:Is this it [http://guweb2.gonzaga.edu/faculty/cronk/chemistry/CHEM445/lectures.cfm?L=12] ? (full scheme)
:Is this it [http://guweb2.gonzaga.edu/faculty/cronk/chemistry/CHEM445/lectures.cfm?L=12] ? (full scheme)
:<small>You need to be familiar with "electrophilic attack on enolates" eg [http://www.chem.ucla.edu/harding/notes/notes_14D_enolates.pdf] http://www.google.co.uk/search?hl=en&q=enolates+reaction&btnG=Search&meta=</small>
:<small>You need to be familiar with "electrophilic attack on enolates" eg [http://www.chem.ucla.edu/harding/notes/notes_14D_enolates.pdf] http://www.google.co.uk/search?hl=en&q=enolates+reaction&btnG=Search&meta=</small>
:<small>There's a useful acetone/'acetone enol form' electron density map here http://www.chem.ucalgary.ca/courses/350/Carey/Ch18/ch18-1.html which shows why enols can react with electrophiles at carbon</small>
:This '''[http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=stryer.figgrp.2793]''' is a far better visualisation of the critical step, showing the role of Mg2+, and the mechanism too. (Don't even ask me how they worked that out - I haven't got a clue - probably isotope labling experiments or something).
:This '''[http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=stryer.figgrp.2793]''' is a far better visualisation of the critical step, showing the role of Mg2+, and the mechanism too. (Don't even ask me how they worked that out - I haven't got a clue - probably isotope labling experiments or something).
:(Note that the [[enol]] based attack isn't as simple as usual since there is a OH next to the keto - making the intermediate structure ''approximately'' a 1,2 dihydroxy alkene... see that technically the reaction forms a bond between two carbonyl carbon atoms!!)
:(Note that the [[enol]] based attack isn't as simple as usual since there is a OH next to the keto - making the intermediate structure ''approximately'' a 1,2 dihydroxy alkene... see that technically the reaction forms a bond between two carbonyl carbon atoms!!)
Revision as of 12:39, 12 September 2009
Welcome to the science section of the Wikipedia reference desk.
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.
I've noticed that occasionally there are lots of dead flies on my outdoors plants (I've noticed them on basil, oregano, and thyme in particular). I occasionally just found one and thought, "well, I guess this just happens to be where he kicked the bucket," but the other day I found dozens of them. They look like normal houseflies, more or less, and are just sitting on the plants, clinging to them, dead. They aren't wrapped in a web or anything so obvious. What causes this? Is this some sort of fly life-cycle thing? Or some odd predator? Or what? We don't use pesticides on the plants. Any thoughts? Imagine a bunch of flies just sitting on a plant. Now imagine they happen to be dead. That's what it looks like. --98.217.14.211 (talk) 00:38, 5 September 2009 (UTC)[reply]
Houseflies and blowflies do not feed on green parts of live plants, but they may feed on the honeydew secretions of aphids if those are present on the plants. There may be some kind of pathogen (virus, bacterium, fungus, parasitic worm eggs, whatever) present in the honeydew and infecting the flies. I've never seen such a thing, so I have no good guesses as to what it may be. If it is a fungus, you would see fruiting bodies emerging from the dead flies. Otherwise, you'd just see dead flies. Sorry. --Dr Dima (talk) 02:59, 5 September 2009 (UTC)[reply]
"Actually happened" means nothing here, you're just asking about time zones, which are all about your reference point. From our article, "At 02:39 UTC on Monday July 21 (10:39pm EDT, Sunday July 20), 1969, Armstrong opened the hatch, and at 02:51 UTC began his descent to the Moon's surface." As always, it was a different day somewhere in the world... --98.217.14.211 (talk) 00:54, 5 September 2009 (UTC)[reply]
OK, no one's picked up on this thread for more than an hour, and a Google search for "there is no time in space" shows a shocking number of people who appear to be serious, so I guess I'd better spoil my own joke here. It's a line from a very famous episode of The Twilight Zone, the one where the aliens' guiding book, To Serve Man, turns out to be a cookbook. --Trovatore (talk) 02:22, 5 September 2009 (UTC)[reply]
As to the original question, if it is July 20 in the US, it can be July 20 or 21 in Australia or New Zealand, but not July 19. Eastern Standard Time in the US, for example, is 15 hours behind Eastern Standard Time in Australia. In all Australian and New Zealand time zones both the landing and the moon walk took place on July 21. --Anonymous, 05:22 UTC, September 5, 2009 (it's September 4 in western US time zones, but September 5 in Australia and New Zealand).
The accumulation of helium in the core causes a gradual increase in the rate of fusion and gravitational self-compression, as helium is denser than hydrogen. Higher temperatures must be attained to resist this increase in gravitational compression and to maintain a steady state.
It makes sense - to get two hydrogen nuclei to fuse you have to overcome their mutual repulsion. In the sun, it's gravity & pressure that does that - so there is some region in which the suns gravity is strong enough and some other region where it's not. As the sun transforms hydrogen into denser helium - it gets gradually denser - the region throughout which hydrogen fusion can occur gets bigger - so more hydrogen fuses and you get more energy. I don't have a reference - but it sure makes good sense. SteveBaker (talk) 02:46, 5 September 2009 (UTC)[reply]
It's not just that there is more volume that can fuse, the part of that volume that was fusing before is now fusing faster because it is hotter/denser. --Tango (talk) 10:42, 5 September 2009 (UTC)[reply]
Here's a response I got from Yahoo Answers: "As the core of the sun converts hydrogen to helium, it becomes more hydrogen-depleted. This forces the core to contract and increase pressure in order to fuse enough hydrogen to counteract gravitational collapse. As it contracts, however, the density gets higher, increasing the gravitational pressure; so the core must also get hotter by fusing even more hydrogen than before in order to maintain the equilibrium. So the sun gets hotter over time. As you say, it is 30% hotter now then when the earth first formed." --Halcatalyst (talk) 14:14, 5 September 2009 (UTC)[reply]
It's incomprehensible! Why (to pick just one part of this so-called explanations) does becoming more hydrogen-depleted force the core to contract? We're told it does that in order to fuse enough hydrogen to counteract gravitational collapse! Why? Is it sentient or something? That's not an explanation - it's a mess. SteveBaker (talk) 18:47, 5 September 2009 (UTC)[reply]
It is pretty common to explain things using anthropomorphic language. The only problem with the explanation is that it isn't very detailed, but that isn't necessarily a bad thing - you have to get the level of detail right for your intended audience. The OP probably doesn't know much astrophysics and probably doesn't want to read the equivalent of a textbook on the subject, so you need to miss a few things out. If you want a slightly more complete version of the sequence of events (they are actually all happening at once, of course), here goes: The hydrogen gets depleted, which reduces the amount of fusion happening, that reduces the radiation pressure so gravity is stronger and the star contracts. As with any gas, it heats up as it contracts. That extra heat and density increases the amount of fusion taking place, increasing the radiation pressure and bringing the star into a new equilibrium. This new equilibrium position is hotter and, therefore, brighter. --Tango (talk) 19:02, 5 September 2009 (UTC)[reply]
It doesn't seem right to me intuitively that a reduction in the reaction rate would lead to a higher equilibrium reaction rate, although I guess it's not impossible. I would think the increase in the molar mass is mostly what drives the volume down and the temperature up. Rckrone (talk) 05:50, 6 September 2009 (UTC)[reply]
My question is this: how can the atomic mass of oxygen-16 be 15.9949146 u when both a proton and a neutron have masses greater than 1 u? Shouldn't the mass be greater than 16 u? Nkot (talk) 01:56, 5 September 2009 (UTC)[reply]
Black hoes can evaporate due to hawking radiation...so would it be possible for a particle to pass the event horizion but have the black hoe evaporate before the particle is absorbed into the singularity? like the particle gets sucked past the "point of no return" but the black hoe evaporates right after?
Also, if you have two black hoes close enough to each other such that their event horizons overlap at a single point...(perhaps the black hoes could be orbiting one another, so they dont fall into each other), what would happen to a particle that traverses a path where the gravitational forces are neutralized by the opposing black hoe--that is the black hoes are possitioned such that their event horizions are both tanget ( at the same point) as the path of the particle. What would happen to the particle, since it is effectively has passed the event horizions of both black hoes? XM (talk) 04:54, 5 September 2009 (UTC)[reply]
In my experience, hoes don't evaporate, absorb particles, or neutralize gravity, regardless of their color. They also don't have event horizons. Or did you mean black ho's? If you did, I have to tell you that they don't do any of these things either. :-D 98.234.126.251 (talk) 05:37, 5 September 2009 (UTC)[reply]
At first I thought XM's L key didn't work, but then it would have been back hoe. The question seems legitimate aside from the... typo?... so I guess I'll answer it. The short answer to all of the questions is that anything that passes the event horizon will hit the singularity because that's how the event horizon is defined. If you escape then by definition you didn't cross the event horizon. So the real question is whether an evaporating hole has an event horizon at all, and the answer to that seems to be yes, at least if you believe the Penrose diagram that was in Hawking's original paper (reproduced here). Hawking himself seems to have decided that he was wrong and there is no horizon or singularity really, but that view doesn't seem to be very popular right now. If the event horizons of two black holes overlap then you really have one black hole with one event horizon, and the no-hair theorem implies that it will quickly become spherical. There's only one singularity so there's no ambiguity about where the particle ends up. -- BenRG (talk) 18:45, 5 September 2009 (UTC)[reply]
Forget about the amusing if racist typographical error! This user has asked a question that is so profound I wonder why I didn't think of it myself! Because if the hole (presumably a very small one) evaporated quickly enough, the particle would have "seen inside" the black hole and then "escaped" as the event horizon moved past it, and this would violate "cosmic censorship" which states that not even information about the inside of the black hole can escape-some physicist think this is just as well since new laws of physics might enter our universe should a naked singularity ever occur.
My answer is that since the particle would effectively be moving near the speed of light (or above(?)) when inside the event horizon, and that the horizon shrinks according to how much hawking radiation it can shed, then there is no reason to suppose that the horizon can catch up with the particle. I could be wrong-can someone do the maths?80.2.195.218 (talk) 13:24, 10 September 2009 (UTC)[reply]
Hey, that's a fascinating possibility -- that information could actually escape from a shrinking black hole. Unfortunately I can't crunch the numbers on this -- my area of expertise is hydrocarbon/petroleum chemistry, not astrophysics -- but it does make sense intuitively that if a black hole is shrinking, then matter/information could escape from it. Note also that the "cosmic censorship" hypothesis has so far only been definitively proved for non-shrinking black holes. (I wonder if Hawking is logged in right now, maybe he could crunch the numbers for you, he's a genius at this kind of stuff.) Well, clear skies to you! 146.74.230.106 (talk) 21:29, 10 September 2009 (UTC)[reply]
leaving groups, chemical attacks, MOs, and stuff.
Sometimes I get the feeling that in some electron movement mechanisms, incoming electron density "pushes" existing electron density in the other direction ... so say in esterification an alkoxy oxygen's lone pair is "pushing" away the pi electrons onto the carbonyl oxygen, and the pi electrons happen to be mobile. So in fact, before that carbon converts to sp3, it's like the alkoxy oxygen's lone pair is participating in the pi system and pushing the existing bonding electrons away. (Okay this might be an unacceptable classical picture though.)
It was mentioned to me in passing half a year ago how nucleophile electrons attack antibonding orbitals ... and in fact I had totally forgotten about antibonding orbitals (as a mechanism for nucleophilic attack) until it was mentioned again. I guess the classical picture is kind of misleading, because the typical picture of SN2 seems to be of say, the better nucleophile "pushing" away the leaving group through backside attack, forming and breaking bonds simultaneously that way. Do those nucleophile electrons in fact, contribute to an antibonding orbital, that somehow negates the bonding between the LG and the carbon? If so, how does an antibonding orbital convert into a bonding orbital (presumably that as the LG leaves a bond must be formed simultaneously)? Is bond breaking and formation truly simultaneous? Will one aspect be slightly ahead?
Another thing that doesn't get discussed much is sp2-sp3 (and sp2->sp3) transitions ... it almost seems to be a factor (if you could somehow get stuff to happen during the transition, and how much it affects equilibrium. It seems to me that possibly in such transitions you could have stereochemistry-affecting effects. What does the hybridisation of the two central carbons look like when the bromonium ion is attackihg an alkene for instance? 1/2 sp3, 1/2sp2?
Also, is acid catalysis always a first order reaction? It seems to me that for example, in acid-catalysed esterification of carboxylic acids where the C=O bond is being restored and kicking out the LG, that the electron density (coming out from oxygen's previous lone pairs) pushes electron density away from the C-OH bond, and meanwhile, this also pushes electron density from the OH (part of COOH) onto a nearby "proton". That is, I'm thinking you don't have to wait for acid to protonate the leaving group, and THEN wait for bond displacement to occur -- acidic protons act like terminal electron acceptors that stabilise the LG. Which kind of mechanism would acid catalysis support? (I'm referring to protonation, such as OH protonation, to create a good LG, not protonation to increase the carbonyl carbon's electrophilicity.) John Riemann Soong (talk) 07:44, 5 September 2009 (UTC)[reply]
First of all I'm going to recommend another book if you haven't already read it, it's "Mechanism and Theory in Organic Chemistry" (Thomas H. Lowry, Kathleen Schueller Richardson) which covers the first three paragraphs of your question in very gory details.
also Transition state - should be the best search term - eg for google books.
First paragraph - the alkoxy (why is it not alchoxy, alchemy, alchohol, k?emistry) O lone pair does just what you say - interacting with the pi system - specifically the C p orbitals. For RC(=O)OR' + R''O- the transition state is close to sp2 C bonded to O- , R , R' with the R''O bonding via the p (pi) orbital.
I don't think your use of the term 'antibonding orbital' is right in all the contexts you use it specifically the ester example Doh. It is right, silly me (deleted stuff below)
Second paragraph:
In a pure Sn2 reaction bond breaking and formation is simultaneous, however in practice if one of the leaving groups is better than the other, asymmetry may develop.
Both views are right. ('classical' and molecular orbital - they're describing the same thing, the same 'movement' of electrons (or electron density) )
Yes - the nucleophile feeds electrons into the leaving groups antibonding orbital - weakening the bond. As the bond is weakened the leaving group will move away from the molecule (electron repulsion).
Curiously the antibonding orbtital for the leaving group is the bonding orbital for the attacking nucleophile (and vice versa) - it's like a see-saw.
The antibonding orbital for the leaving group is the mirror image of the bonding orbital, this might need a little more explanation if you're not familiar with it, both can be anti, or bonding orbitals depending on which side a group is attached.. In the absense of groups on either side both orbitals can be considered plain p orbitals of the same energy, did that make any sense?
More: once a leaving group has left (in Sn2) the bonding orbital it was using, becomes the new antibonding orbital - it relates to the mirror image thing.
Third paragraph:
Roughly yes, someway between sp2 and sp3. For the stereochemistry the vaguely similar epoxide ring opening reactions are a good source of data (epoxides being more stable)
Fourth paragraph - do you mean that reaction of the intermediate in esterification is a concerted reaction in which the OH leaves as it is protonated (1 step) rather than being protonated then leaving (2 steps) - this is quite likely (though depends on absolute conditions)
Yeah, that's what I meant. Basically I'm wondering whether this will allow protonation to occur more often, even though the pKa of a protonated alcohol is kinda low. I wonder this cuz I'm interested in improving the efficiency of longer-term esterification, in weakly or moderately acidic solutions [with acetic or phosphoric acid for example] that must be edible later on. Thus, if I had a pH of 2.7 (or say I somehow got my hand on food-grade phosphoric acid and lowered it to 2 or something), would protonation of the OH group occur more often than the general 2-step depiction would indicate? That is, as the carbonyl's pi system electrons are entering the antibonding orbital (is my conception correct? sp2 bond is being reformed though) of the C-OH bond, this makes the OH group more basic ... making it a terrible leaving group, of course. But wait! There's a hydronium ion nearby (as well as various other protons on HOH's) which basically neutralise the OH group's basicity as it forms, in a sort of concerted reaction. John Riemann Soong (talk) 15:35, 5 September 2009 (UTC)[reply]
Basically yes. (at least there's nothing above that is obviously wrong)
To expand a bit there's a good example of trans-esterification where a chemical buffer (acid base) mixture is used (in dehydrating conditions of course) - in an even more concerted reaction the acid part of the buffer "stands by" to protonate the leaving group (-OR) as it leaves, and the basic part of the buffer "stands by" to deprotonate the attacking alcohol as it attaches to the carbonyl - higher concentrations of buffer help here - but the reaction goes suprisingly well for what may be an effectively neutral solution. Note how both the presence of the acid and base help here. This helps to reduce the activation energy - though the chances of getting all the molecules in the right place mean the reaction is quite slow nevertheless (High concentrations of buffer help). The same method can be used for straight esterifiaction.
The only tricky with the above method is choosing weak acids and bases that will not get involved in the esterification reaction (ie not acetate, or boric acid) Triethyl amine (does not react with carbonyl permanently) is a good choice for the base. Have you heard of tertiary amine carbonium ions? see
R3N+C(=O)R'
They're useful (and often overlooked) intermediates in esterification reactions, especially when you want to avoid any strong acid - you can prepare them before hand from the amine and acid chloride in suitable conditions.
I would like to learn a bit more about the symbols used in this type of molecular diagram, example In this particular example, the wriggly line is the symbol that I don't know about. There surely is an article on wikipedia, but I can't locate it. Can somebody please supply a link? Thanks! --TrogWoolley (talk) 10:24, 5 September 2009 (UTC)[reply]
It relates to the stereochemistry, the atom attactched to squigly line can be in one of two positions (up or down with respect to the paper) - it's a variation of the Natta projection
Here's an example with the two types ('forward', and 'back') - they are the triangular line , and the dotted lines, if a squicgly line was used it would mean that it could be either.
Talking of such things - if you see a percentage above or near the squiggle especially something like "40% R" , or "30% S" this gives the ratio of the two enantiomers - labled "R", and "S" (the R and S should be explained somewhere in one of the linked articles)83.100.250.79 (talk) 19:06, 5 September 2009 (UTC)[reply]
I'll tell you what. Give yourself a half-inch cut on your arm and pour Listerine into it. Then cut your other arm and pour salt water into it. Tell us which one hurts the most! --TammyMoet (talk) 15:04, 5 September 2009 (UTC)[reply]
Even the effect of Listerine in the mouth is only temporary. Saltwater as a mouthwash may not be as potent as Listerine, but it will make your mouth and throat feel better for a little while, certainly better than plain water will. Baseball BugsWhat's up, Doc?carrots16:57, 5 September 2009 (UTC)[reply]
Salt water has applications to oral hygiene other than germ-killing. The salt water interacts with the gums and keeps them in an isotonic equilibrium, minimizing fluid loss. Listerine, or other antibacterial fluid, probably does not satisfy this property. Nimur (talk) 17:31, 5 September 2009 (UTC)[reply]
I think a salt water mouthwash would help kill bacteria. In the past food was preserved by salting. I imagine that salt shrivels up the germs by osmosis. It may, I guess, be less harmful to the healing socket wound than the more aggressive chemicals in mouthwash. I think, for example, that when you use an antiseptic cream on a wound, that it kells off a layer or two of healthy cells. 78.146.76.67 (talk) 11:44, 6 September 2009 (UTC)[reply]
Eating a packet of Ready Salted doesn't sterilise my mouth; is there any real evidence that a saltwater mouthwash would? Especially since the salt will be gone within minutes (unlike preserved food, which remained salted)? I'm not really convinced. Vimescarrot (talk) 17:11, 6 September 2009 (UTC)[reply]
Be sure not to use salt substitute which is potassium chloride instead of sodium chloride. Potassium chloride will damage cells and keep the wound from healing. DISCLAIMER: This is not medical advice. Please consult your doctor to obtain medical advice. --Taxa (talk) 17:14, 6 September 2009 (UTC)[reply]
Also, mixing potassium chloride with sugar and giving it a tap with a hammer can cause fire, so if you do use salt substitute, make sure you clean your mouth out before you eat anything with sugar in it. I think it was potassium chloride, anyway. The stuff you get in water purification tablets.....? --KageTora - (영호 (影虎)) (talk) 11:52, 7 September 2009 (UTC)[reply]
Saltwater rinse recommendations are empirical -- they are not shown to reduce bacterial load any more than tap water, and especially not in the fraction of a minute that rinses are typically used. Even chlorhexidine requires a concentration of 2% and 10 minutes to kill some of the more terrible periodontal pathogens. DRosenbach(Talk | Contribs)01:34, 8 September 2009 (UTC)[reply]
Please do not write in block capitals, it is harder to read and is interpreted as shouting. I'm afraid I do not understand the question. "Free space" is an idealised concept, it doesn't actually exist in reality. --Tango (talk) 13:46, 5 September 2009 (UTC)[reply]
I believe your question should have been worded as 'why is it that when an object is in orbit, why is it in a permanent state of freefall?'. The reason is that the Earth's gravitational pull is dragging on the object bit-by-bit, so that every time the object goes around the Earth it gets just that little bit lower. This is why the Shuttle et al., have to compensate every now and then by firing off their booster rockets to get it back into position. There is no such thing as a permanent orbit. Either something starts coming back down or it starts floating away. Even the moon is getting further and further away from us. --KageTora - (영호 (影虎)) (talk) 14:10, 5 September 2009 (UTC)[reply]
I am sorry, but that is completely wrong. The need to boost is because of atmospheric drag. If you are in a high enough orbit that there is no significant atmosphere, orbits are essentially permanent. The Earth's orbit around the Sun, for instance, will last essentially unchanged until the Sun dies. The Moon is moving away because of tidal forces, but it will never leave orbit entirely. Orbits work because the gravitational pull is always towards the centre of the body and the object orbiting is moving perpendicular to that direction. This means gravity changes the direction of the motion without changing the speed, so it ends up going around the body at constant speed. (This is for a circular orbit, a general elliptical orbit is a little more complicated, but the basis concept is the same.) --Tango (talk) 14:18, 5 September 2009 (UTC)[reply]
You may be making the common mistake of thinking that astronauts are in microgravity because they're out in space, which is 90% wrong: they would be subject to about 90% of the amount of gravity as they are on the ground if they were just standing still at their usual altitude. The reason they're in microgravity is because they're in free fall, as above. Newton's cannonball is an intuitive way to understand that astronauts are continually falling. --Sean15:15, 5 September 2009 (UTC)[reply]
In fact, when it's time to return to the ground, the spacecraft fires a rocket tangent to the orbit (adding no momentum towards the earth). After losing some of the angular momentum, the spacecraft just "falls" back to the ground, indicating that it is still completely under the control of Earth's gravity. Nimur (talk) 17:36, 5 September 2009 (UTC)[reply]
I'll offer some further explanation - astronauts appear weightless because they are "falling" at exactly the same rate as their spacecraft, so relative to each other they don't move at all, hence appearing to float. This is because all objects in a given gravitational field accelerate the same amount. The gravitational force is proportional to mass, but acceleration is given by F=ma (which implies a=F/m). That means the mass cancels out when you calculate acceleration, meaning acceleration does not depend on mass. --Tango (talk) 15:31, 5 September 2009 (UTC)[reply]
To be perfectly clear, "Free fall" does not necessarily imply that the spacecraft is plummeting towards earth. Plummeting is only one kind of free fall. If your "sideways" speed is high enough you can "fall" for ever without ever getting any closer to the ground. (You can sort of visualize this as falling towards the earth, but going sideways enough that you miss it entirely. ) APL (talk) 04:15, 6 September 2009 (UTC)[reply]
Houseflies aren't silent. They're quite noisy. But given that their flying speed is almost the same, any difference in their loudness would probably be attributed to their larger wingspan. Dougcard (talk) 04:14, 6 September 2009 (UTC)[reply]
There may be more to it than just size... It's true that a hornet is much more noisy than a wasp or a bee. On the other hand, the average fly is much larger--but in many cases more silent--than a typical mosquito. To confuse things further, mosquitoes (and probably many if not most flying insects) can apparently operate in more than one flight mode, some of which may be more silent or noisy than others. Just think of a helicopter that generates different noise levels while moving or in hovering position. Also, many sources of noise, in particular from oscillating or rotating objects are quite directional, so the perception of noise may depend on the relative position of the observer with respect to the emitter. Last but not least, we may be more sensitive to noises that are perceived as threatening (fear of mosquito bite) than to those that are considered benign or inconsequential. Michel M Verstraete (talk) 20:32, 9 September 2009 (UTC).[reply]
Maybe - key factors here are the sampling frequency, and possibly you will need a separate comparator.
The power supply will as standard match a single DC voltage, but for audio you want it to match an AC voltage.
What the thing does is monitor the output voltage, and compare it to a standard, if the voltage is less than the standard then the supply is switched on. What you need to do is disconnect the wires that compare the output sample the with the standard, and connect it so that the output sample is compared with an audio input (suitably scaled) - this is where the comparator comes it - the comparator compares the two voltage and supplies a switching voltage to the power transistors.
There's a chance that the whole set up is so well integrated that you won't be able to access the comparator voltage input, (ie a single chip solution) - most likely you will be able to use the recitfied power supply, and switching transistors, but will need a separate comparator. You will probably need to disconnect any smoothing capacitors from the output, (but not imput).
Also note that the switching frequency of the amp will limit the frequency responce, and that there might be a huge amount of switching noise. (a small capacitor in the output stage helps fix this, as does any inductance a loudspeaker has)
Some one who knows more about the specifics of computer smpa supplies will know how adaptable they are.
This should be a useful read [1] - note it says the switching frequency is typically 33kHz - that isn't really enough for high quality audio, but should be good for a subwoofer, or active woofer.83.100.250.79 (talk) 18:33, 5 September 2009 (UTC)[reply]
Another alternative is to use the powersupply as a powersupply for a conventional amp (replacing the old transformer) - extra capacitors in the output stage are a good idea here, the average computer power supply has the potential to outperform the average audio amp power supply if you can isolate the switching noise.83.100.250.79 (talk) 18:41, 5 September 2009 (UTC)[reply]
The smps is designed to deliver current at a given DC voltage. An AC current output will be needed to drive a loudspeaker i.e. positive and negative currents with average zero. Connect the smps output via a resistor to ground. Calculate the resistor value R ohms to drain I amps = half the available maximum current. THe resistor will dissipate IxIxR watts. Connect a large value electrolytic capacitor, typically 470µF 50V or more, from the smps output to the loudspeaker; this blocks DC current into the speaker. Inside the smps you need to access the reference voltage which is often given by a zener diode. Your AC audio signal will be added to the DC reference voltage. Depending on the circuit you can probably arrange that using a resistor and an input coupling capacitor. Cuddlyable3 (talk) 21:39, 5 September 2009 (UTC)[reply]
You read correctly. The arrangement described is a Class A driver that wastes power. A push-pull or Class B driver is difficult to construct with a smps that is designed only to "push". Cuddlyable3 (talk) 17:57, 6 September 2009 (UTC)[reply]
How about if there are two 12V supplies (I believe there are usually 2 12V rails, but not sure if they are powered separately, or have the same rating), both fixed to 6V (for O input) then one can go up whilst the other down, the speaker connected across the two outputs, I suppose the outputs would need to be connected to ground via a capacitor - getting complex..83.100.250.79 (talk) 18:12, 6 September 2009 (UTC)[reply]
neutron decay products
Why is it possible to have a giant neutron (as in neutron star or Black Hole) but not a giant proton and electron as products of a giant neutron decay? --Taxa (talk) 19:03, 5 September 2009 (UTC)[reply]
A neutron star is not a giant neutron, but many neutrons, with a skin of atomic matter. Protons repel each other, so it is very hard to make a collection of tons of pure protons. In fact the energy needed to bring them together will exceed their rest mass. I think there was a reference desk Question about this a year ago. Normally you will get a combination of electrons and protons - that is hydrogen, after all a neutron decay makes protons and electrons as well as the neutrino. Graeme Bartlett (talk) 22:13, 5 September 2009 (UTC)[reply]
Okay, so a behive of neutrons... What happens to the neutrons that decay? Do they turn into protons and electrons and if so where do they go? Do they follow the path of least resistance and get shot out in jets someplace? --Taxa (talk) 17:03, 6 September 2009 (UTC)[reply]
The pressure from gravity forces electrons and protons together into neutrons since neutrons can pack much closer than normal atoms can, which is how a neutron star forms. Neutrons in the core of a neutron star are much more stable than free neutrons and if they do decay the resulting protons and electrons would likely soon be forced back into neutrons. Neutron_star#Structure mentions the core might have electrons and protons mixed in. In the crust where the pressure is lower and normal matter can exist there's probably some transition back and forth. Rckrone (talk) 20:31, 6 September 2009 (UTC)[reply]
discrete solutions of 1H schrodinger
I've been trying to solve this using (unnormalised) formula similar to
Wavefunction = eΣan rn
And ignoring the angular component (eg taking δ2/δθ2 fn(r,θ,μ) to be zero ) ( because I'm not very good/lazy)
Which works and gives me solutions, mostly of the form
an+2(n+2)(n+3)=k1an+1+k2an
where k1,k2 are numbers (I believe there is a name for polynomials of this type but have forgotten - can anyone remember?)
