heat pump for heating and am looking at the manuals. The 12 kW heat pump works best when there is a 35 litres/minute flow through it (it then heats the water by about 5C each sweep through the exchanger). If the flow rate falls below this, the manual says it becomes inefficient and cuts in and out. The circuit I wish to heat is already partly structural to the house and would not take that flow rate (indeed I am guessing even approaching that flow rate would be rather noisy), but the heat loss on that circuit is more than 5C. So is there any reason I cannot just "short" part of the outflow from the heat pump directly into the inflow for the heat pump mixing with the return? It seems a little too obvious as a solution...obviously the later part of the heating circuit will be a little cooler from having a high temp drop across them (thats ok, the later parts of the circuit will be the larger surface area ones), but if the heat pump is happy I know the actual heat is being transferred into the house a reasonable efficiency, right? --BozMo talk 07:22, 20 May 2012 (UTC)[reply]

@@ Line 323: / Line 323: @@
 :It would help if you told us what L, Q, R, & r signify.  Operating a geiger tube at a voltage high enough to cause a gas discharge (as indicated by a glow) is harmful to it.  Normally, you just set the voltage to the value stipulated by the gieger tube manufacturer. This is at a "plateau" of maximum sensitivity that occures just below the glow point. Ratbone[[Special:Contributions/60.230.203.253|60.230.203.253]] ([[User talk:60.230.203.253|talk]]) 07:23, 20 May 2012 (UTC)
 ::Maybe; I didn't actually observe this; it's just a theoretical problem. Q = charge on the wire, R = radius of the tube, r = radius of the wire, L = length of the wire/tube. --[[Special:Contributions/150.203.114.37|150.203.114.37]] ([[User talk:150.203.114.37|talk]]) 07:30, 20 May 2012 (UTC)
-::Then the formula given is the relation between the potential difference and charge as for any coaxial pair of conductors separated by a perfect insulator of permitivity e.  It is a concept useful in calculating the capacitance http://en.wikipedia.org/wiki/Capacitance and has no relavence whatsowver to the operating voltage of a gieger tube.  Under operating conditions,gieger tubes are operated with DC voltage, and no current flows in the tube capacitance, except for the very brief recharging current after each detected particle.  Note that the term e<sub>o</sub>, strictly speaking, being the symbol for the permitivity of free space, is in any case incorrect, it should be the permitivity of the gas used (e<sub>0</sub>.k, which will however be sensibly close to e<sub>0</sub>).  Keit[[Special:Contributions/120.145.31.247|120.145.31.247]] ([[User talk:120.145.31.247|talk]]) 09:22, 20 May 2012 (UTC)
+:::Then the formula given is the relation between the potential difference and charge as for any coaxial pair of conductors separated by a perfect insulator of permitivity e.  It is a concept useful in calculating the capacitance http://en.wikipedia.org/wiki/Capacitance and has no relavence whatsowver to the operating voltage of a gieger tube.  Under operating conditions,gieger tubes are operated with DC voltage, and no current flows in the tube capacitance, except for the very brief recharging current after each detected particle.  Note that the term e<sub>o</sub>, strictly speaking, being the symbol for the permitivity of free space, is in any case incorrect, it should be the permitivity of the gas used (e<sub>0</sub>.k, which will however be sensibly close to e<sub>0</sub>).  Keit[[Special:Contributions/120.145.31.247|120.145.31.247]] ([[User talk:120.145.31.247|talk]]) 09:22, 20 May 2012 (UTC)
 ==Domestic Heat Pump connections==