So, so far I have infinite solutions (using infinite range of a0)
But there is more because I want to normalise which I believe is "integral over all space of fn(r,θ,μ)fn*(r,θ,μ)" which I believe that this will only give non infinite integrals for certain values of a0 - and hence give "quantised solutions" (as wanted).
Can anyone give a clue on how to show which of the integrals will be finite. (Even though I have all the parameters from one of the equations above) - and hence can integrate the equation (giving another infinite polynomial) I haven't got a clue how to evaluate the integral... Thanks.83.100.250.79 (talk) 19:31, 5 September 2009 (UTC)[reply]
Can you be a little more specific about what you're trying to do? Is this energy eigenstates of the hydrogen atom? What's the integral you're trying to evaluate? Rckrone (talk) 05:34, 6 September 2009 (UTC)[reply]
As far as I can tell, the OP is asking for help in evaluating integrals of the type
Specifically, he suspects that the requirement that the above integral be finite will lead to only some allowed sets of . For physical reasons, the OP then hopes that some of these sets of will also satisfy the recursion relation he calculated initially and therefore produce allowed wavefunctions for his problem. Martlet1215 (talk) 10:56, 6 September 2009 (UTC)[reply]
That's it, unfortunately my notes are a bit mixed up and now I'm not sure the equations I wrote above are correct - the only way I'll be able to tell is to start again from the beginning. Nobody seems to have noticed an error though..83.100.250.79 (talk) 11:24, 6 September 2009 (UTC)[reply]
Thanks, unfortunately I've been having some technical difficulties - meaning that a correction to the above recurrence relation didn't get inserted, then wikipedia wouldn't let me edit (overnight) - as a result the solution you saw lacked an (n+2)(n+3) term. To add insult to injury my notes are a mess, and as I tried various formula to solve the differential, I've lost track of which one I was using (basically I'm working from memory and a few scraps of paper). I'll need to start again to get this sorted out.
I still can't find the equation I used - I think it may have been similar to wavefunction =f(x)eax where f(x) is a (possibly finite) polynomial. Maybe someone (with knowledge of what works in the long run) could suggest a good choice for a generic function to start with to solving the differential equation.83.100.250.79 (talk) 16:58, 6 September 2009 (UTC)[reply]
Like Rckrone said initially, it would help if you explicitly told us what you are trying to do. Is this the problem of finding the radial part of the energy eigenstates of the Hydrogen atom? Martlet1215 (talk) 10:54, 7 September 2009 (UTC)[reply]
You guessed right in your first post - I can get solutions to the differential equation (radial only time independant schrodinger), using a variety of different formulations eg e^f(x) , f(x) , f(x)e^ax etc , but I can't work out a way to find out which solutions are not good, becuase the wave function cannot be normalised.
f(x)e^ax seems the most promising so far (a is negative) since if f(x) is a finite polynomial the the integratl will be finite. However I'm having difficulty showing the conditions under which f(x) is finite, and still stuck on rejecting other solutions which I can't integrate (normalise) - I probably need a way to tell which integrals will be infinte (since I can reject all those)>83.100.250.79 (talk) 15:03, 7 September 2009 (UTC)[reply]
September 6
Human race vs Human species
More and more I hear people on the radio or in books speaking of the 'human race'. I tried to ignore it but recently I read it in a lot of well-informed science fiction stories and today I heard it on a science program! that was too much. We're a species. How come we call ourselves so often a race? Where did it come from? None of the articles about race and racism I could find on Wikipedia seem to address the problem. Is it merely a slip of the meaning of the word or has it got a deeper meaning? Thank you. 190.244.183.244 (talk) 02:36, 6 September 2009 (UTC)[reply]
I doubt it's "more and more" — it's always been a common phrase. Probably you've just been noticing it more.
I see nothing wrong with the term. Certainly it has nothing to do with racism; kind of the opposite, given that "the human race" includes all "races". But it has a harder, more martial, more mythical sound to it than species, and sometimes that's desired. Species is a coldly scientific word, and doesn't work well when the poetic or emotional or metaphysical aspects are connotations you want to get across. --Trovatore (talk) 02:57, 6 September 2009 (UTC)[reply]
As Trovatore says, it doesn't have any special meaning, it's just the colloquial usage. Words in the normal vulgar lexicon are often very different from their purely scientific meaning, and that's just how it is. Happens all the time. My pet peeve is decimate. ~ Amory (user • talk • contribs)03:32, 6 September 2009 (UTC)[reply]
The usage is very common and not at all new. It appears in "The Time Machine, by H. G. Wells [1898]" and in "Frankenstein, by Mary Wollstonecraft (Godwin) Shelley [1818]". And those are just the first two books that it happened to cross my mind to search. I'm certain that I could find others.
Besides, "Human Species" isn't quite right either. "Human" refers to a handful of species, we just happen to be the last one standing. APL (talk) 04:10, 6 September 2009 (UTC)[reply]
Perhaps the usage isn't completely nailed down, but using "human" to refer to all of genus Homo is common enough that the human article starts "This article is about modern humans. For other human species, see Homo (genus)." APL (talk) 17:39, 6 September 2009 (UTC)[reply]
"Race" comes from a word meaning "line", which could be broadly interpreted to include "species", although it's more normally used to mean "sub-species", i.e. varieties within a species - although, technically, every "variety" within the human species is unique from the DNA perspective, except for identical siblings. Baseball BugsWhat's up, Doc?carrots09:12, 6 September 2009 (UTC)[reply]
The trouble is that we're comparing one fairly vague term "race" with another vague term "species".
The scientific definition of 'species' is generally that two animals are of the same species if they can successfully interbreed. That would make all modern humans be a part of the same species. However, we say that lions and tigers are different species - despite the existance of ligar and tigons. Actually, that's a terrible example because ligars and tigons are generally infertile - but I happen to like the words "ligar" and "tigon"! Dogs and Wolves might be a better example.
Race (biology) has a somewhat better definition in biology (and all humans are of one race under that definition too) - but the use of the term Race (classification of human beings) in everyday speech has a different meaning. The trouble with the everyday meaning is that it takes a somewhat arbitary group of genes (the one for skin color, for example) and promotes that to higher significance than any other group you might pick. Why classify people according to their skin color rather than (say) their lactose-tolerance or their eye color? It doesn't make any kind of scientific sense. What we're really trying to capture is the place where the persons ancestors originated before the mass migrations of relatively modern times has caused that to become a largely irrelevent thing.
What environmental conditions will be required in order for future human evolution to favour an increase in brain complexity rather than a decrease? NeonMerlin[2]06:19, 6 September 2009 (UTC)[reply]
For traditional evolution to work, two things have to happen - firstly, there has to be a genetic mutation or a chance combination of genes that results in some physical change. Secondly, that change has to give the recipients of that mutation an enhanced probability of passing the gene(s) onto the next generation. So IF there is a genetic change that makes some people smarter than others - and IF that change means that they have more children than the average person - THEN the population will tend to become smarter. Sadly (perhaps) it's not clear that our intelligence is genetically determined...and it's far from clear that more intelligent people have more children. Perhaps very mentally deficient people are less likely to have kids - but there is some evidence that things are heading the way of the movie Idiocracy (it's a terrible movie - no need to watch it!)...the idea being that intelligent people are too busy with their exciting jobs to raise kids - where the dumb people have nothing better to do - so they have vast families. That would tend to produce an evolutionary pressure AGAINST rising intelligence.
But that's traditional evolution. Modern humans have a tendancy to break the rules. When there is a genetic mutation that might prevent some couple from having kids - we step in with medical treatments for infertility. If someone is too stupid to hold down a long term job - we don't allow them and their children to starve - we provide social security and food stamps and such - so they survive, despite that evolutionary pressure.
What I think is more relevent here is the concept of 'memetics' - the evolution of ideas. If two individuals have brains of identical make-up - but one is educated to modern standards and the other is brought up by wolves - we're going to say that the latter person is not as intelligent as the former. Education is memetics in action. Ideas that work are passed on by teachers to children - ideas that don't work fall by the wayside. Ideas evolve. Since humans are the carriers of these 'memes' - we can get more intelligent (well, depending on your definition of that tricky word) as time passes. I believe that is the fundamental flaw in the ideas behind Idiocracy - the brains of the kids who come from large families with stupid parents aren't measurably genetically different from the single child of a Nobel award winner - so providing we have an educational system that allows the efficient passing of memes from the brain of the Nobel prize winner into the brains of the kids of stupid parents, our "intelligence" can continue to evolve despite the lack of genetic improvement.
Efficient transfer of memes in a cognitive way, that is. The other six kinds of transmission listed in the article won't make us smarter, they'll just make more memes. 213.122.26.161 (talk) 14:12, 6 September 2009 (UTC)[reply]
APL may not have considered that much mating these days is recreational. Anything that causes intelligent people to procreate more will have the desired effect. Cuddlyable3 (talk) 13:51, 6 September 2009 (UTC)[reply]
Contrary to what SteveBaker says the Heritability of IQ article indicates intelligence is even more determined by genetics than height. Interesting the correlation between twins gets closer the older they are. It also lists a couple of genes which do contribute to intelligence though probably a very large number do, the number of repeats in bits which don't occur in genes also contributes though I didn't see that mentioned in the article. Dmcq (talk) 14:05, 6 September 2009 (UTC)[reply]
And anything I've read about it indicates rich people do tend to have more children and that correlates reasonably well with intelligence. But then so do the poor. It's the in betweeners that don't. Dmcq (talk)
"Practically, however, the upper bound of these correlations are given by the reliability of the test, which tends to be 0.90 to 0.95 for typical IQ tests..." how is the reliability of an IQ test tested? Is there an official prototype person with an IQ of 100, used for calibration, preserved at a constant temperature under a glass jar in Paris? 213.122.26.161 (talk) 14:22, 6 September 2009 (UTC)[reply]
Some people's results on repeated IQ tests vary a great deal, presumably because they were tested on days on which they had lot of intelligence, or an unusual lack of it. So for this method of testing the test to be a good one, it would have to be performed on a person whose intelligence was known to be constant. Even better, take a person whose intelligence varies over time by a known amount, and see if the test can pick up on the variations accurately. 213.122.26.161 (talk) 15:30, 6 September 2009 (UTC)[reply]
We can't have someone we know to have constant intelligence or that has a known variation in intelligence because our only measure of intelligence is the one we are testing. You have to just do the same experiment with lots of people. --Tango (talk) 17:28, 6 September 2009 (UTC)[reply]
Most people aren't much of a fan of the sort of things that would be required to manually affect human evolution. See eugenics. Evolution is a rather nasty business -- rapid "progress" is made only by the lack of reproduction (often death) of the great majority of individuals in a species. Most people aren't that interested in participating in that type of ethical system. Much of what we call "civilization" are safety nets so that just because you are born with poor eyesight or stubby legs or a slightly less-than-average brain, you don't have to actually die. Consider how much money we funnel to genetic diseases that affect a small minority of people—this is an act of humanity, not evolutionary fitness. There's no reason that eugenics couldn't be made to "work", but it would involve implementation on a horrific, probably impossible scale. --98.217.14.211 (talk) 14:37, 6 September 2009 (UTC)[reply]
Forgive me for avoiding the direct question, but I think it is important to point out whenever questions like this are asked that the minimum time it would take for any manipulation to have the tiniest noticeable effect is on the order of 1000 years. Thus, while such questions might be philosophically interesting, it's important to realize that they don't have any practical relevance. Looie496 (talk) 14:42, 6 September 2009 (UTC)[reply]
Well, not necessarily. It depends what you are trying to select for and what measures you are willing to take. If you're talking about one gene, and are willing to sterilize everyone else who doesn't have it, then you could do it in one generation. Again, nobody wants to do that, though. --98.217.14.211 (talk) 14:53, 6 September 2009 (UTC)[reply]
Where did you get that 1000 years from? If you look at dogs or cats selective breeding has made a huge difference in far far less time. Much of their breeding only started in the 19th century. Dmcq (talk) 15:22, 6 September 2009 (UTC)[reply]
What makes you say that? With drastic enough measures the average [something] of the human race could be changed noticeably in a single generation. To pick a horrific example, if tomorrow the all powerful eugenics king declared that white people were not allowed to reproduce, what would that do to mankind's average skin tone? APL (talk) 17:50, 6 September 2009 (UTC)[reply]
Well if you look at Human skin color it seems that people in Europe were probably dark coloured at the end of the last ice age and the white colour you see now mainly came about in the last 12000 years or so, a third of the change may have happened after the Egyptians stated their civilization so within the time of recorded history. And that was without any eugenics king declaring anything. Dmcq (talk) 18:10, 6 September 2009 (UTC)[reply]
A dog generation is about 1/10th of a human generation and usually includes multiple offsping, so dogs can probably evolve over 10 times faster than humans. And concerning the other points, it's true that you can get faster changes if you focus rigorously on one factor and select rigorously for it. So for example, if you sterilize everybody whose brain size is below the mean, you might get a perceptible increase in brain size in a hundred years. But for plausible interventions, it's difficult to get below 1000. Looie496 (talk) 18:26, 6 September 2009 (UTC)[reply]
Generation length is only one factor in evolution speed. One way to get fast evolution in humans would be to concentrate on just a small group. If you choose 100 people (50 male, 50 female - you might be tempted to have more women, but I don't think it would actually be a good idea) that are already above average in what it is you are interested in and you encourage early and plentiful reproduction you can get a generation every 20 years and remove people from the group that don't fit your requirements you could get pretty rapid selection. --Tango (talk) 18:52, 6 September 2009 (UTC)[reply]
The Ashkenazi intelligence article discusses the hypothesis that ashkenazi jews have greater intelligence on average than most other groups and that the difference is due to natural selection during the middle ages. Dmcq (talk) 20:22, 6 September 2009 (UTC)[reply]
Formula for subjective sound fading with distance, and adding
1) You are in a flat agricultural field (for example) with a speaker giving out, say, 60 decibels. What would be the formula for how the sound fades away the further away you stand from the source? 2) You have two speakers the both give out 60 decibels. If you put them close to each other, can you get more sound at any point than 60 decibels? What would you get? I realise I have not specified every parameter, so reasonable assumptions can be made. 78.146.76.67 (talk) 11:31, 6 September 2009 (UTC)[reply]
Calling sound levels "decibels" is a common error. The term 60 decibels is not a measurement unit; it just means a ratio 1,000. Possibly the OP should have posted "60 dB(SPL)" where SPL is the level reference of 20 micropascals (μPa). 0 dB(SPL) is barely audible. 1) In open air if a speaker gives 60 dB(SPL) at 1 meter one hears 60 - 1020 log(r) dB(SPL) at other distances r meters. 1) Two such speakers broadcasting noise can give 63 dB(SPL) at 1 meter. If they broadcast a single frequency tone then the received sound is spread in a complex interference pattern with peaks of 66 dB(SPL) at 1 meter in some directions. The agricultural field is assumed not to reflect any sound. Cuddlyable3 (talk) 12:56, 6 September 2009 (UTC)[reply]
Thanks, can I just check I understand the formula please? Have I calculated correctly: at 10 meters away the answer would be 50 dB(SPL), at 100 metres away it would be 40 dB(SPL), and at 1000 metres away it would be 30 dB (SPL)? And at a billion (thouand million) metres away the noise would be minus 30 dB (SPL) - that cannot be right! 78.146.76.67 (talk) 13:31, 6 September 2009 (UTC)[reply]
As dB is not an absolute measure, 0 dB has no particular physical meaning and is not the same as "no signal". Negative dB simply indicates that the ratio is increasingly smaller, and that your signal is increasingly less audible. — Lomn14:07, 6 September 2009 (UTC)[reply]
This is a suprisingly difficult thing to answer.
The energy transmitted is proportional to one over the square of the distance - that's nice, simple physics. If you double the distance you are from the speaker - you get one quarter of the sound energy. Easy!
However - you used the word "subjective" in your subject line and "decibels" in your question - and herein lies the tricky part.
The human ear doesn't report a doubling of sound energy as a doubling of percieved loudness...it depends on the duration and frequency of the sound waves as well as the amplitude. But very roughly - the sound has to have 10 times more energy behind it in order to sound twice as loud.
A decibel expresses the ratio of two sound levels - not some absolute number - there are many variations on the use of the term - read: Decibel#Acoustics_2 (or the entire decibel article if you're brave!). Typically, audio equipment is described in terms of 'dbA' - which is ten times the log-to-base-ten of the ratio of the sound energy produced to the quietest sound a human can hear. So a difference of 10 decibels sounds like a doubling in loudness - and a 60 dbA speaker can sound like 6 doublings in volume - 2x2x2x2x2x2 - 64 times louder than the quietest sound you can hear. A 70 dbA speaker would produce ten times as much sound energy - and sound twice as loud.
Incidentally - this is why listening to loud music can be so damaging to our hearing. The damage is done by the energy transmitted into our ears - but ten times the 'safe' amount of energy only sounds twice as loud.
Putting two speakers close together does indeed double the amount of sound energy - but a mere doubling of energy doesn't sound twice as loud to a human. There is also an ikky complication if the two speakers are playing anything like the same sound - interference. If both speakers were playing the exact same perfect sine wave then as you walked around them, the sound would get louder and quieter because at some places the sound waves would add together and in others they'd cancel out.
Thanks. You can imagine the speakers are giving out (different) traffic noise. I'd be very interested to know the formula for "adding" the noise of two different speakers which are at different distances. Also, as I wrote above, I think I must have misunderstood the formula given for the fading of noise with distance. (I'm ultimately trying to estimate the noise at particular coordinates from a non-straight road - probably best to model as several point souces of noise which correspond to individual cars on the road). 78.146.76.67 (talk) 13:46, 6 September 2009 (UTC)[reply]
If you do that, don't forget to take account of the fact that traffic slows down around tight corners - resulting in more vehicles tending to be there than on the straight parts - but also that the engine noise will vary with speed. If big trucks are to be considered, you'll need to consider the noise they make when engine-braking. You probably need to calculate the sound energy emitted by each point-source - use the inverse-square-law to calculate the total energy at any given point - then convert that into dbA's using the equation in the decibel article. SteveBaker (talk) 15:17, 6 September 2009 (UTC)[reply]
A more or less direct quote (after poor translation) from my physics book: The total sound intensity at a specified point is the sum of the intensities caused by single sound sources. This result can be can be applied to, among other things, combining the intensities of different kinds of noise sources. It is not valid if the waves have a phase relation. It should be noted that aside fading with the squared distance and reflection/absorption by ground and other surfaces, the sound fades due to absorption by air. The correct approach to this problem depends on whether you want a safe upper limit or accuracy. --194.197.235.240 (talk) 15:22, 6 September 2009 (UTC)[reply]
If you're serious about this there is professional acoustic modeling software that's used by architects, but I don't know anything about it. For a point source the 1/r² law implies a 20 dB drop for each factor-of-10 increase in distance; for example 80 dBA at 1 meter becomes 60 dBA at 10 meters and 40 dBA at 100 meters. When combining multiple sources it's the energy that adds linearly, not the perceived loudness (ignoring interference, which should be safe to do in this case). If L is a dB measure of loudness then the correct formula for combining them would be , but in most cases you can approximate that pretty accurately by . Aside from reflection off solid surfaces, the 1/r² law can be modified by temperature gradients above the ground, which can act as a converging or diverging lens. I suppose there wouldn't be much reflection or temperature variation out in a tilled field, but I'm not sure. -- BenRG (talk) 15:30, 6 September 2009 (UTC)[reply]
Sound normally fades faster than the inverse square law in a field because the sound is refracted upwards. If you have a clear sky and a cold night though the air will be colder near the ground so the sound is refracted down and it fades more slowly than inverse square. Interestingly sound is not in the main 'carried' by the wind downwards, what happens is that the air in the wind travels faster further away from the ground so making sounds refract down more downwind and refract up into the sky upwind. Dmcq (talk)
Thanks for all the above, especially to BenRG for his formula for combining several sound sources. I seem to recall reading on one of the Wikipedia articles that I read a day or two ago that subjective noise decreases with 1/r and not 1/r**2. Perhaps it would be easier if I considered the sound energy for the fading-with-distance and sound-adding calculations, and then after calculating the sound-energy for the X.Y points I was interested in, I then converted this sound energy into dB(SPL). In that case would the sound energy from different sources of traffic-noise simply add? I assume they would decay 1/r**2. How do you convert sound-energy into dB(SPL) and back please? 78.146.254.202 (talk) 09:19, 7 September 2009 (UTC)[reply]
Traffic noise goes down as 1/r (or in fact a bit faster than that normally because of refraction). The traffic noise is along a line source. There is a considerable amount of refraction reducing the noise because the air normally gets colder with height. A tilled field in sunlight would make the problem worse because it is dark and absorbs heat. That is why sound hardly travels at all on sunny days but you can hear foghorns from many miles away at night. Dmcq (talk)
I'm very interested in traffic noise going down by 1/r rather than 1/r**2 - the former should mean that sound from further away is less attenuated, and would make a big difference to the results of calculations. Do you have any sources, preferably online, that reinforce that idea please? I'm not sure that the two articles above do. Since I've been brought up from a schoolboy to believe that things attenuate by 1/r**2, I need to have a clear explanation for 1/r rather than 1/r**2. Thanks. 89.242.115.9 (talk) 20:23, 7 September 2009 (UTC)[reply]
1/r2 comes about like this: If you imagine a point source of sound (or light or heat or almost anything with waves) out in the middle of nowhere - and imagine it producing one short BEEP! of sound. That sound radiates outwards in all directions equally so at any given moment the sound wave is the surface of a sphere that's growing at the speed of sound. Well, the initial burst of sound energy is spread out over that entire surface - so the amount of sound energy at any given point on the surface is equal to the initial sound energy divided by the area of the sphere - which is 4 pi times the radius squared. Hence - the sound at any point at a distance 'r' is proportional to 1/r2. However, we've been talking about sound ENERGY - and as I explained above - the human ear isn't a very linear device. The formula in decibel and A-weighting (which weights the results based on frequency) relates the percieved VOLUME to the incident ENERGY - and it's not a simple relationship...but I suppose if you imagined that the ear was percieving that volume was proportional to the square of the amount of energy - then volume would be proportional to 1/r - but I don't think that's true.
At any rate, this is a horribly complicated business. Firstly, you don't know your source data to enough precision. Just how much noise does a vehicle make? You have everything from a prius running in all-electric mode - up to an 18 wheeler doing engine braking - and everything in-between. Then you have the number and mix of vehicles on your road - the effects of corners and hills changing their engine speeds and causing them to bunch up in some places and spread out in others. You don't know the distribution of frequencies within those sounds. Then you have the effects of refraction and reflection of the ground - and absorption of sound by the ground, trees, etc - which is different for sounds of different frequencies and different frequency components within a single sound. When you get to the end of all of that - then you have to consider that humans don't perceive sound in a simple way - you have to turn incident energy into dBa numbers using the A-weighting approach. There are so many unknowns in that set of variables that it's going to take a really expert person to get the error bars down to something manageable - or else you're going to have answers that have a couple of orders of magnitude of error associated with them! SteveBaker (talk) 11:25, 8 September 2009 (UTC)[reply]
That's assuming a steady stream of traffic. Why not try google with something like 'traffic noise attenuation with distance'? You'll get some more complex formula that do an accurate job up to a hundred meters or so - after that atmospheric conditions make a large difference. It is a popular study. Dmcq (talk) 00:30, 8 September 2009 (UTC)[reply]
Defunct Sciences
Hello. I'm looking for a list (or group) of studies which were once considered scientific but are now considered (by the vast majority) as defunct. An example I could think of is alchemy; does anyone know any others?
I wonder if reading lines on your palms or bumps on your head would qualify? (Now, I'm seeing Daffy Duck disguised as a swami - "No bumps? We make some!" [pulls out hammer and whacks Porky over the head a few times]). Baseball BugsWhat's up, Doc?carrots14:09, 6 September 2009 (UTC)[reply]
I would like to ask if there is or if it is possible to create a list of chemicals (not only chemical elements) by density? Of course, list of only those chemicals, that have an article in Wikipedia (and when the density is given in the article).
There are vastly too many known chemical compounds to list...millions of them at least. You need to narrow your criteria before there could be any chance of finding such a list. (eg: If you wanted the 100 densest or the 100 least dense, maybe we could help). SteveBaker (talk) 15:00, 6 September 2009 (UTC)[reply]
Since you are restricitng your request to chemicals on Wikipedia, the answer is -- maybe. You will need to either get a Wikipedia database dump and run a program on you own computer, or write a "bot" and get permission to use it. In either case, you will process all articles that include the {{chembox}} template. You do this by going to Special:WhatLinksHere/Template:Chembox. After you get your list the first time, you can look at the "what links here" page occasionally to see if thre are any new chemicals, and update your list. As steve baker said, there are a LOT of chemicals, so if you want to place your in Wikipedia itself, you will need to use multiple subpages. -Arch dude (talk) 16:53, 6 September 2009 (UTC)[reply]
You can save yourself a lot of effort because density is related to the average atomic weight of a compound. For liquids and solids this produces one set of data, for gases another set of lighter densities.
Thus without even looking I can tell you that C6H12O6 (glucose) has a very similar density to H2O (water) , C2H5OH (ethanol), diamond is higher (C). Fe2O3 (rust) is higher still, and PbO (lead oxide) is very high.
I didn't say I had such a list - just that we'd stand a better chance of finding some kind of list for you if you could narrow the criteria. SteveBaker (talk) 02:42, 8 September 2009 (UTC)[reply]
Children with diarrhea have a mean whole gut transit time of 5.5-7.3 hours, and 14.1-15.5 hours during the recovery period.[4] Another study in ill preterm babies found that those fed small amounts of milk had a WGTT of about 32 hours at 3 weeks and 21 hours at 6 weeks.[5] Here's more about baby poo than you'd ever want to know:[6] That study of healthy children found the transit time in under threes is about 30 hours, and in those aged 3-4 it is 36 hours. Fences&Windows02:38, 9 September 2009 (UTC)[reply]
LiFePo4
Sometimes when I find a site with a fairly low price for LiFePo4 the cells are not encased or does not include a charger. Where can I find encased technology of good or best quality with a charger at the lowest price? -- Taxa (talk) 16:58, 6 September 2009 (UTC)[reply]
LiFePo4? Am I to understand that it includes polonium? If so, where did you find those cells, especially at a "fairly low price"? Polonium is very expensive, mind you. 98.234.126.251 (talk) 06:01, 7 September 2009 (UTC)[reply]
Do a search on both spellings and see if you get any results that include polonium. You probably will not since Li and Fe as part of the term modify its sufficiently to make the distiction between Po4 and PO4. -- Taxa (talk) 01:01, 9 September 2009 (UTC)[reply]
Well, the article Lithium iron phosphate battery has a list of manufacturers at the bottom -- maybe some of them might let you order the batteries (and charger) by mail-order catalog at wholesale prices. (You know that for the very lowest prices, buying straight from the manufacturer is usually the best option when possible.) But then again, maybe they don't do that. The only way to find out for sure is to check the company websites. FWiW 98.234.126.251 (talk) 06:21, 7 September 2009 (UTC)[reply]
guys I have a project on Low noise amplifier .Plz can smebdy suggest a very basic tutorial starting right from the basics with circuit diagram
What do you consider "basic"? I would suggest Chapter 11 in Gray & Meyer for a theoretical overview, and I think Planar Microwave Engineering has an entire chapter on LNAs. Are you simulating this design in SPICE, building the circuit out of discrete components or designing for an ASIC, or simply studying the topic for general overview? The design methodology for LNAs will vary dramaticallly depending on what your actual needs are - first of all, what are you amplifying? Voltage amplifiers and current amplifiers have different topologies for best noise characteristics. Next, what are your power, bandwidth, operating frequency? These will also help you select a topology. How do you define "low noise" (e.g. 80 dB, 130 dB, spurs, noise floor, intermodulation, total harmonic distortion, etc.) You might be best served by a basic introduction to amplifiers in general (specifically electronic amplifier), and pay close attention to the relationship between different figures of merit and the noise performance. Then, you can "zoom in" on low-noise topologies and see some tricks to get the noise figure lower. Nimur (talk) 20:17, 6 September 2009 (UTC)[reply]
Here's a tutorial from National Semi, on Youtube. They're pitching their own tools, though, so you have to parse through some commercial product placement; but they have Bob Pease hosting, so it's probably worth it. He "can't tell you how to make a low noise amplifier", but he's going to "talk about how not to make a high-noise amplifier" with his classic awesome hand-drawn notes. Nimur (talk) 20:32, 6 September 2009 (UTC)[reply]
Actually that probably isn't right - looking at the methods for gold refining it looks like the urea is being oxidised by the nitric acid (catalysed by nitrous acid present in aqua regia). Looking... (seems this only works in dilute HNO3)
In concentrated acid it would be easy to assume that everything is converted to N2 , CO2 and H2O , in which case an equation would be fairly easy to get. Is that what you had in mind.?83.100.250.79 (talk) 21:55, 6 September 2009 (UTC)[reply]
Interesting. It could be that the nitric acid was not very concentrated after all (it's 1/4 of a.r.) - but N2? That's weird. IIRC only really dilute nitric acid does that.Tim Song (talk) 18:20, 8 September 2009 (UTC)[reply]
NO2, NO, and NO2+ / NO+ would all be likely to react with the amine group on urea
That's just a speculation, NO+ would make N2 by the same process, both being electrophilic attack on the NH2 ... finally subsitution of H2O at the C=0 ... giving CO(OH)2 (carbonic acid)
Cl2 may also react with urea (as do hypochlorites). There are examples of nitrous acid reacting with urea[8], so NO+ should react as well. The real reaction will be more complicated.83.100.250.79 (talk) 21:03, 9 September 2009 (UTC)[reply]
Relativity and Existence
The star Beta Comae Berenices is about 30 light years away. I was born about 25 years ago. Does this mean from the perspective of Beta Comae Berenices I do not exist? TheFutureAwaits (talk) 22:28, 6 September 2009 (UTC)[reply]
You do, they just won't know it until another 5 years or so. I doubt they're watching your house. But they could be watching TV broadcasting signals. For them, Disco is still all the rage here, as far as they know. Baseball BugsWhat's up, Doc?carrots22:39, 6 September 2009 (UTC)[reply]
I'm not sure if that's right. If that's true then you're saying everything beyond a person's immediate influence is predetermined as it has already happened but they have yet to experience it. TheFutureAwaits (talk) 22:48, 6 September 2009 (UTC)[reply]
"Predetermined?" Well, let's take an example closer to home. At the ballpark, the concession stands have the game on TV. However, due to FCC concerns and such, there is a slight delay, maybe 7 seconds. So if you happen to be watching the game through an exit portal and also watching the TV, you might see the batter hit the ball somewhere. Then you turn to watch the TV, and 7 seconds later you'll see the batter hit the ball. So it's "predetermined" in the sense that what the electromagnetic waves are carrying has already happened. That's going to be the case, whether the delay is 7 seconds or 30 years. Baseball BugsWhat's up, Doc?carrots00:44, 7 September 2009 (UTC)[reply]
Not really...on the earth, the time would be so instantaneous that it truly is action/reaction, rather than having something happen to you then 5 minutes later you experience it. Ks0stm(T•C)00:00, 7 September 2009 (UTC)[reply]
It just depends on how you define existing yet. In the reference frame there that's stationary relative to us (and probably in the frame of the star itself) your birth is in the past even though someone there couldn't receive information of the event for another 5 years. I don't know what you want to call that. However there are some reference frames at that point in which your birth is still in the future (frames moving very quickly away from Earth). Deciding in what situations you qualify as existing isn't necessarily a easy question. For example a proponent of four dimensionalism would say that you exist regardless of where they happen to be in time and space. Rckrone (talk) 23:54, 6 September 2009 (UTC)[reply]
If an explosion equal to 100 trillion666 megatons happened 30 light years away now, would we even know about it, let alone exist? So maybe you don't exist there yet, but it's inevitable that you will. Like the above poster said, it depends on your point of view/philosophy and your reference frame. Sagittarian Milky Way (talk) 00:25, 7 September 2009 (UTC)[reply]
This is entirely a "Tree falling and no one listening debate" and it centers on the fact that people are confusing an event with information about an event. You were born 25 years ago. Information about your birth has not yet reached Beta Comae Berenices. That doesn't mean that according to (hypothetical) people at Beta Comae Berenices you do not exist, it just means they don't know that you exist. There's lots of things that have happened that you have no knowlegde of. It doesn't mean they didn't happen. --Jayron3203:57, 7 September 2009 (UTC)[reply]
Something to keep in mind is that there is a distinction between events that are in your light cone and events that aren't. A past event in your light cone is unambiguously in the past regardless of reference frame. Any event outside your light can be in the past, present or future depending on your reference frame. For example TheFutureAwaits' birth is outside the light cone of someone who is on Beta Comae Berenices right now. If they're stationary relative to Earth, TFA's birth is 25 years in the past as it is to us here on Earth now (although they can't yet know about it). But if that observer were to suddenly accelerate away from Earth to faster than (5/6)c, then TFA's birth would suddenly become a future event. That's not something that can happen with the Giants game I missed last week. Rckrone (talk) 04:44, 7 September 2009 (UTC)[reply]
No, I think your obfuscating the reality of the situation with fancy physics; your still confusing the actual event (TFA's birth) with the receiving information about his birth, which can be played around with. By accelerating at (5/6) c, what you are doing is essentially outrunning the information, thus dilating the delay of that information reaching you to infinity (by keeping the information outside of the "light cone" forever). However, that doesn't change when the event happened, it just changes when the information arrives. Still the same tree, still the same sound, your just running away before you can hear it...--Jayron3204:54, 7 September 2009 (UTC)[reply]
Yeah. You're right. Darn that whole "Lack of a priviledged reference frame" thing. Sometimes the tree makes a sound before it even lands... --Jayron3205:46, 7 September 2009 (UTC)[reply]
If the article is incorrect or confusing, we need to edit it to make it more correct and clear. Always remember that while simultaneity is not absolute, causality is always preserved. The hypothetical observer outside the light-cone could not have been impacted by the OP's birth yet, because of causality; but they could have predicted it (assuming they had sufficient prior information) and therefore be aware of it (speculatively) before confirmation arrives. If their predictions were wrong, they would have no way of knowing until (at least) after the light cone from the OP's birth "arrives" at their location. Nimur (talk) 15:20, 7 September 2009 (UTC)[reply]
One mistake people sometimes make in these things is thinking of objects as points in spacetime. Objects are lines in spacetime. The situation is like this:
| |
X * here-and-now
BCB -> |\ |
| \|
| * birth of TheFutureAwaits
|
The event marked X is the earliest point on BCB's worldline where your birth or subsequent events could have a causal effect. So, from the perspective of Beta Comae Berenices when? "Right now"? Meaningless. At the time the light we're now seeing from it was emitted? That's about 35 years prior to X. Same cosmological time as determined by CMB anisotropy? Around 5 years before X. I can't think of any other reasonable question to ask along these lines.
Forget the relativity of simultaneity. Different coordinate systems assign different coordinates to the same point. That's all there is to the relativity of simultaneity. It's not "wrong" as such, just meaningless. It's a statement about coordinate systems, and coordinate systems don't exist. -- BenRG (talk) 12:15, 7 September 2009 (UTC)[reply]
"Now" does have meaning in the context of a particular frame. (I would point out there's a difference between "meaningless" and "not particularly useful". I would agree that it's often not a particularly useful way to think about the relationship between two events.) In this case I was using "now on BCB" to mean the point that's at the same time as here, according to the Earth frame. I wasn't as clear about that as I could've been. Rckrone (talk) 17:42, 7 September 2009 (UTC)[reply]
But cause and effect travel at, at most, the speed of light. What good is existance if it doesn't make a mote of a difference to the most sensitive detector? (at least temporarily, or, if you're receeding fast enough, indefinately/forever) You can't even be aware of the existance of an event without it causing an effect (otherwise, how did you notice?) TheFutureAwaits exists, to an omniscient observer outside spacetime. But so do the 2012 Olympics (probably). I think we can agree, using the usual Plain English sense of the language, the 2012 Olympics don't exist yet. There is a very high probability that there will be a day, that they exist. That means, for now, they don't exist. That doesn't stop from people from planning or predicting events that don't exist because we're trying to make them exist, or finding out what will exist before it does. And for high likelyhood things (like Greenwich sun meridian crossings?) it's only natural to list tomorrow's and 2010's as fact, but technically they are just as much fiction as Moby Dick until they occur, in the flesh. Interestingly enough, on the last one Ockham's Razor says shut up. Sagittarian Milky Way (talk) 01:16, 8 September 2009 (UTC)[reply]
I think any attempt to naturally generalize plain English language, which is based on a classical picture, to the relativistic case is going to be problematic. In the classical case, events fall into only two categories: past and future (ignoring border cases), and so there are a bunch of equivalent ways we can define the concept of events that haven't happened yet. Some of them are as events that can't affect us now, events the we can still affect, or events that are forward in time from where we are now. In the relativistic case those three aren't equivalent (and the last one depends on your reference frame which makes it kind of problematic). I don't think there's a natural way to resolve the ambiguity without just deciding on some new definitions. Rckrone (talk) 06:57, 8 September 2009 (UTC)[reply]
Would these tomatoes growing at a sewage works be safe to eat?
I just saw this article. Would the tomatoes be safe to eat? I understand that the sludge is eventually turned into compost which is presumably safe to grow food plants in, but what about these plants that are growing directly on the waste? Spin Dryer (talk) 23:27, 6 September 2009 (UTC)[reply]
Food has been grown with feces as fertilizer for thousands of years. That's not a problem. Pathogens won't go through them. More worrisome are things like heavy metals that might also be in waste water. --98.217.14.211 (talk) 23:57, 6 September 2009 (UTC)[reply]
That's the story that was told to me by someone who worked at a sewage farm. At the place he worked, they apparently had some very large, very healthy-looking tomato plants with nice-looking fruit growing in unexpected places. The staff were told that the tomatoes were unsafe to eat (and not to even think about picking them) due to the possibility of contamination by heavy metals and other chemicals from the various cleaning products that people routinely flush/put down the sink. --Kurt Shaped Box (talk) 09:24, 7 September 2009 (UTC)[reply]
People don't actually flush that much nasty stuff down the sink. The main problem with heavy metals is purely 'natural': they bioaccumulate. In other words, are enriched upwards in the food chain - which we're at the top of. So our feces would normally contain many, many times more heavy metal than the corresponding amount of plant-mulch or cow dung. --Pykk (talk) 19:23, 7 September 2009 (UTC)[reply]
In the US, people put all kinds of stuff down the sink. Maybe not so much things like Mercury, but there are all kinds of traces of medications in the water. Googlemeister (talk) 13:33, 8 September 2009 (UTC)[reply]
I think it's important to understand (given the OPs second comment) that AFAIK the danger comes from contamination. It's difficult to grow something in sewage and ensure you don't get any of it on the stuff you are growing. Washing is not guaranteed to get it all off and you may only need a small amount and I presume it may even penetrate the surface somewhat. anyway. I expect this is particularly a problem with leafy vegetables but low hanging fruit like tomatoes will still have some risk. Nil Einne (talk) 09:08, 7 September 2009 (UTC)[reply]
Certainly bugs can penetrate fruit skin. My father was a ship's doctor in the Royal Navy in WW2 and had to deal with a Dysentery outbreak caused by sailors eating water melon stored by traders underwater in the then infected Sweet Water Canal. The sailors seemed to think if they washed the melon and cut the skin it was good enough to sterilise the interior. --BozMotalk09:28, 7 September 2009 (UTC)[reply]
Hmm, even if true it hardly proves that the pathogens got through the unbroken rind. Enough of them might have stuck to the rind after washing, and then when the sailors cut into it, they carried them into the meat on the knife. I don't know for sure that the pathogens can't get through, but I have to say it seems a little unlikely to me given the thickness of watermelon rind and the size of ameobae, combined with the existence of a plausible alternative explanation. --Trovatore (talk)
The answer is on the image description page - "The Space Shuttle Challenger crewmember remains are being transferred from 7 hearse vehicles to a MAC C-141 transport plane". Nanonic (talk) 23:34, 6 September 2009 (UTC)[reply]
According to recent news reports (one example), the Emperor Norton Bridge has been closed because of cracks in the cables that support the deck. Is it accurate to say that there's currently a problem with the "suspension" of this suspension bridge, or would that be a bit of a colloquialism? Nyttend (talk) 02:59, 7 September 2009 (UTC)[reply]
To answer your question, no you would not be correct, because according to that article they "found a crack in one of the eyebars on the side of the structure." I interpret that to mean the actual bridge part, along where the original work was planned. If there was a crack in the cables, I think you'd be correct in saying it, not like Suspension (vehicle) of cars but rather in the actual hanging of the thing. ~ Amory(user • talk • contribs)04:19, 7 September 2009 (UTC)[reply]
Actually, that wasn't the article I consulted; it was something on the Road Runner website. Here is a story from the San Francisco Chronicle, stating that the link "helps hold up the east span". Nyttend (talk) 03:20, 8 September 2009 (UTC)[reply]
I understand an eyebar to be part (or indeed all) of the vertical suspension members of a suspension bridge. Thus I think it would be fine to say that the bridge. Saying there is a problem with the "suspension of this suspension bridge" is probably a bit unhelpful; I'd tend to use the phrase "problem with a vertical suspension link", or somesuch. However, since the eastern span of the bridge, which is where the fault was found, is a combination of "a double-tower cantilever span, five medium-span truss bridges, and a 14-section truss causeway", I'm not convinced that the explanation necessarily relates to suspension at all - though I'm not satisfied that I understand the design of the double-tower cantilever span ... is is cable stayed? --Tagishsimon(talk)21:27, 8 September 2009 (UTC)[reply]
I see that the first image in eyebar explains quite well the space of eyebars in truss sections. So now my view is that although the eyebars are under tension and so can be thought of as suspension members, to describe them as such without more context would probably serve to mislead. --Tagishsimon(talk)21:31, 8 September 2009 (UTC)[reply]
Radioactivity in the early Earth.
From what I know radioactive heating was a major source of heat for the early Earth, and still is today to a lesser extent. My question is, was the radioactivity billions of years ago (when the Earth formed) enough to cause harm to humans if we lived then. Would it be more/less then the fallout from a nuclear blast? —Preceding unsigned comment added by 66.133.196.152 (talk) 06:20, 7 September 2009 (UTC)[reply]
From what I know, the sources of radiation were buried deep under the surface (as they are today) and heating the Earth from within. And since ionizing radiation is strongly absorbed by rocks (from what I've heard way back when, a layer of granite just 1 foot thick can absorb gamma rays almost completely), the radiation at the Earth's surface would be negligible, since all of it would've been converted to heat. The heating from the radiation, on the other hand, would be a whole different story -- since it was strong enough to melt rocks, it would surely burn any living creature to a crisp. 98.234.126.251 (talk) 06:35, 7 September 2009 (UTC)[reply]
Ehhh, if I lived billions of years ago, I'd be more concerned about the radiation I'd receive from the sun due to a lack of ozone rather than radioisotope decay emerging from the earth's core. Primitive earth was probably devoid of ozone, so heavy amounts of solar radiation would be kill a human pretty quickly.
You should also consider that there's no one set quantity of fallout that's released from any nuclear bomb. The fallout quantity depends on the yield, which can vary. Dougcard (talk) 06:39, 7 September 2009 (UTC)[reply]
Much less than a nuclear blast by the time there were humans. That comes trivially from the maths of half lives. Out of interest Rutherford showed without background radiation the world would cool completely in about 10 million years. The fact that the core is still molten etc at 5 billion years is because of radioactive decay generating heat which has nowhere to go. --BozMotalk09:21, 7 September 2009 (UTC)[reply]
It's also worth pointing out that Ernest Rutherford was an atomic physicist, not a geophysicist or planetary scientist. Modern theory have proposed a variety of mechanisms for cooling other than blackbody radiation, including nuclear decay heating and latent heat of fusion from liquid-to-solid crystallization. So, I would take Rutherford's estimate with a grain of salt; modern geophysical theory seems a little more reliable. A common misconception is that all physicists know everything, and are good at back-of-the-envelope calculations. If only we could put the most famous theoretical physicists on the Wikipedia Science Reference Desk to see how their domain-specific expertise translates into general-purpose problem solving skill... Nimur (talk) 15:28, 7 September 2009 (UTC)[reply]
If the radiation is hot enough to melt metal and rocks, it is far more radioactive than the radiation in a nuclear blast. Nuclear radiation is bad for humans and other living things (as they say) but it's not much on a geophysical scale. --98.217.14.211 (talk) 20:07, 7 September 2009 (UTC)[reply]
Not really true I am afraid. It keeps rocks molten because it heats them incredibly slowly over thousands of years but the heat is locked in. At any instant the level is way less than a nuclear blast--BozMotalk21:07, 7 September 2009 (UTC)[reply]
If we only look at 40K, its concentration in potassium was about 12 times as high 4.6 billion years ago. A typical human is about 0.2% potassium by mass, for 75 kg body mass that is 150 g potassium which equals about 3.8 mol. 4.6 billion years ago about 0.14% of the potassium would have been 40K, i. e. 0.0055 mol or 3.3*1021 atoms. About 89% of the decays of 40K is by electron emission (beta decay; the rest of the decays is by electron capture which I will neglect for this calculation), and the half-life is 1.277 billion years, which means the effective decay constant is 1.54*10-17 s-1. So there are 1.54*10-17*3.3*1021 = 5.0*104 decays per second, each with an energy of 1.311 MeV, resulting in 1.06*10-8W, or 0.33 J/year. This results in 4.5 mSv of equivalence dose per year - that (alone) won't do much harm (it is less than I had hoped for). Do the same calculation for radon (resulting from the uranium (especially 235U) decay chain) and you should get a more dangerous results as the amount of 235U was about 93 times of what it is now. Icek (talk) 22:14, 13 September 2009 (UTC)[reply]
An additional note on radon: According to this, a human receives 1.4 mSv/year of radon on average today. That means you would have receives 130 mSv/year 4.6 billion years ago, which is already above the limit for nuclear plant workers. Icek (talk) 22:19, 13 September 2009 (UTC)[reply]
A further note on the 40K calculation: The received radiation power and therefore equivalence dose power would be lower because not all the energy is carried by the electron in beta decay (some by the neutrino). And regarding the electron capture, it causes some X-rays because the electron is captured from an inner shell and an outer electron will fall to the free space on the inner shell, thereby emitting the energy difference as electromagnetic radiation. Icek (talk) 22:24, 13 September 2009 (UTC)[reply]
drug/biochemical for feeling refreshed
I understand that adenosine is one of the causative agents of drowsiness and sleepiness. Is there a chemical that gives you the feeling of being refreshed, like after a long nap or a really, really good sleep? And of course I'm not talking about a simple adenosine antagonist like caffeine, since I quite suspect that adenosine antagonism doesn't give you that refreshed feeling.
If this chemical exists, in what natural sources could I find it, and is it capable of passing the blood-brain barrier? If this chemical were say, used as a study drug, would there be any long-term ill effects? John Riemann Soong (talk) 07:52, 7 September 2009 (UTC)[reply]
There is no drug that causes you to feel refreshed in the sense of completely substituting for sleep. There is, however, a wide class of drugs that make you feel awake and energized, including amphetamine, cocaine, and other so-called "psychomotor stimulants". Looie496 (talk) 15:27, 7 September 2009 (UTC)[reply]
But those are adenosine antagonists, which I specifically excluded! And I want to exclude compounds that give you a "buzz" or make you feel charged with energy, which is not how I feel when I wake up from a long nap (relaxed). Is the feeling of refreshedness perhaps due to the fact that adenosine is still active (but exiting the system), with various neurostimulants coming in, which induces an interesting competitive effect? John Riemann Soong (talk) 15:57, 7 September 2009 (UTC)[reply]
The hypothesis I favor (which comes from Giulio Tononi) says that sleep actually produces structural changes in the brain. While we are awake, synaptic connections get steadily stronger as we store memories, and this change gradually unbalances the brain's activity patterns. During sleep, a process takes place that gradually weakens the synapses. If that's correct, then there is no obvious way for a drug to duplicate the process. (And by the way, psychomotor stimulants are not adenosine antagonists, they are dopamine agonists, a whole different class of drug.) Looie496 (talk) 19:51, 7 September 2009 (UTC)[reply]
How much noise do moving cars and other vehicles make?
Also depends heavily on the make of the vehicle. I remember the first time I saw a Prius, it was cruising through a grass parking lot using only its electric motor. It came up behind me and passed slowly by - surprised the hell out of me when a large white object entered my peripheral vision because I hadn't heard a damn thing! 218.25.32.210 (talk) 02:04, 8 September 2009 (UTC)[reply]
U,S, and other government websites should have some figures, since in the U.S. associations advocating for blind people are demanding that electric cars make noise so they can hear them coming, and I have read statements by spokesmen for companies planning to manufacture and sell electric cars related to this. See [12], [13],[14]. Maybe they could just require them to put in one of those giant window shaking 200 watt sound systems and play brain thumping noise/music all the time. Edison (talk) 15:50, 8 September 2009 (UTC)[reply]
The following web site [[15]] exhibits a scale with the noise level of a wide range of sources, including a car (65-70 dB) and a diesel truck (85-90 dB). I assume that these would be relatively normal, modern vehicles passing by. I have measured levels of the order of 120 dB for a car with a broken exhaust pipe that sounded more like a very loud (and unpleasant) jet engine at a distance of 2-3 m. At that level, noise can be damaging to your ear especially if endured for sustained periods of time. Michel M Verstraete (talk) 20:06, 9 September 2009 (UTC).[reply]
Decibels or dB are units of ratio and nothing more so it is meaningless to call a noise level X dB. The cited web site encourages that common error by more or less implying that setting 0 dB = 0.00002 Pa or 0 dB = 10-12 W is implicit. That is wrong and we should know this as well as the OP who properly quotes a reference for sound levels, as in dB(SPL). Cuddlyable3 (talk) 13:39, 10 September 2009 (UTC)[reply]
OK, that's true. Any decibel scale implies the existence of a reference level, so a given number of decibels is never an absolute measure of some intensity but an assessment of how much stronger or weaker that intensity is with respect to the implied or explicit standard. This being said, our own Wikipedia article on Sound pressure states that 'The commonly used reference sound pressure in air is p_ref = 20 µPa (rms), which is usually considered the threshold of human hearing'... Michel M Verstraete (talk) 21:37, 10 September 2009 (UTC).[reply]
Bouncing soap bubbles, electrical?
Bubbles colliding with each other in the air typically bounce off of each other -- although they can sometimes hit in a way that causes them to join and assume a more minimal double-bubble configuration and sometimes, more rarely, the two will join as a single larger, and most minimal, sphere.
Soap bubbles typically break if they touch dry wood, metal, stone, plastic or skin. They break if they touch sheer silk fabric or smooth nylon but they typically bounce off of wool and raw silk seems to be excellent for bouncing them as well.
Are these exceptions all examples of electrical repulsion? Is there some other explanation (i.e. the fine hairs of the wool are just not enough contact to pop the bubble, the lanolin in the wool and natural oils in raw silk are a factor?)
I am an entertainer with a show that I call Bubble Magic. I've read C.V. Boys classic work on Soap Bubbles and other science literature but I've never seen a scientific explanation for the apparent repulsion between bubbles and wool or bubbles and other materials. 79.193.226.55 (talk) 10:14, 7 September 2009 (UTC)[reply]
What a good set of questions. It is possible for bubbles to bounce off surfaces without contact or electrical forces just from the mechanics of the air between them. Clearly, static and other electrical forces (even van der Waals) could play a part, so could fibre geometry and surface absorbency. I think the film structure of the bubble is fairly well know (with hydro-phobic tails pointing outwards). One thing to try would be to compare dry cleaned wool with natural wool. There is a huge difference in fire resistance (dry cleaned wool is quite hard to burn, natural wool is easy to burn) so I think dry cleaned wool probably has little lanolin. Have you tried other waxy material? --BozMotalk11:21, 7 September 2009 (UTC)[reply]
I have tried other waxy materials including paraffin, bees wax, and even pure lanolin with no success (the bubbles popped) but the lack of "wetting" does seem to be a factor and so it may contribute in some way. Sheer silk (my Hawaiian shirts) seem to grab and quickly drink the drop at the bottom of the bubble ... then it pops. Once a bubble has popped on, even a good wool surface (navy pea coat) that wet spot is no longer a good place to bounce one.
But the two bubbles approaching each other in the air have a very different surface geometry than does the fine-haired wool carpet so it's difficult to see one answer to the question.
The hydrophobic ends of the soap molecules at the surface, I believe, have no great charge to them ... maybe the nearby water molecules attached to those soap molecules that are now all aligned with the same face (charge) aimed outward? What can you say about the "mechanics of the air between them"? 79.193.226.55 (talk) 11:35, 7 September 2009 (UTC) —Preceding unsigned comment added by 79.193.226.55 (talk) 11:34, 7 September 2009 (UTC)[reply]
In the complete absence of any electrical forces I do not think that two bubbles would ever be able to squeeze out the thin layer of air between them. They would deform, the air would squash and they would spring apart. Even in classical fluid dynamics the equations suggest a smooth ball cannot touch a flat wall because of the air cushion, only surface irregularities allow it to happen. But I suspect there will always be electrical forces. --BozMotalk12:08, 7 September 2009 (UTC)[reply]
For your show of course you should be able to charge up bubbles positively and negatively. Any static charge would dominate. All that requires is a metal hoop to blow through, standing on a nylon carpet and having one earthed boot. A bit of judicious shuffling should do the trick. --BozMotalk12:11, 7 September 2009 (UTC)[reply]
Thank you for that! Simply compressed air could account for much of what we see with or without electrical forces. I'll take that with me as I continue to watch what they do in various situations.
I have charged bubbles and seen dramatic effects as a result. I often perform in science centers and when I can, I get a hold of a Van de Graff generator. If you blow a stream of bubbles to it they will act as though they are being pulled in ... until ... the first one pops and then, suddenly the others reverse course! They take on the charge (presumably delivered to them by the splash of the first one) that is now the same charge as that of the dome of the Van De Graff ... these retreating bubbles invariably chase me since I'm the closest available place to discharge their excess ... I will sometimes dodge them for a few steps but then they catch me like little kamikazees. I'll try the trick with the carpet and boot ... 79.193.226.55 (talk) 12:53, 7 September 2009 (UTC)[reply]
I expect you could get a good effect with a piece of PVC pipe rubbed with a piece of wool. The trick of an uncharged bubble being attracted to the charged object, picking up (or giving up) electrons, then spraying ionized droplets on the other bubbles, which then chase uncharged objects or people, is brilliant. Edison (talk) 21:58, 7 September 2009 (UTC)[reply]
Prions have no genetic material as far as we know, thus, the notion of "growing" is only partially applicable. Prions (specifically, the misfolded prion protein, or PrPSc) induce changes in the folding of otherwise normal cellular protein (PrPC); thus, their propagation in a culture would depend on the abundance and availability (for direct contact) of the PrPC, and the stability of the PrPSc under culture conditions. I don't think anyone has identified an efficient system for detecting PrPSc in simple cell culture (in the way that a shell vial assay {is that really a red link??!!} may be used for detecting viruses). Attempts have been made to develop in vitro tests for prions, such as PMCA and various immunoassays (e.g. PMID19060956). --Scray (talk) 20:24, 7 September 2009 (UTC)[reply]
Not a real dictionary word (yet or probably ever).. But [16] "The test method was developed by British Rail and Mebon Paints and the term Prohesion is derived from a key concept of this development namely Protection is Adhesion."
Sometimes, and yes, respectively. Beta particles are nothing more than electrons (negative or positive), so beta decay absolutely must consider the mass of the electrons. (The inverse process, electron capture, also involves electrons.) Tables of atomic masses always include the mass of electrons for neutral, ground-state atoms. TenOfAllTrades(talk) 15:27, 7 September 2009 (UTC)[reply]
Well, it sort of includes the mass of those electrons. Considering that electrons have roughly 1/1856th (give or take) the mass of a proton, and most atomic masses aren't often reported much past the 4th decimal place, so electrons don't often show up, unless we're dealing with hyper-accurate atomic masses (which are available, but not often used in standard calculations). --Jayron3218:53, 7 September 2009 (UTC)[reply]
I understood the original question somewhat differently... First of all, nuclear forces are relevant only to account for the stability of the nuclei, not to explain the interactions between these nuclei and the electrons of a given atom. Second, there are two main classes of nuclear reactions: fission and fusion. Fission is a process where single individual heavy nuclei (e.g., Uranium, Plutonium, etc) break down in pieces as part of a stochastic process, which can sometime be accelerated by external stimulation (e.g., by neutrons). In some of these cases of disintegration, beta particles (electrons) may be released. Fusion requires multiple light nuclei (e.g., Hydrogen, Helium, etc) to merge together: the old nuclei are now bound together by nuclear forces to form a new, singular nucleus. This process only takes place at very high temperature where materials are already either strongly ionized or in a state of plasma. I am not aware that electrons play any role in stimulating (or hindering) such nuclear reactions. In any case, as already pointed out above, the actual mass of electrons in neutral atoms is minuscule compared to the mass of the nucleus. If these various answers do not answer your question, please clarify or elaborate on what you are looking for. Michel M Verstraete (talk) 19:47, 9 September 2009 (UTC).[reply]
What matters in the reaction is not the total mass of the atoms. It is change of mass that is important (At least if what you are trying to do is to calculate the enery released). In that case the mass of the electrons (beta decays and electron capture reactions), is important to get an accurate answer. Dauto (talk) 15:57, 11 September 2009 (UTC)[reply]
Insect Identification
I was wondering if anyone would be able to identify this insect. It was squashed and captured on a piece of tape.
(Possibly) Pertinent Information:
I found the insect crawling on my desk.
The body measures approximately 3mm in length, and 2mm in width.
I am located in Toronto, Canada. Today's temperature is 21 Celsius, relative humidity @ 73%.
Images:
I will refrain from embedding the images as they are quite large and may break the page.
However, the body of the German cockroach measures at least 1.3cm long (according to the wikipedia article). The body of the sample I captured measures around 0.3cm long.
Impact of the type of cigarettes smoked on the decrease in life expectancy
I have a hard time finding numbers concerning the decrease in human life expectancy caused by smoking which numerically examines the differences between the type of product used. I am particularly interested in the difference between average-yield cigarettes with filters and unfiltered and/or high-yield cigarettes. I found an AP article concerning a study conducted in India which claimed that the decrease was roughly 20 years for long-term smokers of unfiltered "bidi" type cigarettes. One must however consider that bidi smokers usually don't have access to healthcare which as good as that commonly available in countries such as the UK - hence the difference might be much lower in the UK, for example. I suppose that the decrease caused by unfiltered cigarettes in "developed countries" is in the range of 11 to 16 years. Thanks for your help. —Preceding unsigned comment added by Binrapt (talk • contribs) 16:57, 7 September 2009 (UTC)[reply]
Careful there. The lack of health care cuts both way. Life expectancy in India is about 10 years below life expectancy in the UK. So people will die earlier even without cigarettes - in other words, any particular harmful custom has less lifespan to take away in the first place. If you die at age 3 from amoebic diarrhea, you cannot die of tobacco-caused vascular diseases at age 60. --Stephan Schulz (talk) 20:13, 7 September 2009 (UTC)[reply]
Anemia/obesity connection
Don't worry, this isn't a request for medical advice; I'm just curious. Is there any kind of anemia that is associated with / is exacerbated by / tends to occur with obesity? +Angr20:29, 7 September 2009 (UTC)[reply]
Well, at the risk of sounding as though I'm not taking the question seriously, you do see these 600lb+ guys on talkshows who seem to forget just how much food they consume on a daily basis when asked directly - claiming that they 'don't eat more than anyone else' and that there must be some sort of genetic/glandular cause for their obesity... --Kurt Shaped Box (talk) 22:29, 7 September 2009 (UTC)[reply]
I also misread it exactly the same way. It is like a magic trick where the audience is misdirected by expectation or "attentional set." Edison (talk) 15:37, 8 September 2009 (UTC)[reply]
It's plausible. I vaguely remember reading about malnutrition and obesity, since people who don't manage their diets well probably aren't eating the right stuff either. But you proably knew that already. 98.14.222.125 (talk) 23:39, 7 September 2009 (UTC)[reply]
No. Type II diabetes/metabolic syndrome is associated with increased ferritin levels, which is a surrogate marker for iron overload, which is negatively associated (weakly!) with anaemia. See [17]. --NorwegianBluetalk20:31, 8 September 2009 (UTC)[reply]
It's been suggested, but it doesn't seem to be the case: "Overweight and obesity were associated with changes in serum iron, TS, and ferritin that would be expected to occur in the setting of chronic, systemic inflammation. However, overweight and obese persons were not more likely to be anemic compared with normal-weight persons."[18]Fences&Windows04:08, 9 September 2009 (UTC)[reply]
Snapping a street light by repeated poking with finger
Is it true that you could snap a street light - I mean a street light on a long mostly vertical pole or tube - by repeatedly pushing it gently with your finger at its resonant frequency? 89.242.115.9 (talk) 20:32, 7 September 2009 (UTC)[reply]
I once saw a group of three college guys "tackling" a sequence of metal light poles by all three simultaneously bumping the pole with their shoulders (one high, one low, one in the middle). This resulted in a wave of energy travelling swiftly up the pole, and a sharp mechanical snap when the weave reached the top, knocking out the light. They scored 4 out of 4 that I saw. There was not a huge amount of force, but the transient shock at the top was impressive. There are legends (doubtful) of Nikola Tesla shattering bridges or buildings with mechanical resonators. I would not rule out the possibility of say a 2 pound force repeatedly applied at a resonant frequency causing resonant motion of a metal light pole, if it is a modern flexible one. An old cast iron one or one on a wooden pole? No way. The designers should have damped out such resonance, to avoid large amplitude resonance in wind, but see Tacoma Narrows Bridge (1940) for a famous example of failure to deal with resonance in a bridge. Edison (talk) 21:56, 7 September 2009 (UTC)[reply]
Well, what is "snapping" and what is "gently"? You can certainly excite it to fairly violent movement with repeated pushing with one hand (and if I can do it with my hand, Big Van Vader can do it with a finger). It will not snap the pole, but it may shed some of the glass components on top. --Stephan Schulz (talk) 21:58, 7 September 2009 (UTC)[reply]
The idea that you can break anything by gently stimulating it at its resonant frequency is deeply flawed. The biggest problem is that each time you 'poke' the object, you have to put in more energy than the object can dissipate during the next cycle of oscillation. If you don't put in that minimum amount, the oscillation cannot grow. Tesla was a clever guy - but he had a tendency to be taken in by simple ideas and to blow them out of all proportions - then to lie about his results, reporting what he THOUGHT they should be - not what they actually were. He had enough successes to become famous - even revered - but in many ways he was a complete nut-job. His theories on resonance are an excellent example of that. The basic idea is true - there are things that will resonate to destruction (the canonical example being the opera singer shattering a glass with nothing more than her voice)...but that doesn't translate to being able to destroy massive objects with tiny forces. SteveBaker (talk) 02:30, 8 September 2009 (UTC)[reply]
Tesla also thought you could transmit electricity through the earth, as I recall. That's not a very practical idea. Edison got a few things wrong too. Both of those guys are examples of the principle that in order to get good ideas, you have to have lots of ideas - some of which will be wrong. Edison, for example, promoted DC while Tesla promoted AC. Guess who won that battle. Baseball BugsWhat's up, Doc?carrots15:59, 8 September 2009 (UTC)[reply]
When I was a kid, we used to knock out lights in parks and playgrounds by kicking the pole behind the maintenance hatch. We didn't know then what exactly was going on, but I suppose there was a relay attached to the inside wall of the pole opposite the hatch, and a sharp kick would break the circuit for just long enough that the mercury vapour lamp would then need to restart, which takes a few minutes.--Rallette (talk) 06:09, 8 September 2009 (UTC)[reply]
Not sure if they had it when Rallette was a kid but don't modern street lights have a collision detection mechanism that cuts the power when it's hit by something (ie. car)? --antilivedT | C | G11:18, 8 September 2009 (UTC)[reply]
It would be quite possible to design a metal lamp post which was quite tall, which was strong enough to support itself and a large wind loading, but was underdamped, if that made it a buck cheaper and it was not a specified quality which was tested for by the purchaser. The same folks who put lead in childrens' products, antifreeze in tooth paste, make cribs which strangle babies, build school buildings which collapse in low level earth tremors, and put melamine in pet food, might just market underdamped lightposts. Edison (talk) 15:45, 8 September 2009 (UTC)[reply]
I was thinking of poking the metal pole when it is at the end of its movement, like pushing a child's swing at the end of its swing. The pole would break through metal fatigue, as when you bend a piece of wire such as a paperclip back and forth. 78.147.7.217 (talk) 21:34, 8 September 2009 (UTC)[reply]
What's the technical term for a fight between two animals where neither combatant is fighting to its full capability in order to ensure that neither is killed or permanently maimed (with the rationale that 'if I go all out on him, he'll go all out on me and we'll kill each other')?
I've been watching gull videos on YouTube again tonight and I've discovered more examples of the previously-discussed 'tug-o-war beak wrestling' - e.g. here and here and I had a moment of realization that that purpose of these conflicts may actually be to ensure that neither bird can attack with its beak (because, let's face it - two gulls could rip each other up pretty badly if they really went at it), whilst both attempt to overpower the other and get him into a position where he can be 'submitted' and forced to disengage and fly away - all with minimal bloodshed. Kind of like a gull version of our submission wrestling, where the object is to get the other guy to tap out, rather than put him into intensive care. I don't know if this is obvious to anyone else - but it never really clicked with me until now...
I've looked for a WP article discussing this behaviour (in animals in general - not just gulls) but all I've found is references to human combat. Any ideas, guys? --Kurt Shaped Box (talk) 01:55, 8 September 2009 (UTC)[reply]
The term that occurs to me is "ritual combat", and I'm pretty sure that's what ethologists call it, however our ritual combat is a redirect to an article that doesn't mention animals. Looie496 (talk) 02:28, 8 September 2009 (UTC)[reply]
Well, it's probably a form of intrasexual selection, though that isn't the name of the combat itself. It's of note that almost all intraspecies fighting is of the nature you describe—it is rarely meant to be fatal, and instead is about posturing, submission, etc. (Interspecies is a totally different question.) There is a lot of discussion of this in On Killing (specifically, that there are other options than "fight" and "flight"—"posturing" and "submission"). --98.217.14.211 (talk) 02:50, 8 September 2009 (UTC)[reply]
Perhaps relevant aphorism: I read once that -- due to the large difference in how serious their weaponry is -- when fighting for mates, "doves are hawks and hawks are doves". --Sean16:22, 8 September 2009 (UTC)[reply]
Funnily enough, I saw a video on YouTube of two hawks fighting last year - it was less violent than in the gulls. One hawk had the other one pinned on its back and proceeded to gently 'beak tap' it until it gave up. I suppose the perception that "I've got you at my mercy and I am now in the position to deliver a fatal blow if I so wished" is enough to bring the fight to an end. It looks to me as though that's what happened in the second gull video - when one bird had the other by the neck/throat. --Kurt Shaped Box (talk) 20:17, 8 September 2009 (UTC)[reply]
Based on the animal videos I've seen, the same thing goes for wolves, lions, and other large predators. So yeah, I suppose that's because they're more likely to kill one another during ritual combat than other animals. Funny thing, though, in Jack London's Call of the Wild, dogs and wolves do fight one another to the death (and I mean dogs vs. dogs or wolves vs. wolves, not wolves vs. dogs). That's a pretty odd thing, considering that he actually did spend a long time in the far north and prob'ly witnessed enough dog-fights to become an expert on the subject. 98.234.126.251 (talk) 05:59, 9 September 2009 (UTC)[reply]
I think Dawkins mentioned this in The Selfish Gene, though I can't find a definite attribution. IIRC, his example was in rattlesnakes, who rarely, if ever, strike one another when fighting over a mate. A single bite would almost certainly be deadly, yet both combatants refrain and therefore increase their risk of losing, but increase their odds of living. In a way, this is a kind of invisible Green-beard effect. That's part of the gene-centered view of evolution, though, so other theorists will have different explanations. Matt Deres (talk) 16:37, 8 September 2009 (UTC)[reply]
Just like us people, we gotta pick fights every once in a while to blow off steam, else we might explode. (BTW, the same mechanism might be operating on an international scale as well -- and in that case, efforts to achieve "world peace" would actually do more harm than good.) 146.74.230.106 (talk) 21:36, 10 September 2009 (UTC)[reply]
Space shuttle landing stuff
In the article, Space Shuttle Columbia disaster, in the section "Destruction during re-entry," it gives a sort of play by play of the events of the landing up to and including the disintegration of the orbiter. Like this:
* 8:53:26 (EI+557) – Columbia crossed the California coast west of Sacramento. Speed: Mach 23; altitude: 231,600 feet (70.6 km; 43.86 mi).
Anyway, is there anything similar out there for one of the sucessful landings? I've looked but I can't find anything. I'm really amazed that the Columbia made it across California in a minute and a half, and I got so into looking at the data that I nearly forgot that this mission was going to end in disaster. It seems that this list was made for some commission looking into the accident, so maybe no one has made anything for one of the many times that things went right.
Thanks for your time. I always appreciate the effort that goes into anwering these questions.
The ill-fated mission didn't result in the shuttle re-entering any faster or slower than it should have done - so the timeline for a successful mission would be pretty similar. However, they don't always come in on the exact same track - so there would be differences of location - if not of speed. http://spaceflight.nasa.gov/realdata/sightings/ offers data for an upcoming landing - in order that amateur 'shuttle-spotters' will know when and where to look. I don't think that's precisely what you want but the interactive applet has a gazillion options for finding out where pretty much any NASA vehicle is at any time - or when it will cross a particular state. That MAY enable you to figure out what you want to know. SteveBaker (talk) 10:58, 8 September 2009 (UTC)[reply]
You may want to login into the NASA system and ask them that question directly: See 42. You may have to wait for 2 weeks to get an answer, but you should get reliable information. Michel M Verstraete (talk) 19:19, 9 September 2009 (UTC).[reply]
The NASA TV channel worked for me last night - at 8:11pm central time the shuttle and space station zipped across the Texas skies right on cue. The sun had only just set and the sky was still pretty bright - yet they were easily the brightest objects in the sky - crossing from horizon to horizon in just a few minutes. The shuttle was about a hand's-breadth in front of the space station and you could easily tell them apart with the naked eye. With binoculars, you can see the big solar panels and really get a feel for just how gigantic the ISS has become. Very impressive...also a complete waste of money - but that's another story! SteveBaker (talk) 13:43, 9 September 2009 (UTC)[reply]
Thanks for all the responses. I have gotten ever closer to the info I want, and if I put it together I should have something interesting. Maybe I should share it? —Preceding unsigned comment added by 169.231.32.17 (talk) 08:59, 10 September 2009 (UTC)[reply]
Hello.
I am wondering if anyone can direct me to a manufacturer of equipment for quickly drying a powder that is highly heat sensitive. The powder must remain below 30 C. Our current method is; after reprecipitating the product using water to dissolve and acetone to crash out for purification, we air dry in a humidity-reduced room, then dry under vacuum in pans until the product is a consistent weight. Some method of drying the product more quickly without heating it is necessary. A small amount of water hydration will remain, but the excess moisture needs to be removed. Thank you in advance for your help. 134.217.112.15 (talk) 15:49, 8 September 2009 (UTC)[reply]
Agitate the pans, and heat them to 30 C (or whatever safe temperature is below the decomposition temperature of your product). The evaporation of water from the material will tend to chill the remaining material, reducing the rate of further evaporation (even under vacuum). Agitation ensures that water remains uniformly distributed over the exposed surface (maximizing evaporation), and will encourage the breakup of water-retaining clumps. Sifting through screens or shaking in the presence of inert metallic or ceramic balls will have a similar effect. TenOfAllTrades(talk) 22:32, 8 September 2009 (UTC)[reply]
You could put it in a round-bottom flask and dry it under vacuum in a warm-water bath, with constant agitation; that's the method we use in our lab for drying heat-sensitive materials. Caution: drawing too much of a vacuum could chill the flask so much that it condenses moisture from the air and rehydrates the material. You should experiment with different suction pressure / water temperature combinations to see which one gives the shortest drying time without condensing moisture from the air. 98.234.126.251 (talk) 05:36, 9 September 2009 (UTC)[reply]
Have you investigated Fluidized bed techniques? Although these seem more commonly used for other purposes, it seems to me that they might also be efficacious for this one. Try searching the technical literature using this term. Actually, some of TenOfAllTrades's procedures may approach this area. 87.81.230.195 (talk) 19:58, 9 September 2009 (UTC)[reply]
Perfectly legal, I came here first in the course of researching companies to replace our current drying equipment. I would like to keep the product being made confidential but it is a reagent used in medical diagnostics. The equipment we use consists of a vacuum oven, which we do not heat, and acetone/dry ice-cooled moisture traps. This process takes 1-2 weeks for each batch to dry, which we are trying to find a way to speed up. Agitation while drying under vacuum would probably be best, but we are looking for specific manufacturers in the US which provide this type of equipment. I am sorry if the first post did not make that clear. To answer some questions above: we cannot heat the flask/oven due to the heat-sensitive nature of the product. We do currently use vacuum drying, but do not have equipment to agitate the product while doing so. The product is water soluble, and very hydrophilic, so drying is complicated by the tendancy of the product to absorb water from the air. Again, we are simply looking for suggestions of manufacturers who make vacuum drying equipment which agitates and possibly grinds the product while drying. Thank you. 134.217.112.15 (talk) 14:13, 10 September 2009 (UTC)[reply]
What you're looking for is a rotary evaporator -- unfortunately I don't remember off the top of my head which company made the one that's in our lab, and I didn't take a close look at the company logo. (In the lab, all of us just refer to it as "the rotavap machine".) Maybe if I have time, I could google some chemical equipment manufacturers to see if they make this kind of equipment. Please stand by, and keep this channel open. 98.234.126.251 (talk) 05:47, 11 September 2009 (UTC)[reply]
Apparently, people with a humanistic background tend to consider Aristotle as the [] of science. On the other hand, people versed in the natural science consider Galileo the first scientist.--Mr.K.(talk)16:07, 8 September 2009 (UTC)[reply]
It depends largely on what you define "science" as being. Strictly speaking, Aristotle bears little resemblance to a "scientist" by a primarily modern definition (he explained how he thought the world worked, and paid some attention to natural history, but he exhibited nothing like an empiricist ethos). Galileo looks a LOT more like a "scientist" than Aristotle did, though calling him the "first" is problematic (he was not totally novel, just very famous). In any case, the "science" of the Early Modern (Scientific Revolution) period looks a lot more like modern science than does Ancient "science", though even the Early Modern stuff looks pretty quaint compared to, say, 19th-century science (which is for all intents and purposes "modern", even though many of the things we currently associate with science are firmly rooted in the late 20th century). --98.217.14.211 (talk) 16:31, 8 September 2009 (UTC)[reply]
It really depends on what tradition you are following. There were lots of early Arabic chemists, for example, doing real "Science" as before 1000 (see Muhammad ibn Zakariya al-Razi (aka Rhazes) in the 900's and Geber in the 700s. Even in Europe, Roger Bacon set out the basic precepts of the scientific method in the 13th century, 250 years before Galileo. If ANYONE is the father of modern western scientific thought, Roger Bacon is probably it, but even then I wouldn't make that assertion too strongly. --Jayron3221:04, 8 September 2009 (UTC)[reply]
The greeks frowned on doing experiments - believing that you could simply think up all of the answers from first principles. That kind of thing works great for mathematics - but sucks when it comes to the physical sciences. Aristotle wasn't a scientist by any modern definition. SteveBaker (talk) 21:16, 8 September 2009 (UTC)[reply]
I agree generally, but there are important exceptions. Hippocrates (who might actually be a group of people, but it doesn't matter) was an amazing observer of empirical facts; and Archimedes was certainly open to experimenting. Looie496 (talk) 23:47, 8 September 2009 (UTC)[reply]
Aristotle seems a pompous fool. No one person delayed the rise of the scientific method more than "the philosopher," as Aristotle was known many centuries after his time. Galileo was an empiricist and scientist. Edison (talk) 02:51, 9 September 2009 (UTC)[reply]
Unfortunately, Plato helped justify anti-science book burnings in later centuries by early Christians. Democritus, for example, was quite a materialist but hardly any of his writings has survived. 98.14.222.248 (talk) 22:13, 10 September 2009 (UTC)[reply]
For woman, does size matter?
I don't know many couples where the man is shorter than the woman. A well-known exception is Sarkozy and Bruni. Is that so important that the man is taller? On the other hand, I know several couples where the woman is making more money, is more intelligent, has more culture than the man. Mr.K.(talk)16:09, 8 September 2009 (UTC)[reply]
I know plenty of couples like that - assuming Mr.K. means an interest in the arts, etc. by "having culture". Why do you ask? --Tango (talk) 00:33, 9 September 2009 (UTC)[reply]
I was just interested in seeing some actual examples. But you may be right — if we define culture as meaning an interest in the arts, then surely some examples can be found. While that may be the most common and basic definition of culture, it is by no means the only definition of culture. As we well know, culture often embraces the decidedly un-cultural. Maybe I'm nitpicking or being pedantic. But much of what passes for culture challenges our definition of culture. Bus stop (talk) 00:55, 9 September 2009 (UTC)[reply]
I agree that culture is far more than just the arts, but in the context of a person "having culture" I can't think of anything else that could be intended. --Tango (talk) 01:19, 9 September 2009 (UTC)[reply]
Two facts come into play here. First, because men are generally taller than women, the great majority of couples would have a taller man even if pairing were completely random. Second, there is (with many exceptions) a broad statistical association between tallness and social dominance, and a broad statistical tendency for men to be dominant (in the technical sense of the word) in relationships. Looie496 (talk) 18:31, 8 September 2009 (UTC)[reply]
Males are typically taller, but I think even that is possibly a RESULT of the thousands of years of selective breeding, not the cause of it. I think since women are the ones that give birth and breastfeed they are vulnerable at those times and so males play the role of the provider and protector. Males and females instinctually look for signs of health and virility in a partner, basically how suitable they will be for breeding with. A lot of what we find "attractive" these days stems from those ancient subconscious instincts. Some of those factors now get distorted, warped and manipulated by modern society since we no longer need to rely on "basic instincts" for survival. I think height is definitely one of the factors used as a indicator of health and vigour specifically in males. It's the iconic superman, fit and upright. In females the main factors are typically youth, breasts and hips, height is not necessary to make a good mother. In many other species the female is the bigger of the 2, I don't think there is any reason why female humans couldn't be the bigger gender on average if we bred that way for thousands of years, I think that baby nursing is the thing that prevented that from happening. Now I think there is definitely more of a trend away from the "classic" large hourglass shaped mother figure. It used to be that unless a woman had big hips she was much less likely to survive her first child birth, these days with modern paediatrics pretty much any woman can give birth relatively safely, so being skinny and tall is no longer a breeding liability. To specifically answer the question, no i do not believe it is "important" but it is a very strong left over instinct for a lot of humans. There will be expeptions to the rules as there always are, and the rules will slowly shift over time, but anyone claiming it is "important" is doing so based on reasons of purely human construct. Just like the belief that it is important to marry into your race. Vespine (talk) 23:08, 8 September 2009 (UTC)[reply]
"...signs of health and virility", yep. I once knew a couple who got married and between them probably maxed out the weight limit on a typic elevator. There's really no accounting for attractiveness. There's no absolute rule. Baseball BugsWhat's up, Doc?carrots00:13, 9 September 2009 (UTC)[reply]
Of course I wasn't suggesting there were any absolute rules, but that's a far cry from claiming there are no observable trends. Otherwise advertising wouldn't so heavily rely on sex to sell things. And as the above points out, in the scheme of things it makes perfect sense for two obese people to marry. The scenario which does not follow the trend is if a skinny person marries an obese person, but even the fact that this happens sometimes doesn't disprove the "trend", deviations from the norm are perfectly expected in any biological system, what would truly be strange is if there were no deviations from the norm. And I use the term "norm" in the purely statistical meaning of "average", not as an antonym of abnormal, that's an entierly different discussion. Maybe it is because people associate "norm" with good and "outside the norm" with bad that they are reluctant to admit the observable trends, but I think that's a fallacious association. "The norm" is not intrinsically superior to "outside the norm", it just means "average". Vespine (talk) 23:53, 9 September 2009 (UTC)[reply]
Aversion and fear
If you are afraid of something - dog, plane, spider or whatever - and you avoid any contact with or even approaching it, does your fear get bigger? If yes, why does it grow? Logically, if you don't have any contact with something, shouldn't the tendency be to loose any feeling towards it? --Mr.K.(talk)16:10, 8 September 2009 (UTC)[reply]
I don't know if it would grow, but there's a good evolutionary reason why it wouldn't lessen over time (or at least not quickly). Let's say you come across an animal you've never seen before in the woods, and it nearly eats you. Even if you don't come across another one for years, if one day you do, it's important for your survival that you vividly remember your terrifying experience and stay away or else this time you might be lunch. That fear instinct is very important, so it can be hard to overcome even if we know rationally that a certain fear is illogical. Rckrone (talk) 18:02, 8 September 2009 (UTC)[reply]
The thing is that for somebody with a phobia, you think about the thing even if you don't have contact with it. A person with a spider phobia can lie in bed imagining that a spider might be about to crawl on them, and the idea can be so vivid that it is as fearsome as the actual event. This probably isn't a functionally useful phenomenon; it comes from having an overactive fear system. Looie496 (talk) 18:15, 8 September 2009 (UTC)[reply]
On the contrary — 'refreshing' the memory that way may be one of the mechanisms by which a person retains (evolutionarily-useful) responses to seldom-occuring but dangerous aspects of their environment (per Rckrone). I forget the original author, but I once heard an anology along the lines of: "Consider a small mammal living in the forest. Every time he sees a shadow move, he jumps. Nine times out of ten, it's the wind in the leaves, but one time out of ten it's a lion. The reflex action is still useful, because the penalty for jumping when the leaves twitch is much smaller than the penalty for not jumping when a lion pounces." Obviously, one can be too sensitive, to the point where one is leaping at shadows all the time and not eating; that's going to take an evolutionary penalty. But it's not clear where to draw the line between useful wariness and harmful paranoia. TenOfAllTrades(talk) 19:38, 8 September 2009 (UTC)[reply]
It could also be what psychologists call superstitious behavior. Running and hiding when you hear a noise averted the supposed danger. You run and hide everytime you hear a similar noise, and, lo and behold you angain "avoid being eaten," reinforcing the connection between fleeing the noise and staying uneaten. The counteracting principles are habituation and experimental extinction (of the learned response, not the wee fleeing creature). In humans, Cognitive behavioral therapy is a modern technique for lessening such self perpetuating associations.Edison (talk) 22:28, 8 September 2009 (UTC)[reply]
I think that's basically a different way to saying the same thing that I did. Concerning evolutionary utility, that's valid to a degree: there is data showing that phobias tend to focus on things that it is useful for monkeys to fear, for example snake and spider phobias are much more common than gun phobias. But it's important to keep in mind that not everything evolution produces is functional -- because each individual is a random mix of genes from the population, every so often you're going to get a combination of things that don't fit together very well. Looie496 (talk) 23:55, 8 September 2009 (UTC)[reply]
Having no fear can get you killed. Just ask Steve Irwin. Fear of spiders and snakes makes basic sense. These are two types of creatures that can hurt you, and if you avoid them you're less likely to get a venomous bite. The same principle, by the way, applies to speaking in public. Baseball BugsWhat's up, Doc?carrots00:10, 9 September 2009 (UTC)[reply]
Irwin did not die because he had no fear. Stingrays are not a creature that are particularly worthy of fear. I bet more people are killed by horses each year then stingrays. Googlemeister (talk) 14:17, 9 September 2009 (UTC)[reply]
Standard therapy for phobias of all kinds is gradual supervised contact with the feared object. Exposure builds up "evidence" that the item is harmless, thus outweighing the fear (based on a real experience or not) that it is dangerous. Constantly avoiding the feared object gives the person a one-sided view, reinforced by each occasion of deliberate avoidance. (If you don't leave home because you believe there are man-eating tigers in your (Northern Hemisphere densely populated) street, you will never find out that there are none). - KoolerStill (talk) 15:04, 9 September 2009 (UTC)[reply]
how to pump a longboard (skateboard)
i have heard that it is possible to "pump" a long board. in doing so you can basically move your longboard forward without ever having to push with your foot. when i heard this was possible i searched around the internet trying to find a good explentation on how to do this. i found websites but when i tryed i waas unsuccesfull. i figured if i knew exaclty why this is possible i would understand it better and might be able to pump my longboard. so exactly why is it possible? does it have to do with your momentum or is it something else? it would also help if someone could sugest a really good web site where i could get further explinations on how to preform this.
thanks —Preceding unsigned comment added by 76.235.215.200 (talk) 17:45, 8 September 2009 (UTC)[reply]
It is just momentum shifting, like using a swing or twirling a hula hoop. I was going to go into a detailed description, but thought I'd check to see if Google was nice and found one on ehow here (can't link straight to ehow - it is blocked). -- kainaw™18:11, 8 September 2009 (UTC)[reply]
Note: this is presented as observation and not advice, since falling is fairly likely while practicing a new technique. It reminds me of a way of skating along while keeping the iceskates parallel, but shifting from one edge to the other. On the board if the axis of the board points to the right, then a push to the right by dropping a bit and rising back up, will result in some forward resultant. Flip the board a bit to the left and pump left. Edison (talk) 22:23, 8 September 2009 (UTC)[reply]
Asian (Indian) elephants and African elephants
Often in the media, we see Asian (Indian) elephants being used as a beast of burden (and other positive interactions with humans). However, I have never seen the same for African elephants. Is there a reason for this? Are Asian elephants easier to "domesticate" while the African elephant can not? —Preceding unsigned comment added by 69.77.185.91 (talk) 18:36, 8 September 2009 (UTC)[reply]
(after ec)According to this - "When the Belgians in colonial times intended to stub the jungle of Belgian Congo, they employed Indian mahouts, who successfully caught and trained the African elephants according to the Asian method. Nowadays safaris on the backs of African riding elephants become more and more popular in Africa". This article says that despite the common belief in Africa that they're untamable, the African Elephant is "with patience, quite trainable" and also notes that Hannibal battled the Romans with AEs. Well, I don't suppose that he would've have done that unless they could be reliably trained not to panic and run amok when the javelins started flying and indiscriminately trample both friend and foe on the battlefield... --Kurt Shaped Box (talk) 19:58, 8 September 2009 (UTC)[reply]
There also may be the consideration as to whether the benefits of African Elephant's extra strength and work capability when compared to other beasts of burden justified its huge daily food requirements. --Kurt Shaped Box (talk) 21:03, 8 September 2009 (UTC)[reply]
Well, throw enough spears at any large animal and it's eventually going to lose its cool - but the point that I was trying to make is that unless the elephants could be conditioned not to behave like this for the most part, they'd be absolutely useless as war mounts and a complete liability on the battlefield. The fact that it was even possible to lead them into battle in the first place shows that they can be trained... --Kurt Shaped Box (talk) 15:02, 9 September 2009 (UTC)[reply]
The point I was making was that although they can be trained to some extent, they frequently have been a liability on the battle-field. In contrast, I don't think (arguing somewhat from ignorance here) horses have been criticised as much for this. AlmostReadytoFly (talk) 15:32, 9 September 2009 (UTC)[reply]
Agreed there. A panicked, out of control horse isn't going to accidentally (or deliberately) kill/maim 15 guys in the space of a minute or so as it tries to flee. As a side issue, I wonder how rogue elephants on the battlefield were dealt with in ancient times? Did the soldiers try to kill the beast, or did they get out of its way as fast as possible and let it run off into the distance (then perhaps attempt to catch it again later once it had had time to calm down)? --Kurt Shaped Box (talk) 20:21, 9 September 2009 (UTC)[reply]
Opposing armies probably let them run away, but the army using them wouldn't want their own troops crushed/killed. The War elephant article says "The driver, called a mahout, was responsible for controlling the animal. In many armies, the mahout also carried a chisel-blade and a hammer to cut through the spinal cord and kill the animal if the elephant went berserk." 12.34.246.72 (talk) 15:38, 11 September 2009 (UTC)[reply]
Be careful not to confuse "domestication" and "taming". Strictly, a species is only domesticated when it is bred in captivity and the captive-bred population shows physical and/or mental modifications from the wild population. By contrast, wild-breeding animals can be tamed, and trained, to the extent that their species' and their individual innate natures render them amenable to being so.
In all history, only the long-vanished Indus Valley Civilization (aka the Harrapans) are known to have successfully domesticated the (Indian) Elephant. Otherwise, elephants have been successfully bred in captivity only rarely, and most of the (mostly female) working elephants in India and elsewhere have been caught and tamed, although some have been born to tamed females allowed to mate with wild males. Our articles on Domestication and on the Asian Elephant are guilty of some loose and inconsistant language in this regard. 87.81.230.195 (talk) 19:28, 9 September 2009 (UTC)[reply]
As a separate point, note that although African elephants were used as war elephants in the Mediterranean sphere (to which Asian Elephants were also imported), these were usually the North African Forest Elephant, a species or sub-species now extinct (not least because of this exploitation) which was somewhat smaller and probably more tameable than the surviving African Forest Elephant and even larger African Bush or Savannah Elephant. 87.81.230.195 (talk) 19:41, 9 September 2009 (UTC)[reply]
Biochemistry question: What does oxygen do in this case?
After oxidating glucose into CO2 and water, oxygen helps it become stable because of the oxygen even though the accumulative G0 value is largely NEGATIVE...how's that possible? —Preceding unsigned comment added by 93.183.239.108 (talk) 19:25, 8 September 2009 (UTC)[reply]
Per the lead paragraph of our article on energy density, "hydrogen has a higher energy density per unit mass than does gasoline, but a much lower energy density per unit volume." However, for the strictest possible interpretation of "most efficient manner" (that is, matter-antimatter annihilation), they'd be identical, as mass is the only consideration in such a case. — Lomn00:43, 9 September 2009 (UTC)[reply]
I think the OP meant the most amount of work you can extract in the most efficient manner from those 2 - otherwise efficiency doesn't mean much (as Lomn pointed out). --antilivedT | C | G01:14, 9 September 2009 (UTC)[reply]
The work done by an engine typically comes from heat energy produced in some exothermic reaction, so you could do work by annihilating hydrogen or gasoline just as you could by burning them (and a lot more of it). Rckrone (talk) 03:11, 9 September 2009 (UTC)[reply]
Which is why we engineers say: "What a bloody stupid answer, the OP is CLEARLY not talking about matter/antimatter or direct matter-to-energy conversion."...this is merely confusing an otherwise clear answer - which is "Hydrogen, by mass. Gasoline, by volume." SteveBaker (talk) 13:34, 9 September 2009 (UTC)[reply]
Steve, I'm not sure if I should take offense at this or not (particularly as an engineer). I led the original answer with precisely "hydrogen by mass, gasoline by volume", and then added antimatter as the addendum. — Lomn16:27, 9 September 2009 (UTC)[reply]
Then the question devolves into a request for clarification about the intended meaning of "efficiency" - if the OP meant thermodynamic efficiency, then Lomn's theoretical answer is valid. If the OP meant "practical, realizable, efficient combustion in an engine, in 2009", then clearly an analysis of matter-antimatter reactions is irrelevant. Again, the practical considerations start to become convoluted by details other than the raw thermal energy released from oxygen-fuel combustion: does manufacturing and carrying around the necessary, custom equipment to perform hydrogen combustion count against the overall "efficiency" score, or are we again only talking about the net energy released? Fuel efficiency will suffer, because 1 kg of hydrogen can not be stored as easily (ergo, as "efficiently") as 1 kg of gasoline. The OP did not state explicitly, but I think their question was implicitly asking for the molar heat of combustion, normalized by the mass-per-mole, for oxygen-fuel combustion of 1 kg each. The answer to this question was helpfully provided and referenced below, by 83.100.250.79. Nimur (talk) 17:50, 9 September 2009 (UTC)[reply]
I would combine a unit mass of hydrogen plus unlimited oxygen for the one case, and the same unit mass of gasoline plus oxygen for the other case. Chemists are welcome to chime in with respect to the energy liberated in each case. Naively, I would bet on hydrogen. Edison (talk) 02:40, 9 September 2009 (UTC)[reply]
And just for fun semtex exploding gives about 4 MJ/kg. But of course the clever thing about semtex was not the energy yield but the fact is couldn't be detected by a sniffer dog... --BozMotalk13:20, 9 September 2009 (UTC)[reply]
Nerves and eyes
Pathways from eye to brain, with midline-crossing pathways in blue, and ipsilaterally projecting pathways in red.
Most parts of the body are connected to the "other" side of the brain, so the right leg is handled by the left brain, et cetera... I know this is different for the eyes, but are they entirely the same side, do the muscles go with one and the optic nerve with the other, or some combination? SDY (talk) 02:17, 9 September 2009 (UTC)[reply]
Through the marvel of hemidecussation, the left half of each retina goes to the left hemisphere and contrariwise. It is said to be a sharp split at the center of each retina. Not sure how the fovea centralis splits, but presumably the same. This shows up in the Michael Gazzaniga experiments with split brain patients. If an object is to your left, it is processed by the right hemisphere, and the left hemisphere knows of it only if the corpus callosum has not been severed. Optic fibers cross over at the optic chiasma. In movies of split-brain patients, the hemisphere controlling speech would produce verbalizations which were mere guesses regarding visual stimuli presented to the other hemisphere, but the non-speaking hemisphere would cross-cue by shaking the head, whereupon the speaking hemisphere would change to the answer the nonspeaking hemisphere had seen. We seem to have two brains, each quite capable on its own. Edison (talk) 02:33, 9 September 2009 (UTC)[reply]
(ec) Our article on the optic chiasm explains this, although not very well. Basically, both eyes send some input to both sides of the brain. The way I always remember it is each side of the brain gets input from the opposite side of the world. The right side of the brain gets input from the left side of the world; the left side of the brain gets input from the right side of the world. If you work it out, the consequence is that the left side of the retina in both eyes projects to the left side of the brain, and the right side of the retina in both eyes projects to the right side of the brain. This arrangement may seem bizarre at first, but it has the very useful property of causing the visual input to each side of the brain to come from the same place where tactile input to that side usually comes from, and where motor output from that side goes. Looie496 (talk) 02:41, 9 September 2009 (UTC)[reply]
(ec again, yikes) The "final common pathway" for eye movement is a set of small brainstem nuclei, but huge areas of the brain are involved in eye movement control. By the way this picture is a beautiful illustration of the left-right pathways, too bad we don't have one as nice. Looie496 (talk) 02:53, 9 September 2009 (UTC)[reply]
It might be useful to note here that while one-sided numbness and paralysis caused by stroke is evidence of a defect in the opposite side of the brain, blindness in either eye caused by stroke is associated with a defect in the same side of the body as the eye.Mrdeath5493 (talk) 07:12, 9 September 2009 (UTC)[reply]
They do have that effect. Most of the mass of ordinary matter (~90%?) comes from the strong force, and most of the rest from kinetic energy of the quarks. All other contributions including the Higgs interaction are a small (~1%?) correction on top of that. I think the W and Z bosons also get small masses from QCD (the Higgs mechanism article says they do). However the masses you get this way don't match what is observed, and the Higgs field was introduced to make up for that difference. The Standard Model attributes all mass discrepancies to the same Higgs field, though you could explain them in other ways (like additional Higgs fields or, in most cases, explicit mass terms). I'm not sure where the idea came from that the Higgs field is the source of all mass. It would be better to say that the Higgs field is the source of all the mass that we don't understand. John Baez points out that of ~25 adjustable parameters in the Standard Model, 22 describe the Higgs or its interactions with other fields, so it's probably hiding a lot of beyond-Standard-Model physics. -- BenRG (talk) 17:53, 9 September 2009 (UTC)[reply]
The answer to that question really depends on whose mass you are talking about. As BenRG pointed out above, if you are talking about the mass of a bound state (not a fundamental particle), like a proton or an atom, the energy of the vector bosons responsible for the bonding does indeed contribute to the mass of the bound state. On the other hand, if you are talking about the mass of the fundamental particles (quarks, leptons, W, Z, and the higgs itself), the vector bosons do not contribute. The reason for that is the fact that the ground state (state of minimal energy) of the vacuum of the theory correspond to a non-vanishing higgs field. It is said that the higgs has a non-vanishing vacuum spectation value (VEV). The vector fields MUST have vanishing VEVs (otherwise Lorentz invariance would be spontaneously broken) and therefore they do not contribute to these masses. Dauto (talk) 15:50, 11 September 2009 (UTC)[reply]
Pass the veggie dogs please!
I've been a vegetarian for 14 years, for ethical reasons. My family had dogs when I was growing up, but I never had my own dog as an adult until now. (The arrangements have been made; she'll be coming to live with me in a few weeks.) I would like her to live on a vegetarian diet as well. What I'm wondering is this: Can a dog truly be as healthy on vegetarian dog food as on regular dog food, or is there still some legitimate controversy about that? How much more expensive is it to keep a dog on a vegetarian diet? The dog is about three years old and has been eating a non-vegetarian diet up to this point; would that present any particular problems or challenges? - Lydia's Alpha-to-Be —Preceding unsigned comment added by 71.104.238.248 (talk) 03:37, 9 September 2009 (UTC)[reply]
Well, some people at least maintain that some dogs can thrive on vegetarian or vegan diets. Watch out for dilated cardiomyopathy due to L-carnitine or taurine deficiency. Never own a cat. - Nunh-huh04:04, 9 September 2009 (UTC)[reply]
Yes, I'm well aware that cats are carnivores and aren't biologically equipped to eat a plant-based diet, and if I ever did happen to have a cat in my life, I would respect that and buy something meat-based. However, since dogs are omnivores (like humans) I've heard that they can thrive on a vegetarian diet. Of course, the typical diet of canines in the wild has more animal protein than that of primates in the wild, so I don't imagine it would be a strict equivalence. Of course I would talk to the vet before making the final decision on Lydia's diet (and I could always make a change one way or the other based on her health, our finances, new advances in dog food, etc.), but I just wanted to get an idea early on, while we still have a couple months to figure out exactly how she's going to fit into our lives. Thanks for your answers! - Lydia's Alpha-to-Be —Preceding unsigned comment added by 71.104.238.248 (talk) 06:47, 9 September 2009 (UTC)[reply]
Dogs are indeed omnivores - not herbivores - their primary diet is meat - although they'll eat vegetable matter if they are hungry enough and there is nothing else available. Humans are also omnivores - but our primary diet isn't (or at least, shouldn't) be meat. So we do much better on an artificially herbivorous diet than dogs do. I don't think it's fair on the dog to try to make it be what it isn't. Trust me - dogs have no ethical desire to avoid eating meat. Even our 10 year old labrador - who is the gentlest, most mild-mannered dog you'll ever meet - has absolutely no compunction in chasing down a squirrel, ripping its head off and feasting on it's still warm-and-twitching body. If you don't like that dogs behave like this - then the remedy is for you to avoid owning one - rather than trying to change it's ethics to match yours. Dogs eat meat and chew bones...it's what they do...it's by FAR their greatest joy in life...it's what they live for. SteveBaker (talk) 12:58, 9 September 2009 (UTC)[reply]
I agree with SteveBaker. While an ovo-lacto-vegetarian diet won't kill a dog (the way it will a cat), it's really unfair, maybe even cruel, to the dog to force it into a dietary lifestyle it's incapable of understanding or appreciating. If you were still at the point of merely considering getting a dog, I'd advise you to get a rabbit or guinea pig or some other thoroughly vegan pet instead. As it is (your getting a dog is now irreversible), please allow your dog to eat what makes her happy, the same way you eat what makes you happy. +Angr14:03, 9 September 2009 (UTC)[reply]
Agreed. I very much doubt that domestic dogs, cats and whatever else really care enough about their owners' moral, ethical, religious or socio-political views to make personal sacrifices. --Kurt Shaped Box (talk) 14:40, 9 September 2009 (UTC)[reply]
On the other hand, How much meat do pet dogs usually get anyway? Aren't most cheap dog foods little more than vitamin enriched, meat flavored, corn meal? A quick look at some ingredients lists leads me to believe that some of the popular brands are mostly grains with some lard mixed in for flavor. If the brands of vegan dog food on the market have found some good replacement for the lard flavor (have they?), then how's the dog even going to know?
(Personally, I still wouldn't try it with a cat. The risks are a lot higher, and I frankly don't trust quality-control on pet foods at all. But what do I know? I've never had any particular inclination to make myself a vegetarian, let alone a pet.)APL (talk) 14:51, 9 September 2009 (UTC)[reply]
As pack animals, dogs will be loyal to you no matter what you put them through - but you have a responsibility to return the favor by keeping them happy. Dogs are the least vegetarian animals I can think of! The evidence of their carnivorous tendencies is everywhere. It's not about dog food...most of the dry food doesn't look like it's ever been need an actual animal. That's not the point. Dogs need to chew to keep their teeth in good shape - the various plastic toys and fake bone-like products don't seem to interest them much. That means that they need real bones and chewable rawhide toys and things like (my dog's favorites) pig ears just to keep their teeth healthy. Our dogs occasionally get those two foot long cow bones - which they'll (hilariously) laboriously haul around the house and try to hide from each other as they gradually gnaw this 10lb bone down to nothing over a period of weeks. One of those will entirely consume their waking hours for almost that entire time - it's hard to imagine how they are not utterly passionate about that behavior. When dogs sleep - you can see their feet twitching and they make tiny dream-barking noises as their jaws move and their cheeks puff out...then you see their jaws moving with little dream biting and chewing motions. There can be little doubt that they even DREAM of hunting small furry animals. Truly - find a good home for your dog and get a rabbit (I recommend an English rabbit) - they are very cool: you can train them - treat them like a cat or small dog - and feed them vegan stuff and they'll be very happy and healthy.) SteveBaker (talk) 15:55, 9 September 2009 (UTC)[reply]
Anecdotally, a friend of mine in Scotland, who had grown up on a working farm with dogs and was in addition a bright and unsentimental woman, fed her Alsatian and Golden Retriever on a vegetarian diet, carefully balanced, with appropriate added supplements, and bones to gnaw. My friend was not herself a vegetarian, and I can't remember her specific motivations, but the two dogs (whom I often looked after) thrived physically and mentally, and their faeces were markedly less smelly than is generally the case with meat-fed dogs. I agree that one should not do this without specialised knowledge and/or vetinarian advice. As it happens, another Scottish friend, my landlady for a time, kept a wild rabbit (rescued as a kit) as a perfectly contented and agreeable house-pet in her upstairs flat, though I agree an outside hutch might be considered more usual. 87.81.230.195 (talk) 19:00, 9 September 2009 (UTC)[reply]
The key there is probably the "and bones to gnaw" part. If bones are large, fresh and uncooked (required to avoid choking risks) - the dogs get a good deal of nutrition and 'meaty stuff' from the marrow inside the bones. If she had cut off the bone supply - I'd imagine the problems would begin right there. On a farm, the dogs may well have been 'self feeding' too - catching rabbits and squirrels, rats & mice - that kind of thing. SteveBaker (talk) 20:15, 9 September 2009 (UTC)[reply]
On the subject of dogs and bones, do you remember a time a few years back when 'they' were saying that dogs should never be allowed to gnaw on bones - due to the risk of sharp bone splinters being swallowed and causing internal injuries? I don't know if that was just one particular school of thought on the matter (I've never owned a dog, so I wasn't really paying it a huge amount of attention) but I definitely remember 'dog experts' on TV saying it several times. --Kurt Shaped Box (talk) 20:37, 9 September 2009 (UTC)[reply]
Wasn't that strictly for poultry bones? And seriously, if bones were regularly killing dogs, wouldn't we have ended up evolving dogs that did not engage in such dangerous behavior? 65.121.141.34 (talk) 21:16, 9 September 2009 (UTC)[reply]
The issue is with thin cooked bones. Cooked chicken bones are certainly not good - raw is fine though. Large bones are OK cooked. Dogs have evolved to gnaw bones - that's why they have not evolved toothpaste. SteveBaker (talk) 02:38, 10 September 2009 (UTC)[reply]
I would say that your primary ethical consideration here is to that one living creature you are bringing into your life, not to every other living thing on Earth whose life you wish to spare also. This is especially so since a dog will give you unconditional love and devotion and do anything to be part of your "pack". Dogs will eat pretty much anything, and most are also incredibly stoic, so they won't show signs of dicomfort until they have gotten pretty sick. While it certainly may be possible to create a perfect vegan dog food, the danger I see is in subtle deficiencies building up over time, such as the cardiomyopathy mentioned above. She won't tell you about her discomfort until it's too late. The safest way to go IMO is to use the highest quality of food you can afford, and don't worry about the meat components too much. Meat should always provide the best balance of nutrients for a dog, it's what they've been eating for thousands of years. It's true that they'll find certain types of grass and spend 20 minutes eating them (fibre? micronutrients?), and it's true that they'll eat any kitchen scrap you give them (mostly because they've been staring at you while you cook, trying to figure out how to get up on the counter where the human food is) - but if they can catch a squirrel, they'll ignore the corn cobs.
And if your dog is used to one type of food, you should be very careful if you're changing over to another, their guts will be set up to process the food they're used to. Franamax (talk) 20:17, 9 September 2009 (UTC)[reply]
Start by giving the dog the same diet as it has been used to - enough trauma with the move itself. Then using the many thoughtful suggestions above, gradually alter the dog's diet to one that is more in keeping with your beliefs. Abrupt diet changes can cause intestinal distress resulting in diarrhea and you wouldn't want that along with all of the other distress of a having a new household member. hydnjo (talk) 21:18, 9 September 2009 (UTC)[reply]
Making the dog of a vegan eat only a vegan diet, which it would not choose if given a choice, is a bit like baptising the dog and feeding it the eucharist. Or declaring the dog is an athiest, or a Tory. The moral choice of avoiding meat and animal products just will not give the dog the moral glow it gives the human. My father would feed his hunting dogs steak which had gone out of date in the freezer of a restaurant, but they also ate cornbread he baked for them. They would also eat green plants they found in their prowls. The vet's advice is the thing to give heed to. Edison (talk) 22:50, 9 September 2009 (UTC)[reply]
I think this, and all similar sentiments above, is missing the point. Sure, the dog won't have the same moral objection toward eating meat that a vegetarian or vegan might have. But this is irrelevent. Some dogs don't apparently have moral objections to mauling children, but we don't let them get away with that while saying things like "well we might feel morally happy about not mauling children but the dogs don't share our viewpoints". Very few, if any, ethical vegans or vegetarians are going to hold the belief "it is wrong for me to cause the suffering and death of animals but it is fine for everyone else". Most ethical vegans or vegetarians will think "it is wrong for anybody to cause the suffering and death of animals" and they will acknowledge that it happens "in the wild" but view this as an unfortunate result of evolution. Someone who owns a dog has every right and is, indeed, behaving quite admirably, to want to give their own dog a healthy lifestyle with minimal impact on the lifestyles of other creatures. Maelin (Talk | Contribs) 06:40, 10 September 2009 (UTC)[reply]
When a dog urinates on a fire hydrant, he is not committing vandalism, he is just being a dog. Don't force your moral beliefs on your pet. It will not understand. Googlemeister (talk) 13:17, 10 September 2009 (UTC)[reply]
Seems to me that any moral or ethical vegan wouldn't even consider the concept of "owning" a dog. Is it OK to own a slave, so long as you feed it only carrots? "Someone who owns a dog has every right and is, indeed, behaving quite admirably..." - that's quite a statement of beliefs right there. Maybe this would be better addressed at the Humanities desk? As far as SciRef goes, the general advice here seems to be that meat is the better way to go. Taking ownership of an animal and then forcing that animal not to ever eat another animal as an expression of your love for animals? Ackk - something missing in that equation. Franamax (talk) 11:18, 11 September 2009 (UTC)[reply]
How to charge the batteries and measure current?
Hello friends!, I know a lot of methods still available to measure the current stored in a Battery. But i need a simple method or atleast understandable. I have 4 AA size 1.2V, 850mAh NiCd batteries. I designed a charger circuit with charging voltage of 5V and charging current of 240mA for 4 batteries in series. So that each battery gets 240/4 = 60mA. Since my battery has 850mAh rating, now I don't know how long do I have to charge or how much charging current is actually needed?. I need atleast 4.8V,250mA for my main project ciruit to work.And my batteries with the above rating doesn't even supply power for more than 10 minutes even when it's been charged for more than 8 to 10hrs. So with my current battery rating,the backup should atlaeast last for 3 hours?What could be the problem?. In charger circuit, I use a 22 (1 watt)Ohm current limiting resistance in series with the input 5V source to get a current of 240mA. All 4 batteries are connected in series while charging.when I read the voltage of the battery, it shows 1.26V(1 battery) when charged fully and 1.1V when everything was drained!. For my main project, it consumes 250mA current with 4.8V requirement. ...In my case, is the battery actually charging or do I need more current rating for my design.? Please help me —Preceding unsigned comment added by Balan rajan (talk • contribs) 05:29, 9 September 2009 (UTC)[reply]
Current is not stored in a battery, charge is. The rating is for milliampere-hours, which (someone check my math) 1 milliampere-hour = 3.6 coulombs. A battery with a rating of 850 mAh will supply a current of 1 milliampere for 850 hours, or 850 milliamperes for 1 hour, or some combination thereof, such that milliamperes X hours = 850. The time to charge the battery will depend a lot upon how much current the charger itself operates on. Assuming a perfect system, if your charging current is 60mA (240 mA divided by 4 in series, if I am reading that right), then the battery should be fully charged in 850/60 = 14 hours, ten minutes, give or take. It will probably take somewhat longer than that given inherent losses in the system, so lets say 16-17 hours for a full charge. 8-10 hours would not nearly be long enough. Also, you do not indicate what your discharging circuit current is. If I draw 8.5 amps out of a fully charged, 850mAh rated battery, it'll be fully drained in 6 minutes... SO it looks like a combination of a) not enough charging time and b) too high a load on the battery when being discharged. --Jayron3205:47, 9 September 2009 (UTC)[reply]
Although not as explosive it actually takes more a complex design to charge NiCd/NiMH batteries than Li-ion ones. Seeing that you probably aren't too advanced in electronics (all components get the total current if they're series, see Kirchoff's circuit laws), you should probably use an off-the-shelf charger for now. Also, batteries don't have perfect reciprocity, they tend to drain a lot faster under high current (ie. 1mA load might last 850 hours, but 850mA load probably won't last an hour), so you should take that into account too. --antilivedT | C | G11:28, 9 September 2009 (UTC)[reply]
Well, you're probably not charging the batteries at all - that's the problem. The best way to charge NiCads is to use a constant current source, not a constant voltage source, as you're trying. Commercial chargers provide this, which is why they're better. As you describe your set up, you have 4 ~1.2V batteries in series, a total of ~4.8V. That voltage is dropped by a 22 ohm resistor, meaning that the current flowing is I=V/R=0.2/22=9mA. With this current, they'll take about 100 hours or so to charge. As the batteries charge, their voltage will increase slightly and so the current will drop further and the charging will take longer. So we see this circuit just won't work. With a constant current charger this problem does not occur. It's then easiest to work out when they're fully charged by feeling them - they will get slightly warm when they have charged up. --Phil Holmes (talk) 14:38, 9 September 2009 (UTC)[reply]
There is a limit to how fast a battery should be charged or discharged to avoid damage, explosion, or shortened service life. Isn't there some rule of thumb of using the ampere hour rating divided by 10 to get a charge rate for Nicads? That would imply 85 ma for 10 hours to charge the 850 mah cells. The charge should taper off so that the current drops to a mere trickle when the battery is are charged, rather than continuing to raise to voltage to force the same current through a fully charged battery. That would eventually cause loss of moisture in the electrolyte, outgassing, overheating, and explosion unless the venting system was adequately vented. Edison (talk) 22:41, 9 September 2009 (UTC)[reply]
C/10 makes a good charging rate. The problem with trying to reduce the current when the batteries are nearly charged is that it's hard to tell when NiCads are nearly charged. The normal way is genuinely to check whether they're warm. This can be done with a thermistor to automatically cut the charge, or by hand when you expect they will be charged (after about 14-16 hours, if fully discharged). Providing they are checked frequently they will not suffer significant damage. --Phil Holmes (talk) 08:54, 11 September 2009 (UTC)[reply]
I couldn't understand his calculation..."you have 1.2V 4AA batteries in series, a total of ~4.8V. That voltage is dropped by a 22 ohm resistor, meaning that the current flowing is I=V/R=0.2/22=9mA. With this current, they'll take about 100 hours or so to charge. As the batteries charge, their voltage will increase slightly and so the current will drop further and the charging will take longerWhy does he substitue v=0.2v in this calculation?. Applying 4.8V as the charging voltage across 4 AA batteries with a current limiting resistance of 22Ohms in series will charge a battery with V = 1.2V and I = 1.2/22 = 54mA. But he says the battery will be charged at 9mA. Could you explain (Phil Holmes) what it means please?. —Preceding unsigned comment added by Balan rajan (talk • contribs) 09:21, 12 September 2009 (UTC)[reply]
In your original post, you say you have the batteries in series. That means that the negative electrode of one is attached to the positive electrode of the next, so they are effectively stacked on top of each other. Their voltages therefore add up, so the total voltage across 4 cells in series is 4.8V. The full current flows through each battery, in effect one after the other.
You may have meant that they are in parallel - i.e. with all the negative terminals connected and all the positive ones. In this case, the voltage across them would be 1.2V, and any charging current would be shared. However, charging them in this way would also mean that it's unlikely that each will be properly charged. The current will not flow equally through all 4 and you will end up with one getting over-chrged and another not being properly charged. I would re-iterate a previous comment - if you do not understand the concept of series and parallel properly, you probably should not be trying to make a battery charger which could potentially cause the batteries to explode. Buying a commercial one is likely to be cheaper in the long run. --Phil Holmes (talk) 10:01, 12 September 2009 (UTC)[reply]
Glucose oxidation
I'm rephrasing the question because it seems that it wasn't completely clear the last time around.
The oxidation of glucose to CO2 and water has a very large accumulative G0' value and yet glucose is quite stable in the presence of oxygen, why is this so?
(ec)The editor above is absolutely correct. The stability is because at room (or body for that matter) temperature, there is insufficient energy to overcome the activation energy. Gasoline, or even hydrogen mixed with oxygen has incredible potential energy, but the activation energy is such that the mixture can exist for an indefinite period of time. Something either needs to bring the energy above the activation energy, such as a spark provided to the fuel-oxygen mixture, or the activation energy needs to be decreased, such as with the presence of an enzyme. Glucose is oxidized through a series of reactions, using a large number of highly specialized enzymes, each of which not only lowers the activation energy so the reaction can proceed, but also ensures only the desired product results from the reaction. Enzymes not only allow the reaction to proceed at the temperatures present in a biological system, but also prevent the ligands from reacting with or into any undesired chemicals. TheSeeker 4Talk18:14, 9 September 2009 (UTC)[reply]
How can matter exist?
From my understanding of chemistry (mostly learned a few years ago in high school), I do not understand how matter can exist if the protons and electrons are so small in the atoms that make up matter. I skimmed the atom page, but didn't see any real answer, although there probably isn't one yet. How can matter exist if there's basically nothing in the atoms (correct me if I don't know what I'm talking about = ])? EVAUNIT-66618:24, 9 September 2009 (UTC)[reply]
(EC with bugs)Fundemental particles (the quarks that make up protons and neutrons, and leptons of which the electron is one example) are essentially point-particles(dimensionless). However, the interactions between these particles occurs at measurable distances, and the distances of those interactions defines the size of things like atoms and molecules. So, even though the parts that fundementally make up an atom have no size at all, because these particles exist at a definable distance, THAT distance defines the size of the atom, which in turn gives volume to matter. For example, though an individual electron is essentially volumeless, the electrons orbit an atom in the electron cloud, the interactions between electrons and protons, and between electrons and each other define the size and shape of this cloud, and THIS is what gives an individual atom a real volume (see atomic radius. Put a group of atoms together, and they way that they interact with each other defines the bulk volume of matter, and thus you have the world you see around you.
Does the solar system exist? Most of the solar system is not occupied by planets, moons, or other heavenly bodies. Does a galaxy exist? Even though stars can be thought of as large, most of the galaxy (I think) is composed of fairly empty space.
Furthermore, the speed of electrons, combined with their very small orbits, places them almost "everywhere at once." I think this gives the substantiality to matter.
One finds it difficult to walk through walls not because the wall or the person is terribly dense, although the person could be dense in a different sense, but rather because of the likelihood that the fast moving electrons comprising the atoms of the of the human will "just happen" to be in a position in their orbit to encounter the fast moving electrons in the atoms comprising the wall. Bus stop (talk) 18:56, 9 September 2009 (UTC)[reply]
Nonsense! Superman was always perfectly law-abiding. They probably just hadn't passed that law yet, I imagine it was still stuck in committee. (Superman really wasn't that good at politics). Franamax (talk) 19:42, 9 September 2009 (UTC)[reply]
Forgetting the ongoing absurdity of a man flying, "The Mysterious Cube" has to be in the top echelon of the most preposterous episodes in the series. A scientist friend of Superman's convinces him that he could pass through solid objects, through the power of super-concentration. Now, maybe that could be theoretically possible, given that Superman can fly faster than light and see through anything (except Lois' dress). All well and good. Except that when he passes into the ultra-dense material comprising the cube, his suit comes with him. Apparently his super-suit had a mind of its own. Baseball BugsWhat's up, Doc?carrots03:41, 10 September 2009 (UTC)[reply]
Thinks look solid - because even though they are mostly vacuum - they still block and reflect light perfectly well. They feel solid because the electrons in orbit around the atoms of that chair you're sitting on are negatively charged and they repel the negatively charged electrons of your rear-end. Hence things feel solid. The mass of the atom is almost entirely centered in the middle...but they weigh what they should - so they are heavy enough to seem solid. Between those things - matter does indeed behave like it was solid. However, truly "solid" materials - something like a neutron star - which have all of the atoms squished together without all of those gaps - seems like really bizarre 'stuff'. As our article points out - the entire mass of all of humanity - crushed to the density of a neutron star would be smaller than a sugar cube. Neutron stars have so much gravity that they bend light by enough that you can stand 'in front' of one and see all the way around the back of it! SteveBaker (talk) 20:10, 9 September 2009 (UTC)[reply]
The basic reason that matter occupies space is the Pauli exclusion principle. There's a short discussion at Pauli exclusion principle#Stability of matter. Electromagnetism is involved, since it's the electromagnetic force that determines the available orbitals, but it's not an electromagnetic contact force that prevents you falling through the floor—I'm pretty sure of that. Electrons aren't pointlike in any sense that matters here. They effectively fill the available space as far as the Pauli exclusion principle is concerned. The nuclei are relatively well localized with large gaps between them and they contain most of the mass, so in that sense you can say that ordinary matter is mostly almost-empty space, but not in any other sense. -- BenRG (talk) 13:45, 10 September 2009 (UTC)[reply]
Well, indeed, but the article you cite says "They are a relatively short-lived phenomenon, lasting a few tens of thousands of years." A few tens of thousands of years is actually a very long time by our standards. So I'm guessing that in a hundred years another picture of the formation might look almost identical? — Sam 63.138.152.155 (talk) 19:03, 9 September 2009 (UTC)[reply]
For the linked picture of the butterfly (often called by its less romantic name, the Bug Nebula), the explosion which created the nebula occurred roughly two thousand years ago. The expanding cloud is now (mostly) a very, very long way from its parent star, so it feels very little gravitational attraction and will be moving outwards at a (nearly) constant rate. So, figure in another hundred years the nebula will be about five percent broader, and a little bit dimmer. The difference will be perceptible (if you have two photographs, side by side) but not particularly dramatic. TenOfAllTrades(talk) 19:36, 9 September 2009 (UTC)[reply]
The pillars in that photo at right are about 100 Trillion kilometers tall. The dust would have to be moving at an impressive speed for the picture to change very much in 100 years! SteveBaker (talk) 02:35, 10 September 2009 (UTC)[reply]
Oxymuriate of mercury
This substance is frequently mentioned in nineteenth-century medical texts - what's it's modern name? Mercury (II) chloride is one obvious suggestion, but that's "corrosive sublimate", and the "oxy" might suggest it contains oxygen. Thanks in advance. Tevildo (talk) 21:57, 9 September 2009 (UTC)[reply]
To confuse things oxymuriatic acid is actually (probably) chlorine gas. see [20] this is Humphrey Davies discovery that chlorine gas (oxymuriatic acid) is not what it was thought it was.. 83.100.250.79 (talk) 22:27, 9 September 2009 (UTC)[reply]
Drug dealers are usually members of a union or professional organisation, such as the Royal Society of Skag Peddlers. Their websites will probably have statistics on the qualifications of their members.
As far as I know, cocaine isn't worth synthesising from scratch, so the chemistry is pretty basic acid-base reactions and purification. LSD requires a bit of tinkering, and MDMA is a proper synthesis job.
Apparently, most LSD produced in America is made by a handful of experienced chemists in northern California, many of whom have been at it since the sixties. They can get away with it because there are so few of them, they all know each other, and they only make small quantities (because that's all you need, it seems). I once read that federal agents uncovered a drug production lab housed in a grain silo buried underground.
I reckon the overall answer is this: most drug producers don't know loads, just what they need to. But certain drugs need skill to make, and are made by a few specialists.
I would guess, not very. You don't need much formal training is chemistry to just follow a recipe and make the drugs. The training is required to do research and come up with new knowledge (about new chemicals, new ways to make old chemicals, explanations of why certain things are the way they are, etc.). Drug manufacturers don't generally do that kind of stuff. It is similar to how you don't need to be an automotive engineer to work in a car factory. --Tango (talk) 22:42, 9 September 2009 (UTC)[reply]
I'm curious whether the field would weed out those who didn't have some sort of (non-monetary) interest in science (albeit black market science)? Actually my main reason of wondering this was just as Malcolm X thought his bookie would've been a NASA scientist if he had been born white, whether many of the "top" drug makers would've been found in top schools had they been born into more privilege. Presumably, the less skilled you are (or the less you know what's going on), the less your purity and the less your yield. (higher grade => higher prices?) John Riemann Soong (talk) 22:49, 9 September 2009 (UTC)[reply]
A chapter of Freakonomics discusses the retail (not manufacture) of illegal drugs in a Chicago housing project. JT (the area manager for the Black Disciples gang) is a college graduate who could evidently, in other circumstances, have done well for himself in the licit business world, and who runs the gang like a Harvard Business School grad. -- Finlay McWalter • Talk23:01, 9 September 2009 (UTC)[reply]
The lack of safety is probably due to lack of skill, but it could also be because of poor equipment. (Given the amount of money you can make out of producing meth you would think they could get good equipment, but not doing that would be poor business skill rather than poor chemistry skill.) --Tango (talk) 13:28, 10 September 2009 (UTC)[reply]
I just want to note that it REALLY depends on the drugs in question and the type of manufacture. As far as I can tell there are entirely different methods of production and people involved in the production of cocaine, LSD, ecstasy, and methamphetamine, and the skill required is likely quite different as well. My suspicion is that in terms of theoretical knowledge, meth makers are probably pretty low on the list—any old hillbilly seems able to set up a meth lab (albeit not the most safe one). --98.217.14.211 (talk) 23:55, 9 September 2009 (UTC)[reply]
The meth synthesis is easy if you start with pseudoephedrine, which comes from decongestant pills. People who do that generally just follow recipes. I don't know about the other drugs. Looie496 (talk) 00:00, 10 September 2009 (UTC)[reply]
I would think that the best drug makers would be the best technicians as opposed to the best engineers or chemists. The reaction pathways are already known, and recipes to achieve those reactions are known. All that remains is to select the best pathway and recipe in terms of safety and purity of yield, then do it properly each time. The incentives for performance are avoiding blowing up your house and avoiding angry dealers whose customers have all been driven insane by the latest batch. A chemist might understand the reaction energetics perfectly but not notice the crud at the rim of the bucket. An engineer might want to scale the process up or use too expensive of gear for the low-volume element. A technician will perfect the recipe.
That's different from drug marketers, many of whom should probably have gone to Harvard so they could do benign things like devastate the world financial system instead. ;) But I'd hire up a really good lab assistant anytime to cook up the best drugs. Franamax (talk) 01:43, 10 September 2009 (UTC)[reply]
As an aside, is an adrenaline injection directly to the heart really the correct procedure in case of a cocaine overdose? No, I'm not looking for medical advice, I'm just wondering if they do this for real. After all, moviemakers are pretty notorious for shamelessly mangling the facts. 98.234.126.251 (talk) 06:06, 11 September 2009 (UTC)[reply]
Tarantino is a sensationalist semi-amateur hack who has never created anything new. He's the sub-prime mortgage salesman of Hollywood shows. Where is Trovatore, who told me on the other desk that Tolkien is boring? Nay, that movie crap is boring! (Though it's true that Tolkien recycled ancient myths, and Tarantino also recycles familiar movie themes - but Tolkien thought more than 10 minutes ahead). 98.234 as far as I'm aware, direct injection of adrenaline to restart a heart is accepted last-chance technique for non-infarcted heart attack. I'm not sure on exactly how successful it is (though I've seen it done in several movies and it always works there). I'm pretty sure that the miraculously recovered patient doesn't have their makeup and hairdo in perfect shape when they spring back up though. We actually had a mention of this here a while ago, you would need to check the archives. From the last go-round here, I think you need to be pretty precise about where you put the needle, but it seems to be a valid technique.[22]Franamax (talk) 06:22, 11 September 2009 (UTC)[reply]
Clarification: In the documentary "Extreme Ice" the way the exact location of the camera was maintained was to fasten it to rock above the ice flow. This was probably okay for showing what happened each hour for over a year and maybe even in the event of an earthquake because the field of view is so wide. Doubtful even an institution would have use for lengthier time lapse but then we still spend a lot of money to drill holes in the ice to get data that goes back lots of years. I'm speaking theoretically in the event someone wanted data that required the exact location from the object being recorded. -- Taxa (talk) 01:11, 10 September 2009 (UTC)[reply]
I'm sorry, your question is not very clear. What do you mean by "location"? And do you mean "How will" or "How do"? "Will" implies the future, and Wikipedia is not a crystal ball. However, if you mean how astronomers are able to take pictures of celestial bodies as they were 100 years ago, it is because even light does not move instantly. Light moves at a constant rate. The reflected light off of a celestial body may have been reflected off of the body 100 years ago, but is only reaching earth now. Therefore, astronomers looking through telescopes see the celestial body as it was when the light reflection they are viewing was reflected off of its surface. For example, if the sun suddenly disappeared, we, on Earth, would still see the sun for about eight minutes after it actually happened. Intelligentsium00:58, 10 September 2009 (UTC)[reply]
At a distance of 100 light years an astronomer almost certainly sees light emitted not reflected from a celestial bodies, and that is definitely so in the case of viewing our Sun. Cuddlyable3 (talk) 11:44, 10 September 2009 (UTC)[reply]
(ec)I'm thinking this has more to do with a question of whether it is possible to take a comparison picture in 2109 at exactly the same location in space, relative to the sun, given the somewhat eccentric and very slightly chaotic earth orbit. I'm wondering of how much importance to comparative astronomy is the change of location of the camera base due to earth orbit. --Tagishsimon(talk)01:01, 10 September 2009 (UTC)[reply]
Obviously, without a lot of expensive and impractical spacecraft, it's going to be impossible to get perfectly repeatable positioning. However, the size of the error due to the earth going around the sun - and the sun moving around the perimeter of our galaxy are very small indeed compared to the distance between stars. Stars in our local group are moving at almost the exact same speed as us around the edge of the galaxy - so the closest stars are moving the most slowly relative to us. Their relative positions will change very little over very long timescales. For stars at much longer distances, the amount of change compared to the distance between us means that those stars move VERY little relative to us. Take some hard numbers: The sun is moving at about 20 km/second relative to our neighboring stars. Over 100 years, the sun (and therefore the earth) will have moved 100x356x24x60x60x20 kilometers - that's 63 billion km - about six lightyears relative to the other stars in a region perhaps 20,000 lightyears across. So most of those stars won't move more than one part in several thousand compared to us...that's measurable - but not huge. The entire galaxy is of course spinning - we're going to have moved about 60 light years over the next 100 years compared to more distant galaxies...our galaxy is moving at about the same kind of speed relative to the other galaxies closest to us - but now we're talking about looking at objects like other galaxies...those are at distances measured in the millions of lightyears. Truly, our motion over 100 years is negligable to their distances. SteveBaker (talk) 02:26, 10 September 2009 (UTC)[reply]
You may want to check those numbers - a light year is just shy of ten trillion kilometers, so we'd move less than a hundredth of a light year in a century. If we take the diameter of the solar system as twice the semimajor axis of Neptune's orbit (screw you, Pluto, and your Kuiper belt buddies, too!) 63 billion km is only seven times the width of the solar system. That's enough that you could see slight shifts in the apparent positions of a few of the nearer stars, but even the nearest stars are almost a hundred times that distance from us. TenOfAllTrades(talk) 02:38, 10 September 2009 (UTC)[reply]
Yeah - that didn't seem right...oh - I see what happened. I slipped between meters and kilometers halfway through the mental arithmetic! Never trust a software guy without a calculator! Thanks for the double-check! Anyway - the point is made - we move a TINY distance compared to the distances to the objects we're observing. SteveBaker (talk) 02:49, 10 September 2009 (UTC)[reply]
If by location you mean taking a picture of the night sky from the same latitude and longitude, then precession and perhaps crustal movement will likely have more of an effect on the placement of the stars than the solar system's movement through the galaxy. There's also the proper motion of the stars themselves. ~AH1(TCU)23:01, 11 September 2009 (UTC)[reply]
Preventing/Dissuading housecats from slaughtering houseplants...? Hizzelp!
So I've got 4 wonderful little housecats and a ~100sqm apartment (a bit crowded, yes, but they are all street-rescues from urban China). Until recently, I've had houseplants scattered around on high shelves and the occasional wall-mounted planter or hanging basket, but what I'd really like to do is green the place up to the max... which would require plants be placed on counters, tables, and the like.
To date, whenever my cats have been presented with an opportunity to reach one of my plants (say, by careless placement of a chair or box) they have proceeded to go all Paul Bunyan / Mike Mulligan & Mary Anne.
Has anyone had success deterring this behavior? The cats already know it's not acceptable, so only do it when I'm sleeping or at work. Is this driven by the rarity of the occurrence? In other words, if they had a wide selection of targets would the urges eventually be satisfied? Or do I need to be more proactive?
I'd be thinking boredom as the first problem, too many cats, small place. What other interesting things are there for cats to do at your place? If they can't pursue their traditional activities (chasing/hiding/jumping), they will do other and weirder things.
As for deterrence, cayenne pepper sprinkled onto the soil? Water an ashtray and pour a bit of the foul water onto the plant (bet they've never attacked an ashtray!) I think you need to solve the boredom problem first though. Franamax (talk) 01:53, 10 September 2009 (UTC)[reply]
Oh yes, if the cats really do know that wrecking your plants is unacceptable - and think about it, did you really scold the right one at the right time, so that they learned the right lesson? They don't always have very good memory and understanding of causality. The cats themselves form a peer group, so they will be watching each other to see who gets awsy with what. Anyway, when a cat really does think it has got your measure, then specific destructive acts are usually commentary on what the cat sees as the latest thing you did to piss it off. In other words, the cats are engaged in a power struggle with you, so it's best to appease them rather than punish them. ;) Franamax (talk) 02:14, 10 September 2009 (UTC)[reply]
If they're chewing them up, there are sprays made from bitter melon that might work. These should be non-toxic unlike some of the other bitter sprays, so it should be safe for both the cats and the plants. I've never used them myself, but was actually thinking about trying it on electrical cords as a precaution. If they're urinating in the pots, then I've read in many books that covering the dirt with marbles work well. --Wirbelwindヴィルヴェルヴィント (talk) 04:28, 10 September 2009 (UTC)[reply]
Cats actually need to eat grass. They are obligate carnivores, but because they clean themselves, they need to induce vomiting and so they need to eat grass in order to do this. My guess is that your cats are self-medicating. We found that our cats will eat plants shaped like grass (spider plants and the like), but not plants with rounded leaves. So maybe stick to plants like African Violets (Saintpaulia), Begonia rex or Anthurium. Oh and the tip about marbles works: we use largish stones from the garden, you might be able to get some shingle.--TammyMoet (talk) 12:11, 10 September 2009 (UTC)[reply]
I just finished watching a visible pass of the International Space Station and Space Shuttle Endeavour. I noticed that there was a strange glowing haze around the ISS to one side, in the shape of a crescent moon. My guess is that it's some kind of out-gassing or perhaps the remnants of a burn they just did (wouldn't they do some kind of burn shortly after the departure of a shuttle?). I was worried that the station has depressurized and what I was seeing was the atmosphere leaking into space, but NASA would know about that and there is no mention of such a catastrophe having occurred on their website or on NASA TV.
I took photographs of the ISS, but they're long exposure so the station shows up as a bright streak. You can still see a bit of a haze around the station, but you can't see the shape that the haze took, so these pictures don't really show what I observed very well at all.
Whatever it was, it was following the station in its orbit because it kept pace with it as it moved across the sky. It made it appear as though the station had a comet-like "tail". I've seen dozens of ISS passes and I've never seen anything like that.
Ah! I'm too late. Yes, most often, if you see a large gas plume coming out of a manned space vehicle, it is... urine. This is a disaster for scientific instrumentation. A professor of mine was principle investigator on a Space Shuttle mission to study plasma ion effects in the high ionosphere; in at least one experiment, the trailed antenna became so coated in urine ions that it directly affected the scientific measurements. So much for the vacuum of space... Nimur (talk) 08:19, 10 September 2009 (UTC)[reply]
In the Mercury space program an astronaut referred to the little sparkly drops of frozen liquid outside the spacecraft as "the constellation Urion." Edison (talk)
history of the octet rule ... and proving that something is an element
It is the 19th century. You don't have any NMR, or mass spectroscopy. How do you prove that something is an element? Couldn't you say that the substances in your hypothetical compound are too bound together to release them by most known means? For example, in Davy's conclusion that HCl did not in fact contain oxygen, how did he defeat the idea that the oxygen was just really really strongly bonded to the other compounds.
The other thing is that I'm really confused at how people came to realise the octet rule. I mean, not long ago, Descartes was thinking that elements bonded to each other via hooks. How did people discover the chemical bond and the octet rule before the discovery of the electron? How did people propose enantiomers, Kekule structures and structural isomers and acetylation reactions before they had even discovered the electron?
It is clear that in the latter 19th century, people recognised the principles of unsaturation, electrophilic and nucleophilic attack ... but I can't really see how people discovered them if they didn't know the electron existed. How did people discover carbonyl reactions if they had no idea of partial charge and bond polarity? John Riemann Soong (talk) 03:09, 10 September 2009 (UTC)[reply]
The "Octet rule" is a historical artifact from early versions of the Periodic Table of the Elements, especially John Newlands's "Law of Octaves" which observed that, when arranged in order of atomic weight (protons and "atomic number" would not be elucidated for many years) chemical elements would repeat their properties every 8 elements. If you arrange the elements in order of atomic weight, then the 3rd, 11th, and 19th elements (lithium, sodium, and potassium) all display strikingly similar properties, as do other elements spaced similarly. Furthermore, Gilbert N. Lewis and Irving Langmuir devised important atomic models which basically connect this empirical observation with the electronic structure in the atom.
As to your first question, reliable methods of calculating the molar mass of a substance predates understanding of actual atomic structure by almost a century. Since elements cannot be broken down into smaller substances, you just keep doing things like electrolysis or other similar processes until you don't produce any new subtances with smaller molar masses. Once you rigorously show that something cannot break down anymore, you likely have an element. However, some early models of atomic structure were based on the theory that there was exactly one element, hydrogen, and that ALL other "elements" were merely increasingly more complex conglomerations of hydrogen. Its an easy mistake to make, since the molar masses of nearly all elements are simply interger multiples of the molar mass of hydrogen. See Prout's hypothesis for more on this reasonable, but ultimately dead-end theory. --Jayron3204:29, 10 September 2009 (UTC)[reply]
Another general comment is that you don't need to know how something works to know that it does work. The actual nature of a bond ("covalent electrons") isn't needed to recognize that "something" holds molecules together in certain ways, and using certain chemicals, those things can change, and then eventually one sees patterns in what kinds of chemicals cause certain changes. For example, you don't have to know that "unsaturation" means two shared covalent pairs, you just have to see that one "unsaturation" in a molecule absorbs one hydrogen molecule or dihalogen molecule. The names of many old concepts are based on observed behavior, not on the now-known chemical structure: if something is "saturated" it cannot absorb any further. DMacks (talk) 04:36, 10 September 2009 (UTC)[reply]
If you are interested in understanding the nature of the chemical bond, one could do no better than to go to The Source, which in this case is the book The Nature of the Chemical Bond by Linus Pauling. The section on covalent and ionic bonding in EVERY single general chemistry textbook for the past 50 years is basically cribbed directly from Pauling's work. --Jayron3204:39, 10 September 2009 (UTC)[reply]
Did people realise that alkenes had stronger "double bonds" (cf. to single bonds), smaller bond lengths, etc. or only unsaturation? How did they rationalise why C=O was preferred over C=C, for instance? (How did they interpret keto-enol tautomerism)? John Riemann Soong (talk) 13:44, 10 September 2009 (UTC)[reply]
At the time concepts like "saturation" WRT hydrocarbons was developed, people had almost no idea what was actually going on, things like "double bonds" and "keto-enol tautomerization" were unknown at the time. All that was known was that some compounds tended to react with excess hydrogen in the precence of a metallic catalyst, and others did not. This sort of work was going on in the late 1700's and early 1800's. At that time, people had means of deducing the molecular formula of a compound, indeed the basic concept of the isomer was known by 1827-1828 or so. People knew that isomers had to result from difference in molecular structure, however they had no idea what those structures looked like. However, the basics of organic chemistry structure didn't even come into place until 1858, when Kekule proposed the tetravalent carbon atom. Remember, before Kekule was born, chemists knew what isomers were, and could identify the molecular formula of a compound, but they did not even know enough about the structure of these molecules to know that carbon formed 4 bonds. It wasn't until Gilbert N. Lewis in the 1910's when people even had an idea what a double bond was. True experimental evidence the structures of compounds probably only came about in the 20th century, with modern analytical techniques like Mass spectrometry and Nuclear magnetic resonance. --Jayron3205:49, 11 September 2009 (UTC)[reply]
heat of fusion of aluminum oxide...
Need this to solve my thermodynamics problem set, but google is being obnoxious. In fact, the scientific community as a whole is being obnoxious. How the hell are we missing the heat of fusion for such a common substance? And why isn't there an easily-available database somewhere? Why does "aluminum oxide" + "heat of fusion" turn up no useful results? John Riemann Soong (talk) 03:42, 10 September 2009 (UTC)[reply]
This source from google books: [23] does not give the value directly, but notes that liquid Aluminum Oxide is a well studied material, which may mean that enthalpy of fusion data may be availible. I would check the footnotes of that reference for more details. However, it may also be that such data has never been reliably measured given Aluminum Oxides rediculously high melting point. It could just be that its hard to do a reliable calorimetry experiment to find the heat of fusion of a material that melts at a higher temperature than your measuring devices do! --Jayron3204:04, 10 September 2009 (UTC)[reply]
q: how were michelson and morley able to get the initial interference pattern if their did not exist any path difference initially??
I don't think it matters whether the two paths are perfectly identical. Interference fringes will be seen in the field of view and the object of the experiment is to see whether the fringes move when the apparatus is rotated. They don't. Cuddlyable3 (talk) 11:27, 10 September 2009 (UTC)[reply]
I think the question is, if the path distances are equal and the beams are perfectly aligned, wouldn't the phase difference on the detector be essentially uniform across the surface of the detector, and therefore show no fringe pattern? Rckrone (talk) 16:43, 10 September 2009 (UTC)[reply]
And to answer that, the two distances won't be exactly equal, and the wavelengths involved are small enough (~500 nm) that even small discrepencies will cause fringes. In particular, if d is the approximate path length and a is the discrepancy between the two paths then the change in path difference between the center of the detector and a point at radius r is approximately ar2/d2. Just to pick some numbers, with d = 10m and r = 5cm, there would be 1 fringe showing up for each 2cm that the path distances were off by. You could probably set it up so that initially there were no fringes if you wanted, but I assume it's easier to see a change in the pattern if there are some. Rckrone (talk) 17:18, 10 September 2009 (UTC)[reply]
temperature-dependent heat capacity of aluminum
I note that aluminum exhibits a relatively significant temperature-dependence ... a lot of books and internet pages mention this, but won't give me a function. Can someone give me a function? If I extrapolated experimental data onto excel, what minimum degree polynomial would be useful? John Riemann Soong (talk) 06:34, 10 September 2009 (UTC)[reply]
As far as I remember, the reason is as follows. Alumin(i)um has a relatively high Debye temperature (about 400 K), that is, some of its phonon modes have energies well above the room temperature (about 295 K). As a result, at room temperature the heat capacity of aluminium is below the 3Nk value of Dulong-Petit law, and growing with temperature. Please see our article on Debye model, or Ashcroft & Mermin textbook. You can use the formulae derived in the framework of the Debye model. You should also be aware that the heat capacity of a solid may actually exceed 3Nk close to the melting point, and that there is also a (small) electron contribution to the aluminium heat capacity that is not accounted for by the Debye model. --Dr Dima (talk) 08:33, 10 September 2009 (UTC)[reply]
How to calculate the density of a gas
How do you calculate the density of hydrogen at 50 Celsius and 0.5 MPa? This is not a homework question, I need it for work. I suppose you perhaps could use the ideal gas law pV=nRT but dont know how. Thank you! Wikifantast (talk) 06:47, 10 September 2009 (UTC)[reply]
There is no prediction involved, it's a trick - the idea is to try and work out how the trick is done - predicting the numbers is not one of the options. my own take is that it involves mirrors and the balls are simply showing what is reflected onto them from the floor. That's why he does not lift the balls as it would spoil the effect. it's also why we get the misdirection about him not being able to reveal the numbers before the drawn occurs because of legal rights. No legal rights would stop him making a prediction, it's simply because the trick needs him to know the numbers before they 'appear' on the balls - however he does it. --Cameron Scott (talk) 11:52, 10 September 2009 (UTC)[reply]
It's not science, it's 'magic' and from my experience the answer is usually something dull and 'obvious'. One thing to note about Derren Brown is that he often 'reveals' how a trick is done. Don't be fooled into believing that this is necessarily any more the truth than the first trick you saw. (i.e. the explanation of how the trick was done is a lie itself, the 'real' way is not revealed). 194.221.133.226 (talk) 14:21, 10 September 2009 (UTC)[reply]
One technique he is said to use, which might be applied here (I have not seen the specific trick) is to tediously film all the possible outcomes of a trick (say, drawing a specific card), each time without any camera cut, and then show the appropriate "take". Obviously this would not be possible for the 14 million-plus possible outcomes of the Lottery, so some combinatory cutting might be employed, but I also thought that he usually avoided such simple cutting "cheats." 87.81.230.195 (talk) 15:13, 10 September 2009 (UTC)[reply]
I saw this and he didn't predict anything, he only revealed the numbers after they'd been drawn, so even as a magic trick it was quite disappointing. Anyway, he himself will be revealing how he did it tomorrow evening, so you won't have long to wait.--Shantavira|feed me16:44, 10 September 2009 (UTC)[reply]
His programme is subtitled "How to win the Lottery", if he could do that we could all win the lottery and probably only get our stake back. —Preceding unsigned comment added by 86.4.181.14 (talk) 14:50, 11 September 2009 (UTC)[reply]
...and winning the lottery is easy - you just have to buy 14-million-and-something tickets. Winning the lottery and making a profit - that's the hard one ! Gandalf61 (talk) 15:01, 11 September 2009 (UTC)[reply]
And what a disappointment that show turned out to be - he constructed a pseudo-science narrative so he did not have to simply say "it was spilt screen". --Cameron Scott (talk) 21:13, 11 September 2009 (UTC)[reply]
Yeah, very disappointing. The coin tossing thing is well known and has nothing to do with the law of large numbers, which is what the rest of it focused on (albeit a misapplication - wisdom of crowds is one thing, psychic power of crowds is quite another). --Tango (talk) 00:33, 12 September 2009 (UTC)[reply]
It depends on your definition of heart. In the colloquial sense, a heart is an internal organ that serves to pump blood (or other circulatory fluid, as in hemolymph), thereby maintaining circulation. The human heart, along with other mammals, birds and crocodilians is divided into two halves that are synchronized in their beating -- one pumps the systemic circulation while the other pumps the pulmonary circulation. Annelids may have enlarged vessels that contract, thereby pushing blood in the closed tube system, or a number of vessels connecting the dorsal and ventral main vessels that contract -- some would refer to each of these as a heart. So does an earthworm have 5 hearts? Not nearly in the sense of the human heart. There's probably a similar thing going on in cephalopods like octopi. DRosenbach(Talk | Contribs)15:36, 10 September 2009 (UTC)[reply]
"The circulatory system of cephalopods is an extensive system of vessels and capillaries . . . Blood within the vessels follows more or less the same route through the body as in other molluscs, but in addition to a systemic heart, the circuit includes two branchial hearts, which pump blood through the gills." --Edward Ruppert and Robert Barnes, Invertebrate Zoology (quote taken from this web page). Looie496 (talk) 17:26, 10 September 2009 (UTC)[reply]
Extremely fast Blu-Ray recorder required
I want to record rock concerts (with the permission of the artists) and almost instantly burn blu-ray disks for sale to fans after the concert-whilst I am doing this just to promote my favorite acts, completely free of charge right now, I later plan to turn it into a commercial business. I have heard about 8x blu-ray recording, but I need MUCH more speed for this proposition, even if I used an array of burners the crowd would have gone home before the disks were burnt. Is there any technology that could get around this problem on the horizon, such as a burner with multiple lasers for example? Supposing a powered vacuum suction plate clamped the disk tightly to a platter-could a much higher speed (presumably using a higher power laser as well) be achieved without turning the disk into a safety hazard?80.2.195.218 (talk) 12:56, 10 September 2009 (UTC) —Preceding unsigned comment added by 80.2.195.218 (talk) 12:53, 10 September 2009 (UTC)[reply]
You might need to consider a parallel burning scheme if you want to make a business out of this. Unfortunately, fast burning is not scalable. You can invest in several slow burners and have them running simultaneously. Keep in mind the important difference between throughput and latency. These terms apply equally well to the low-level "bits" burning on to the Blu-Ray; and also the high-level, conceptual throughput for your business model. You want to generate the maximum number of discs in the shortest amount of time - so while a faster burner decreases delay per disc, it may be less effective at increasing total discs per unit time than investing in several dozen cheap burners. Nimur (talk) 13:52, 10 September 2009 (UTC)[reply]
I understand that you want to catch the fans before they leave but consider that they won't actually get to play the music until they get home. (Can't your favourite acts pre-record for you?) An idea is that you could give away CDs that give users access to a website where they can download a recording of the concert. CD players are commoner than Blu-ray players and I suppose they all have Internet. For you the advantages would be 1) Easy, cheap no-rush technology. 2) You have time to listen to and edit the material. 3) You can use the website for any kind of promotion, knowing who are the visitors. Cuddlyable3 (talk) 13:55, 10 September 2009 (UTC)[reply]
But not CDs of the very concert at (the end of) which they're being sold, surely? I thought Midge Ure had once done so during an April 2002 solo tour, but in fact the double-live album in question had been compiled during the earlier gigs of the tour. What had me going was that a very similar incident to what occurred at the gig I attended (Midge telling off a fan for singing along at the beginning of "Dancing with Tears (In My Eyes)") appears on the recording, but the recorded event's details differ slightly from my (admittedly unsober) recollection.
Sorry, I misunderstood what the previous poster was suggesting, and thought he/she was just suggesting they sell CDs. Instead it is more like the "voucher" model that I mention below. --98.217.14.211 (talk) 16:50, 10 September 2009 (UTC)[reply]
I think that if the technical problems could be overcome, the OP would find a small but dependable market (given typical post-gig euphoria and the desire for personal souvenirs), but making it cost effective might prove difficult. 87.81.230.195 (talk) 14:50, 10 September 2009 (UTC)[reply]
This sounds... not very feasible. Technically, or as a business model. (What happens if there is a glitch with the machine, or the concert gets out early? Then you've just ended up with 100 BluRays that won't offload any time soon.) You could, of course, sell vouchers -- they buy the disk, you ship it to them the next day. But I imagine that isn't as fun. It doesn't really seem feasible to me unless you lock the crowd up for a little while after the show ends. --98.217.14.211 (talk) 14:14, 10 September 2009 (UTC)[reply]
There are already people who are doing this exact thing with CD's. Certainly my son went to a concert over 2 years ago and bought a live, recorded-on-the-spot CD containing the musical highlights as he left the concert which included still photos of the crowd - which we know had to have been taken during the concert. So the only thing that's novel here is to record a Blu-Ray rather than a CD. The problem is indeed the speed of recording. But a standard drive cannot simply be made to spin faster to record faster! That's an exceedingly naive expectation! There is the matter of the laser delivering enough energy to one tiny spot on the disk to do whatever it does to record a '1' or a '0' there. If you make the disk spin 10x faster, the laser has to deliver 10x the energy per second to be able to burn the disk. It's possible that the chemical processes that cause the disk to change state require a certain minimum time to happen. That means that you quite simply cannot use the same laser diode that an off-the-shelf BluRay recorder uses...you'd need some really fancy industrial laser. That laser has to be at exactly the right frequency - which might mean that you need something rather exotic. Can you even buy such a laser with a sufficiently small spot-size? It would also have to be focussed the exact same way as the original LED in order to deliver the laser light to a sufficiently precise spot on the disk and also to put the energy into the layer of dye that you want to record onto without melting the plastic on the outside of the disk. Would the cheap lenses they use stand up to that amount of energy delivered that quickly? Worse still, you probably have to cool the disk because as it absorbs all of that energy, it's going to get hot. Then, the data rate that you have to supply to the laser would require entirely new electronics - and probably new software. All-in-all, this is an exceedingly tricky technical problem - requiring an understanding of lasers, chemistry, physics, electronics, software and mechanics. I would be EXCEEDINGLY surprised if our OP were to be able to pull off such a system without a large team of experts working for him/her. SteveBaker (talk) 20:23, 10 September 2009 (UTC)[reply]
Supposing there was a way to quickly create a photo-mask http://en.wikipedia.org/wiki/Photomask like the ones used to create silicon chips and then use it to burn a blu-ray disk using a blue laser with a beam the size of the disk? Or could a high powered digital laser projector achieve the same effect with the right focusing lenses? Any heat could be dissipated by a cooling plate with liquid nitrogen. User:Trevor Loughlin—Preceding unsigned comment added by 80.2.200.204 (talk) 05:30, 11 September 2009 (UTC)[reply]
This discussion has now become a question about what the current limiting factors are for modern data transfer to optical storage media (rather than strictly about faster Blu-Ray burners). I suspect that, per Steve's answer, there are so many bottlenecks that you need technological advances on all fronts - mechanical, chemical, electrical, digital/bandwidth, that making a dramatic revolution in one area is still insufficient to speed up the entire process. Trevor's suggestion of a photomechanical transfer seems unlikely to speed the process - having spun photomask polymers a few times myself, I would definitely not call the process "quick and easy". To do it on a one-off basis requires a research laboratory and a lot of expertise; to do it in a controlled, mass-production sort of context requires a clean room. Assuming the technicalities of turning one of those facilities into a standalone box were solved, we haven't even addressed the issue - because in order to photomask, you need a mask - so you'll need to transfer the data to the mask first. Presumably once you have that mask manufactured, you can re-use it; but all told, at least using present technology, that idea would be much much more complicated than simply burning the discs individually. The ultimate goal, which is to parallelize the data transfer, is fundamentally the right direction; but I don't think we're going to see this scale of "all-at-once", gigabytes-of-parallelism data copying until a few more miles down the road of technological progress. By that point, optical storage media will probably be less desirable, less economically competitive, and a whole new array of parallelization options will become available. Nimur (talk) 11:23, 11 September 2009 (UTC)[reply]
Have an array of burners recording at 1x, basically in real time while the concert is happening. When the music stops, close the sessions, and all discs are finished in a few moments. 70.90.174.101 (talk) 08:11, 12 September 2009 (UTC)[reply]
Car tire wear
I have a car that is rear wheel drive, but for some reason, the right front tire seems to wear the fastest. Why would this be? We drive on the right hand side. Googlemeister (talk) 14:37, 10 September 2009 (UTC)[reply]
Are you properly rotating, balancing, and aligning the tires/wheels? I'd put any of those three well above which side of the road you drive on. Also consider damage to the wheel bearings or other components that may cause the wheel to lose alignment even if you're following the basic maintenance schedule. — Lomn14:56, 10 September 2009 (UTC)[reply]
(EC) The most likely cause is that the wheel is slightly misaligned (see Wheel alignment); the exact pattern of wear would indicate to a mechanic the nature of the misalignment. Less likely is that, overall, you make significantly more left turns than right - possible if you habitually follow different "out" and "return" routes, as many people do. 87.81.230.195 (talk) 14:59, 10 September 2009 (UTC)[reply]
If the wear isn't even it could be due to incorrectly inflated tyres. Take it to a local garage and they'll do an alignment/balancing check and get it sorted - also remember to check your tyre pressure regularly (i'm terrible for doing it only before long-ass trips). 194.221.133.226 (talk) 15:14, 10 September 2009 (UTC)[reply]
You could be taking far more left hand turns and if the car is improperly balanced, will put a great deal of pressure on the right front tire. Livewireo (talk) 19:19, 10 September 2009 (UTC)[reply]
This answer seems to imply that the driver is going in circles. :) Statistically speaking, you really should be turning both ways with equal frequency. When you go back home on the same route you came, all your turns are reversed. Franamax (talk) 10:48, 11 September 2009 (UTC)[reply]
Too bad, if you did you could just tell your pit boss to get the g-dam RF wheel into shape. Or else you're trying to sucker us Wikipedians into helping your NASCAR team for free. I get confused on this stuff... :) Franamax (talk) 14:12, 11 September 2009 (UTC)[reply]
Does Otitis lead to high frequency loss in Adults?
Otitis? after treatment is it recommend a audiometer exam be done for adults who had otitus? Finally, can otitus be associated with hearing loss and or tinnitus?
Oh, I think it can be answered without offering any advice. There are three kinds of otitis: otitis externa, affecting the outer ear, otitis media affecting the middle ear, and otitis interna, involving the inner ear. The most common of these is otitis media, which causes a (usually transient) hearing loss, but the poor hearing is caused by a loss of conduction of sound to the inner ear. But high-frequency loss is more suggestive of sensory-neural injury, involving the inner ear, rather than conductive loss. As for the advisability of audiometry: if you are having hearing problems or tinnitus, this is something you should ask your doctor. It's not routinely done after every bout of otitis. - Nunh-huh18:59, 10 September 2009 (UTC)[reply]
Displacement is a vector because in order to describe WHERE something has moved to, you need to say BOTH how far it moved AND in which direction. That's pretty much exactly what a vector is. --Jayron3220:42, 10 September 2009 (UTC)[reply]
Sound has a wavelength, and resonance was known about, so it would have been possible to make back of the envolope calculations to calculate the resonant frequency of a ported loudspeaker with out too many tears.
More relevently sealed box, mass damped and of course horn type loudspeaker where much more likely to be encountered then.
This site may be of interest [25] - note a reflex design from 1930 which may be worth investigating. I'm sorry I haven't really answered your question, but hopefully it will serve as an introduction to a better answer below:::83.100.250.79 (talk) 16:32, 10 September 2009 (UTC)[reply]
An early version of the now widely-used bass-reflex cabinet design was patented by Albert L. Thuras of Bell Laboratories in 1932. The patent shows that the mathematics was already understood at that date. Cuddlyable3 (talk) 20:47, 10 September 2009 (UTC)[reply]
There were excellent engineers designing audio equipment, ranging from microphones to amplifiers to loudspeaker enclosures, at such places as Bell Labs, back in the 1920's. They went far beyond guessing and rule of thumb. For some early Google Book hits before 1950, see [26]. Unfortunately, the books and journals are not old enough to be more than snippet view, but the journals could be accessed at a good engineering library. The research publications from Bell Labs apparently are not in the Google Book search index, but could also be found in an engineering library. The Society of Motion Picture and Television Engineers (back before TV) also worked on well engineered sound systems for movie theaters, and again the journals are in engineering libraries. The Journal of the Acoustical Society of America also has papers on loudspeaker enclosure design dating back to the 1920's [27]. The Proceedings of the Institute of Radio Engineers (predecessor of IEEE) published studies on the design of speaker enclosures. Here are some from the 1930's:[28]. It must be noted that lots of radios and record players back then had beautiful wood outside and crummy speaker enclosures. Edison (talk) 04:12, 11 September 2009 (UTC)[reply]
it have seen that if an ufo is looked anywhere , then it is seen that aliens use zigzag motion to come at the earth, so why can not use this type of maotion to go to any other planet? —Preceding unsigned comment added by Aatmic (talk • contribs) 15:45, 10 September 2009 (UTC)[reply]
If your question is why don't humans travel to other planets in a zigzag motion, then the answer is that it requires a large amount of energy (in the case of our spacecraft, rocket fuel) to change the direction of a vehicle traveling at such speeds. Googlemeister (talk) 15:50, 10 September 2009 (UTC)[reply]
Your question reads to me like you're asking why aliens don't zigzag over other planets as well as Earth. In which case - they probably do. If you're asking why we don't do it to travel to other planets; zigzags are not efficient for travelling from one location to another, as pointed ut by Googlemeister. If they are zigzagging, it's not to reach the planet - it's for some other reason (photography, malfunction - take your pick, the possibilities are limited only by your imagination). 90.208.66.97 (talk) 21:07, 10 September 2009 (UTC)[reply]
I don't know what UFO's you're thinking of, but keep in mind that coming to Earth is a very different deal than going from Earth to another planet. The former requires you to slow down from an orbital speed to a speed suitable for flight, whereas for the latter you'd want to accelerate as quickly as possible to get out of Earth's atmosphere. As it turns out, the space shuttle does zigzag across the sky upon re-entry: it traces S patterns in the sky to lengthen the distance it travels and bring it down to landing speed. --99.237.234.104 (talk) 00:01, 11 September 2009 (UTC)[reply]
What is the name of the machines or prcosses that perform simple function (turning on a light) by using convoluted tasks to perform the action (a ball drops into a cup which tips over causing a basket to rise on a pulley, withsomething attached to flick on the light switch) —Preceding unsigned comment added by 138.64.2.77 (talk) 16:31, 10 September 2009 (UTC)[reply]
On what factors does the resistance of a conductor depend? Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why does the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value? What change will occur in the current through it? —Preceding unsigned comment added by 59.96.142.2 (talk) 17:01, 10 September 2009 (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 aim 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. Refer Resistor for starters. Rkr1991(Wanna chat?)17:55, 10 September 2009 (UTC)[reply]
It sort of behaves as a kind anti-gravity, in that it causes objects (that are far apart) to accelerate away from each other, while gravity causes them to accelerate towards each other. I don't think it is actually related to gravity in any way, though. There is a theory that it is a result gravity itself (not any kind of anti-gravity) behaving differently are large distances, but I don't think it has much support. --Tango (talk) 19:18, 10 September 2009 (UTC)[reply]
Lux and sunlight
i've never hear the term lux. All it is is is measurment of sunlight fraction of earth. how much dimmer than light is nighttime or midnight on Earth? Total blackness would be 450,000 times dimmer than light? Will 100 times dimmer be sunset or sunrise?--209.129.85.4 (talk) 20:46, 10 September 2009 (UTC)[reply]
(after ec)You'd struggle to find "total blackness" anywhere on the surface of the Earth, since it's all illuminated by the Sun during the daytime, and by the Moon and stars at night. "Total blackness" (zero light) cannot be expressed as a fraction of another light on a linear scale. Time to check the Lux article...You also asked "how much dimmer than light is nighttime or midnight" - do you mean how much dimmer than daytime? If so, that will depend on weather (cloud cover would reduce light both at night and during the day), the time of the month (full moon will illuminate more). A location and general weather you're expecting for these answers would help. 90.208.66.97 (talk) 21:02, 10 September 2009 (UTC)[reply]
I was involved in a "struggle to find total blackness" at a company making a test card for TV cameras. We could find no black paint that was truly black so the solution was to build a little box behind the card, viewed through a hole in the card. The inside of the box was lined with black velvet. Cuddlyable3 (talk) 13:37, 11 September 2009 (UTC)[reply]
I noticed with my grandparents and other older people that they sometimes need help getting out of regular chairs. Is this something with the back muscles, such as the spine? Is it generally arthritis - I don't recall my grandparents, at least, having complaints about that, though. Or, is it just a combination of factors? Notice I'm trying to be vague so it doens't sound like a request for medical advice or diagnosis; I'm jsut asking about a general problem. Maybe you can just link an article or two. The article on sitting really didn't answer it. (Maybe I should have looked under "rising" :-) )4.68.248.130 (talk) 23:21, 10 September 2009 (UTC)[reply]
As somebody who's approaching that zone myself, I can tell you that the problem is that your joints tend to stiffen when you sit for a while. Rising from a sitting position requires a substantial amount of leg muscle strength and a rapid change in joint angles. Looie496 (talk) 23:40, 10 September 2009 (UTC)[reply]
Especially if it's a deep, soft chair that you used to be fine with getting in and out of. Hence the market for chairs that slowly push you up and let you straighten your legs out. A great invention. I'll get one when I'm old and crotchety. >:) Baseball BugsWhat's up, Doc?carrots01:57, 11 September 2009 (UTC)[reply]
Hi. I was looking for information & links on the workings of blood glucose monitoring devices, but on a very detailed level. Like "Chemical X reacts with Y" or "These bonds are broken/deformed in the reaction," and I have a particular interest in the reactions/mechanisms on a nano level.
There are a few systems: the basic schema is that glucose passes through a selectively permeable membrane controlling the transport of analytes to a metabolizing enzyme system that generates a response on interaction with the glucose, leading to an electrical signal generated by the transducer and eventually amplified and translated into glucose concentration on the meter screen. Details can be found here. You may also find this of interest. - Nunh-huh00:49, 11 September 2009 (UTC)[reply]
Need help understanding American sports gambling notation...
Redskins (+6.5) over GIANTS // SEAHAWKS (-8.5) over Rams // PACKERS (-3.5) over Bears
I've read the relevant Wiki article, but it just left me more confused. Are the above "lines" discussing the spread or the odds?
For example, if I choose to place a bet on the Redskins in the first example, do I win if they score 7 more total points than the Giants? Or do I just get $6.50 for every $1 I bet on them if they win? If the former is true what happens if I choose the Giants?
Far as I know, those numbers are the "spread". If the Skins win by only 6 over the Giants, then they failed to "cover the spread" and someone betting the spread on the Skins loses, even though the Skins won the game. The spread is a way of "leveling the playing field", meaning, yeh, the Skins should win, but by at least 7, or they didn't cover. The 1/2's are to ensure that you can't "tie" the spread. You can only cover it or not cover it. Of course, you can also lose, in which case you obviously didn't cover. Baseball BugsWhat's up, Doc?carrots01:23, 11 September 2009 (UTC)[reply]
This site [29] doesn't explain the system, but it does give the spreads, all of them as negatives, and it lists the Giants as favorites over the Skins (which is what I would have expected, actually) at -6 1/2. I guess the way to look at that is, take the final score (assuming Giants do win) and subtract 6 1/2 from the margin. If the Giants still come out on top, then they've covered the spread. Baseball BugsWhat's up, Doc?carrots01:43, 11 September 2009 (UTC)[reply]
Not a typo, just how to think of it from the other team's perspective. You'd buy the Redskins and if their score + 6.5 is greater than the Giants, you'd win. I think I finally understand this now... 218.25.32.210 (talk) 01:50, 11 September 2009 (UTC)[reply]
No, not quite, I don't think so. The Giants are favored to beat the Skins. If you bet on the Skins, it's + 6 1/2, which I interpret to mean that you add 6 1/2 to the Skins score at the end and see if it results in a winner for the Skins or not. It's the flip of the same idea - if the Giants fail to cover the 6 1/2 spread, the ones betting the spread on the Skins win. Baseball BugsWhat's up, Doc?carrots01:55, 11 September 2009 (UTC)[reply]
(ec)That's it. If you take Redskins, then (Redskins score) + 6.5 > (Giants score) gives you a win and the opposite gives you a loss. If you take Giants, then (Giants score) > (Redskins score) + 6.5 gives you a win and the opposite gives you a loss. When you win, it's typically referred to as "covering the spread." Hope that helps. Makeemlighter (talk) 01:56, 11 September 2009 (UTC)That's not what I meant. Makeemlighter (talk) 01:58, 11 September 2009 (UTC)[reply]
In your example, betting on the Giants, I think you would actually say (Giants score) - 6.5 > (Redskins score) gives you a win. Hence the minus figure used with the favorite. Arithmetically, of course A - C = B is the same thing as A = B + C Baseball BugsWhat's up, Doc?carrots02:04, 11 September 2009 (UTC)[reply]
(undent). Just a little more on spread betting. It is often (incorrectly) assumed that the spread is somehow set by an "expert" who has studied the teams and determines who should be favored to win. In actuallity, the spread is set by the market. It's pure free market economics. As more people put money down on one team, the spread shifts to compensate until betting is aproximately even between the two teams. The bookmakers are looking for perfect 50/50 betting between the two teams; since they are paying out full on whoever wins (if you win, you double your money, less the vig), they do best when there is approximately even betting between the teams. The "favorite" is thus the one the market favors, and not necessarily the better team. Often, teams with popular followings tend to get better spreads than their skill level would dictate. --Jayron3205:32, 11 September 2009 (UTC)[reply]
It is not quite that they do best when there is even betting, it is that they have the least risk. Businesses don't like risk. They would make more profit if they had a 75%/25% split and the 25% side won, but they would rather guarantee a small profit than have a chance at a large profit and a risk of a large loss. --Tango (talk) 14:48, 11 September 2009 (UTC)[reply]
The "market" definining the betting odds is what tips the experts off that a fix might be in. Such was the case in the 1919 World Series, where everyone expected the Sox to win, but the betting favored the Reds. As well it should have - since the fix was in. Baseball BugsWhat's up, Doc?carrots16:47, 11 September 2009 (UTC)[reply]
Battery leak query
Hi, I have a pinhole leak in a deep cycle acid battery. I was wonderin if their was a common susstance ie tape,glue etc I could use to block it & not be dissolved by the sulphuric acid cheers. —Preceding unsigned comment added by 219.88.82.92 (talk) 01:51, 11 September 2009 (UTC)[reply]
Sulfuric acid is extremely corrosive and rapidly degrades all organic substances (including duck tape and all kinds of adhesives), as well as most metals. The ONLY way to fix your battery is to WELD THE LEAK SHUT (and no, soldering won't do, the acid will eat right through solder before you can blink). So, either get your welding torch, or turn in the battery for toxic waste disposal. 98.234.126.251 (talk) 06:23, 11 September 2009 (UTC)[reply]
Really? Sulfuric acid eats solder instantly? I find that very unlikely; Lead is pretty resistant to sulfuric acid. The all-inclusive statements are probably inaccurate; this company says PVC is "excellent" for storing sulfuric acid up to 60% concentration. I don't think it's a good idea to recommend welding torches and sealed acid batteries in the same sentence, though. Nimur (talk) 13:50, 11 September 2009 (UTC)[reply]
Sorry, can't resist anymore. Epoxy resin.[30] But you should talk to a battery specialist before doing anything. Find a battery shop and call them up. Real life people who deal with that stuff every day will always be willing to give you advice, and it will often be better than the advice you get from the internet. Franamax (talk) 06:46, 11 September 2009 (UTC)[reply]
Of course, all of this advice for repairing the battery may not be worth the time, materials, and effort. It may be better just to recycle the faulty battery and buy a new one. --Jayron3207:06, 11 September 2009 (UTC)[reply]
Rice is 370 kcal/100 grams, and the average person needs something like 2000 kcal/day. Assuming that all of the various micronutrients could be provided in pill form (so we could avoid scurvy and beri-beri and that stuff) a person would need 2000/370*100 = 540 grams of dry rice. Yields of rice probably vary significantly, but here's one data point: [31] which indicated that in 2006, rice in Texas was averaging yields of about 7200 pounds/acre. That's about 16000 kilograms/acre. Putting it all together, from a caloric standpoint, a person eating nothing but rice would consume 540 * 365 / 1000 = 197 kilograms of rice in a year. That means that, given that an acre of rice could produce 16000 kilograms, one acre could feed 81 people (assuming no wastage, and assuming that those 81 people ate nothing but rice for their caloric intake). ANother measure of productivity is Edible protein per unit area of land, of which rice produces roughly 224 lbs/acre or 493 kg/acre. A person needs roughly 50 grams of protein per day, so that's 50 * 365 /1000 = 18 kg per year. 493/18 = 27 people per acre of land could subsist on nothing but rice from a protein point of view. --Jayron3207:01, 11 September 2009 (UTC)[reply]
Er, oops. Lets do that one again. Redoing the math with the CORRECT factor results in 16 people being supported on 1 acre of rice from a caloric point of view, and 5 people being supported on 1 acre of rice from a protein point of view. Blame the fact that it was after 1:00 AM local time...
That's a bit misleading, though, since rice sucks as a protein source -- it's a much better carb source. Probably even a little bit of protein supplementation would push the numbers way up. Looie496 (talk) 01:36, 12 September 2009 (UTC)[reply]
Even if you eat enough rice to get the right quantity of protein, that still won't be your protein needs met. There are various types of protein you need and rice doesn't have them all. You need a pulse (eg. some kind of bean) in addition to a cereal to get the full range. (There are some exceptions - quinoa is like a cereal, but with all the protein, for example - and animal products tend to have all the protein you need, but cereal+pulse is a good general rule for vegetarian meals.) --Tango (talk) 02:06, 12 September 2009 (UTC)[reply]
REQUEST FOR INFORMATION ON UGCA 292 (DDO 125) GALAXY
Hi,
I request you to help me get some information about UGCA 292(DDO 125)GALAXY.
This is a list of galaxies that have been seen with the naked-eye by humans. So, humans can see 14.7 million lightyears into the universe, unless you believe in the rumors about M101 being seen with the naked eye. M101 is 27 million lightyears away.
I've read that the eye of a cat can collect about 20 times more light than the eye of a human. Then, given the brightness of the dimmest galaxies that have been seen by humans, that seems to imply that all the galaxies listed here should be visible to a cat.
Sound weird, but I'd go along with it. If cats can see in rooms where humans are virtually blind, then there is no reason why they could not see many stars not visible to humans. In the same way, humans would be able to see many more stars than a dog could. Birds could probably make out far better colour variations amongst stars than humans can. Myles325a (talk) 03:52, 11 September 2009 (UTC)[reply]
On the other hand cats have 20/200 vision which for a human is considered legally blind. They're good for catching mice but not for anything too discriminating. So yes they collect the photons but what actually would someone with that sort of vision see if they went out on a clear night? The sky would be lighted up but they wouldn't be able to pick out many details besides the moon. Dmcq (talk) 12:03, 11 September 2009 (UTC)[reply]
It is not just the absolute amount of light from an object that determines whether or not we can see it, it is that light relative to the light around it. The smallest a human with 20/20 vision can resolve is 1 arcminute. If the galaxy in question is smaller than that then it will be blurred with the surrounding sky. If that surrounding sky is too bright, you won't be able to make out the galaxy, even with very sensitive eyes. --Tango (talk) 16:41, 11 September 2009 (UTC)[reply]
I just had a quick work out of the figures and as far as I can see they'd collect 20 times more light but it would be spread over 10x10 more area, so they need 5 times the flux from a star as us. Each magnitude needs 2.5 times as much flux so they'd only see stars of about 1.5 less magnitude han a human. In an urban setting we only see down to about magnitude 4 so they'd see just the stars in List of brightest stars but not much more. In ideal conditions they'd see those of about 2 magnitudes more. Dmcq (talk) 18:37, 11 September 2009 (UTC)[reply]
What Australian plants relate to American ones from the time of Gondwana?
As an Aussie, I am interested in the fauna and flora connection between America and Australia, which were once joined together with what is now Antarctica. For example, the New World tomato has its distant relative the uncultivated bush tomato, and I yesterday I read that the well-known Australian plant Monstera deliciosa has its distant cousin in Mexico. The (only?) marsupial find outside of Australia is the American opussum. Interestingly, Australia does not have the cactus succulents of the American deserts, although we have a lot of similar deserts. Would that mean the continents drifted apart after the evolution of marsupials but before the evolution of cacti? Has there been any research done on this? Myles325a (talk) 04:12, 11 September 2009 (UTC)[reply]
There are actually close to 100 species of opossums in the Americas, plus one other marsupial called the Monito del Monte. Anyway, my understanding is that the split-off of Australia happened about 140 million years ago, as described in our Pangaea article, while the main radiation of flowering plants happened around 100 million years ago. So, although flowering plants had probably already existed for a long time, they were still at a relatively limited level of diversity back then. Some of them no doubt reached Australia after the land-split, carried by wind and wave from Southeast Asia or other places. (I'm stretching my limits here, there's tons of research on this stuff but it's well outside my usual domain.) Looie496 (talk) 04:47, 11 September 2009 (UTC)[reply]
Path through Earth
For a physics assignment, I was able to show that the time it takes to circle the earth at an orbit near the earth's surface is equal to the time it takes for a mass passing through the earth to return to it's original position (T=sqrt(4*Pi^2*R/g)). Is this a coincidence or is there a deeper reason behind the equality? —Preceding unsigned comment added by 76.68.245.247 (talk) 05:34, 11 September 2009 (UTC)[reply]
Well, we're not supposed to do your homework for you, but I'm not sure how much to give away before it starts being unfair... To make it easier, imagine the earth as a circle instead of a sphere. Now imagine two points at 12 o'clock, one will accelerate towards the middle then decelerate back to a stop at the 6 o'clock position purely by gravity. The other will orbit half way around the circle at a constant speed and also stop at 6 o'clock. Compare both vertical components of their velocity vectors and you'll hopefully find your answer. The journey back to 12 from 6 will just be the same thing again. Vespine (talk) 06:11, 11 September 2009 (UTC)[reply]
I struggled with that for a few minutes, but yeah, that's beautiful. Both objects are falling freely under the influence of gravity. Is it true then that both objects would be at the same position in the vertical plane? Or would one (through) need an integral function to calculate it's instantaneous position, and the other one (orbiting) use a cosine rule? Sorry to step on the OP's question (and it seems like they already did the homework) but this is interesting to me too! Franamax (talk) 06:39, 11 September 2009 (UTC)[reply]
The question is simply a more complex version of The Monkey and the Hunter problem, except this time the hunters bullet is fired with enough force to make it into orbit and the monkey if falling through a giant hole though the center of the earth. But otherwise, the exact principles that hold true in The Monkey and Hunter Problem hold true here. --Jayron3206:46, 11 September 2009 (UTC)[reply]
The monkey and hunter problem only works in a uniform gravitational field, which this isn't. This behavior actually depends on the specific properties of the gravitational field at work here. For example it wouldn't work for a hollow spherical shell or more generally a sphere with non-uniform density. Rckrone (talk) 07:05, 11 September 2009 (UTC)[reply]
(ec)It can be shown that the gravitational field inside a sphere of uniform density is proportional to the displacement vector from the center. Therefore your acceleration in the vertical direction only depends on your vertical position. Note that both paths initially have zero vertical velocity so they behave the same along the vertical axis. Rckrone (talk) 06:55, 11 September 2009 (UTC)[reply]
I would also have accepted "yes" or "no" as an answer. ;) But yeah, vectors, I'd forgotten about them. I thought Vector was either a running shoe or an energy drink. Why did I ever think that the horizontal component of velocity had anything to do with the vertical? Thanks and d'ohh! :) Franamax (talk) 07:16, 11 September 2009 (UTC)[reply]
Well both are oscilators about the same centre of mass , one circular, one linear, and involve the same force. So there should be a close connection between the two - as to whether both are the same for a good reason...83.100.250.79 (talk) 11:16, 11 September 2009 (UTC)[reply]
Why not consider the motion of the rotating object in two axises - is the equation obtained the same as the linear oscillator? And is there a reason for this? (eg consider circular motion as two oscillators, one sin the other cos)..<hint>try the differential equations for the circular motion expressed as separate for the two axis</hint>
I think a key point here is that the force stopping the object that is going through the Earth from orbiting is the reactive force with the sides of the tunnel. That force will obviously be perpendicular to the motion of the object, and such a force does not change the energy of the object. That means you can calculate everything using energy really easily, since it is constant for both objects. The initial energy of each object will be the same (assuming they start travelling at the same speed, which I think is part of the problem). The angular speed depends on the linear speed and the distance from the centre, the linear speed gives you kinetic energy and the distance from the centre gives you potential energy, which must add up to that total energy. When you work out the maths you should (I haven't actually tried) get that the average angular speed is the same for both objects (ie. it takes the same amount of time for each to get from start to finish). --Tango (talk) 18:04, 11 September 2009 (UTC)[reply]
I don't think conservation of energy simplifies the problem. Also the two objects have different total energy so I'm not really sure what you mean.
As I mentioned before, and as 83.100.250.79 hinted at, the force law here (gravity inside a sphere with uniform density) has a particularly nice form, that of a 3D harmonic oscillator, which is the key to the problem since as a result the equations of motion in each dimension are independent from each other. a = -(g/R)r, which means az = -(g/R)z. A close orbit just above the surface acts the same as an orbit just below the surface, so both paths can be treated as following the force laws in the interior of the sphere. Rckrone (talk) 18:51, 11 September 2009 (UTC)[reply]
If they start at the same place with the same speed, they have the same total energy. Are they not supposed to start at the same speed? --Tango (talk) 19:12, 11 September 2009 (UTC)[reply]
The one that goes straight through has to start at rest for it to work.
On an unrelated side note, I think it's worth pointing out that the Earth doesn't have uniform density and so this wouldn't work on the actual Earth. The one going straight through would make the trip faster. Rckrone (talk) 19:21, 11 September 2009 (UTC)[reply]
Really? But in the limit as the two points on the surface get closer together the distance travelled will be same and the one travelling in a straight line won't have any time to accelerate, the one in orbit will have a speed independent of the distance. I don't see how they can have the same travel time under those conditions. --Tango (talk) 19:31, 11 September 2009 (UTC)[reply]
It doesn't work for any path through the sphere, only one through the center (otherwise gravity isn't the only force acting on the object and so it won't behave like a harmonic oscillator). Rckrone (talk) 19:37, 11 September 2009 (UTC) Edit: No wait that's wrong, it would still behave like a harmonic oscillator, and so the period for any straight path through the sphere is equal, which is an interesting result of its own. Note that that period is the time it takes for the orbiting object to make a full orbit.[reply]
Comment - Tango as I remember the orbiting object has just enough speed to orbit, but the through the earth object starts with v=0 as I remember the period for the orbit is sqrt(4pir3/Gmearth) , and you need to pick the uniform density version of the earth (I think) - which is described in good detail at shell theory (God bless newton) (It's also described by Gausses law but I won't recommend that).
One of the curiosities is that the force inside the solid sphere = -kxdisplacement - which is just like that of a simple spring (eg Hookes law ) - this gives a chance to suggest that the students work out the neat solution by "assume that the force is like a spring" - before actually covering the whole shell theorem thing.. The odd consequence of this is that the adage "simplify = lie to children" - actually turns out to be "not lie to children" since the more thorough analysis gives exactly the same result (until you consider non-uniform earths...)83.100.250.79 (talk) 22:16, 11 September 2009 (UTC)[reply]
Orbital velocity at the Earth's surface (for a circular orbit) is about 8km/s. I can't see how an object can start at rest, roll along a near horizontal path and accelerate up to almost 16km/s at the half way point (which is what would be required for a short journey through the Earth). That's just an absurd speed. --Tango (talk) 22:30, 11 September 2009 (UTC)[reply]
Looking at the equations in harmonic oscillator the k in F=-kx is "Gmearthmobject / r3" - which when placed into the equation for period of oscillation gives the same equation as that given above for the time it takes a thing to orbit the earth. I don't know what is going wrong with the numbers.?83.100.250.79 (talk) 23:03, 11 September 2009 (UTC)[reply]
For any straight path between two points on the Earth, the amount of time it takes for the object to get from one end to the other is the same as the time it takes for the orbital object to get halfway around the Earth. If the path is very shallow, it's also much shorter than the distance traveled by the orbital object, so the straight line object doesn't need to approach the orbital speed. (The max speed is (L/2)*sqrt(g/R) where L is the length of the path. The equation of motion is x(t) = (L/2)cos(sqrt(g/R)t), where x is the distance from the center of the tunnel.) If the path is through the center of the Earth, then the max speed of the straight line object is exactly the same as the orbital speed, which shouldn't come as a surprise. The straight line object under goes a large acceleration, but he change in velocity of the orbital object over the same period of time is even larger, by a factor of sqrt(2) since it's had to make a 90 degree turn. Rckrone (talk) 02:12, 12 September 2009 (UTC)[reply]
If you were to drill from the USA straight through the earth, you would find yourself boring a hole into the bottom of a 13,000 foot deep tub of water called the Indian Ocean. Once you broke through, I wonder how fast that water would push you back through to the USA, and how much of the USA would then be inundated. Baseball BugsWhat's up, Doc?carrots02:51, 12 September 2009 (UTC)[reply]
Assuming you could keep the magma from collapsing in on the tunnel and you could keep the tunnel cool enough that there could be liquid water in it, then some water would shoot out of the tunnel initially, but it would settle at sea level, so unless you started at Death Valley it wouldn't flood anything for long. Rckrone (talk) 03:17, 12 September 2009 (UTC)[reply]
I think this result is less surprising if you kinda sneak up on the orbiting thing. Imagine a gun is fired horizontally - if you dropped a bullet onto the ground at the precise moment that someone fired the gun - both bullets would hit the ground at the exact same instant. Some people find that surprising. Now mentally transition to the classic thought experiment where you fire the gun at progressively higher muzzle velocities so the bullet goes further before it hits the ground each time. At long enough ranges (and at correspondingly high enough muzzle velocity), the curvature of the bullet's path starts to more and more follow the curvature of the earth. until it completely misses the surface and orbits. But at no point did the bullet you dropped behave any differently. If you dig a hole for the dropped bullet to fall into - it'll continue to accelerate at the same rate as the shot bullet. SteveBaker (talk) 04:05, 12 September 2009 (UTC)[reply]
Today, a UTI can be easily treated with proper prescription medication (assuming it's not severe). Though, I've always been curious as to how such infections were treated before modern medicine came into play. Or, better yet, how do people treat such infections where proper medical supplies are not available? 76.121.122.191 (talk) 07:04, 11 September 2009 (UTC)[reply]
People used herbs by and large for medication: things such as rose hips, saw palmetto berries, cranberries, goldenseal, cleavers, marshmallow. And they drank lots. --TammyMoet (talk) 08:42, 11 September 2009 (UTC)[reply]
Many bacterial infections that are now treated with antibiotics would most likely be cleared by the body's immune system. This is how it worked for millenia. Some people (perhaps those with compromised immune systems or specific susceptibility to certain infections) would develop more serious infections, such as a urinary tract infection turning into pyelonephritis or otitis media extending into mastoiditis or meningitis. Bacterial illnesses sometimes enter the blood at which point there can be sepsis. But most minor bacterial illnesses can be dealt with effectively by the body. Antibiotics help to clear an infection more quickly, prevent complications, and keep the infection from progressing in those who might otherwise develop a serious illness. I have no doubt that some of the compounds in "natural products" are useful or bacteriostatic (some people swear by cranberry juice for prevention of UTIs) which is probably why folk remedies have been passed down for ages. --- Medical geneticist (talk) 13:16, 11 September 2009 (UTC)[reply]
Many vitamin therapies have been disproven. I'm not sure it's true that vitamin C would help a great deal in UTIs. I do know that antibiotics are a major reason for the substantial global increase in life expectancies. 98.14.222.248 (talk) 22:39, 11 September 2009 (UTC)[reply]
Okay, I'm having issues with interpreting the constants. Is rho for a particular type of reaction (e.g. base-induced hydrolysis) constant across a range of substituents, or does it change depending what type of substituents you put on it? Conversely, does sigma change depending on the reaction involved? Is it simple as matching constants for the substituents with a constant for the reaction?
The other thing I'm working out is that whole affair when both constants are negative ... I mean yeah, cool, mathematically the equilibrium constant increases, but why does it work out that way in real life?
(Lastly, and my least important question, what happens when you have several substituents?)
Oh, can I confirm that the rho constant for most basic hydrolyses of substituted benzoic acid esters will be positive? John Riemann Soong (talk)
electron withdrawing in two directions (induction only)
Can two electron withdrawers connected (directly or indirectly) to say, a carbon atom, end up having opposing effects on the partial charge on that oxygen? This is purely for the inductive effect, so take meta-fluorobenzene as an example or something (but I meant EWGs in general). So let's say I have some electronegative substituent, X. So let's have a symmetric molecule X-R2-R1-CH2-R1-R2-X. Would the central carbon be in fact less positive than it would be if it were H3C-R1-R2-X?
The other thing is why does the inductive effect usually exceed the mesomeric effect (except for say, alkoxy substituents) when the inductive effect would be presumably weak across so many bonds, and resonance stabilisation/destabilisation would presumably have more influence? John Riemann Soong (talk) 12:08, 11 September 2009 (UTC)[reply]
If I pull on your left hand, you become strained a bit. If Numur pulls your right hand, you're proposing that you become less strained overall. Either we're both pulling in the same direction, in which case you're strained even more strongly that way, or we're pulling in opposite directions, in which case you're being pulled twice as strongly apart. The inductive effect is a vector, so you need to do vector math. Or even do simple partial-charge addition: two partial-positives make even more positive (it's not literally added numerically, but qualitatively they do reinforce each other). Every intro-orgo text I've read has a whole section about dipoles. DMacks (talk) 16:26, 11 September 2009 (UTC)[reply]
Well yes, but carbon tetrachloride has no dipole moment. Directly attached to the carbon, I can see a stronger effect yes, but perhaps a better example is something with sp2 geometry. Let's say, attach an EWG group on one of the sp3 arms of the carbonyl... doesn't that strengthen the C=O bond and make the resonance structure where the has 3 lone pairs contribute less greatly? Suppose you have an aryl ring with some EWG on it. To remove the mesomeric effect contribution (and just look at induction effects), let's put this EWG say on meta position. But wouldn't this make the ring more electron dense, thus increasing the contribution of the resonance structure where the carbonyl carbon forms a double bond with the connecting aryl carbon? John Riemann Soong (talk) 17:25, 11 September 2009 (UTC)[reply]
CCl4 has no net (molecular) dipole moment because the vectors cancel each other out for the molecule as a whole. But looking closer, they reinforce each other to make the carbon more δ+. CCl4 has a central positive surrounded (symmetrically) by negative, not "no partial charges anywhere". Putting an inductively withdrawing group on an sp2 side of a carbonyl makes the carbon more δ+ by adding its own vector to the existing one of the C=O. Our acyl chloride article discusses that as the reason why these types of molecules are particularly electrophilic at the central C compared to other carbonyls. DMacks (talk) 18:59, 11 September 2009 (UTC)[reply]
Yes, but in reactions it matters what the nucleophile "sees". Aren't nucleophiles attracted towards dipole moments? I mean, I'm sure the chlorine atom of perchlorate is extremely electrophilic but its reactivity is stymied by the fact that reactants have little access to that positive charge. John Riemann Soong (talk) 05:26, 12 September 2009 (UTC)[reply]
Let's take this back to the carbonyl question. If the two =O are attached to the same carbon, then yes they DO cancel, which explains the structure of Carbon dioxide which has no net dipole moment. However, in nearly all other situations, the =O's will not be 180 deg away from each other, meaning that they will always work additively. Consider a molecule like Acetylacetone (aka 2,4-pentadione). The two carbonyl groups are so electron withdrawing that the hydrogens on Carbon 3 are measurably acidic, and the corresponding ion is a useful thing. --Jayron3206:02, 12 September 2009 (UTC)[reply]
Fly energy; really quite impressive
I've had one housefly buzzing around my apartment for two days. Yes, it's annoying, but at the same time I've been thinking about the energy required to do what it's doing. There's nothing to eat here. I have no food scraps lying around at all. Just really quite sterile. This things is in constant motion, flying around seemingly nonstop. It would seem superhuman (well, supermusca domestica); the equivalent of a human doing ten marathons. I don't really have a targeted question exactly. I'm just wondering if anyone can comment on the little bastard's incredible energy output.--162.84.164.115 (talk) 12:38, 11 September 2009 (UTC)[reply]
Also, consider how little the fly weighs and how well-optimized it is. It's fantastically lighter and more energy efficient than any sort of machine we could make at the moment. --98.217.14.211 (talk) 00:13, 12 September 2009 (UTC)[reply]
I don't think it's that amazing. You could easily walk around your apartment (stopping periodically as the fly does) for two days without food or water. You wouldn't enjoy it - and maybe the fly doesn't either - but you could definitely do it. And consider that you have to generate heat internally (which means you have to eat literally ten times as much as a cold-blooded animal) - and you have to run your big brain (20% of your food energy goes to running that gadget!). Admittedly, you aren't flying - but proportionate to body size/mass/whatever, the air is thick and gloopy stuff for a fly - for a tiny insect, it's more like swimming in thick treacle than flying - the air supports it quite well - it doesn't take much energy to stay airborne. But that's nothing! According to (for example) this, a crocodile can survive for two years without eating. SteveBaker (talk) 03:22, 12 September 2009 (UTC)[reply]
Please excuse me, but I'm skeptical of the reported observations here. I've never personally seen a fly last that long without running out of steam -- how do you know it was the same fly? Looie496 (talk) 01:31, 12 September 2009 (UTC)[reply]
decreased electrophilicity of diacetyl?
Is diacetyl more electrophilic or less eletrophilic (towards hydration) than formaldehyde? The carbonyl carbons both have methyl groups on them, which can stabilise the partial charge via hyperconjugation, but I'm thinking placing two carbonyl groups next to each other would destabilise them (by synergistically making each their carbons even more positive). John Riemann Soong (talk) 13:30, 11 September 2009 (UTC)[reply]
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not 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 how to solve such problems. Please attempt to solve the problem 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.
Many of people tell about allergy to iodine, chlorine or other chemical element (not substance). Is it theoretically possible to have an allergy to chemical element (not substance containing this element) which is essential to human organism? Renaldas Kanarskas (talk) 15:42, 11 September 2009 (UTC)[reply]
An allergy is when your immune system overreacts to something harmless. A sensitivity to iodine, or something, isn't likely to be caused by the immune system, so isn't an allergy. --Tango (talk) 17:51, 11 September 2009 (UTC)[reply]
Ther is two types of albedo, visual and geometric albedo, but I have no idea about both of them, and what they mean? Article we have don't make sense to me.--209.129.85.4 (talk) 16:51, 11 September 2009 (UTC)[reply]
According to Geometric albedo - it is the ratio of reflected to incoming light ASSUMING the object is a flat surface at right angles to the incoming light rays. The "visual geometric albedo" is the exact same thing but with the narrower definition of the word "light" to mean only light within the visual spectrum. So, if an object looked very white to the naked eye, it would have a high visual geometric albedo - but if it failed to reflect infra-red and ultraviolet light very well then it's overall "Geometric" albedo would be much lower. Worse still, there is the "Bond albedo" which refers not to a perfectly flat surface - but instead to the net reflection from a spherical object such as a planet or moon. SteveBaker (talk) 02:50, 12 September 2009 (UTC)[reply]
electropotential of esters
Can someone get me an electropotential map of an ester versus an electropotential map of a ketone? I don't have a Hartree-Fock calculator right now and google images is being unhelpful. John Riemann Soong (talk) 17:37, 11 September 2009 (UTC)[reply]
Did you mean "electron density ...." - google images gives a few results for this term, but nothing amazing, both cross sections and '95% cutoff' images found.83.100.250.79 (talk) 22:28, 11 September 2009 (UTC)[reply]
Heavy Metals in the Earth's Crust
Why is it uranium or even gold can be found near the surface? It seems like when the Earth first formed from a molten state each element would settle into individual layers ordered by density. Why isn't this the case? TheFutureAwaits (talk) 18:03, 11 September 2009 (UTC)[reply]
To an extent, that does happen, but it isn't perfect. There would be all sorts of convection currents and things that keep moving the heavy elements back towards the surface. --Tango (talk) 18:06, 11 September 2009 (UTC)[reply]
Also note that most of them form compounds with other lighter elements such as Pitchblende a uranium oxide. Others, such as gold, are soluble in high temperature aqueous solutions and tend to be transported towards the surface and deposited from these hydrothermal systems as the temperature falls. Mikenorton (talk) 18:55, 11 September 2009 (UTC)[reply]
Light pressure means the mirror must be pushed backwards by the light. That is going to accelerate the mirror, and the energy for that has to come from the light, so the reflected light must have less energy. --Tango (talk) 19:27, 11 September 2009 (UTC)[reply]
My girlfriend and I made soft pretzels a couple of days ago using this recipe. We decided to put cinnamon-sugar on some of them. However, when these pretzels came out of the oven the cinnamon had turned a deep grayish green, a quite unappealing color! We ate them anyway, and they tasted like cinnamon-sugar pretzels. We suffered no ill effects. What could have caused this though? We've made the same recipe before without incident. Could our cinnamon be bad? Or could it have something to do with the baking soda the pretzels are dipped in before baking? --S.dedalus (talk) 20:51, 11 September 2009 (UTC)[reply]
Lots of colored foods are actually decent acid-base indicators, for example red cabbage and tumeric are two classic examples I can think of, but there are likely others. I am not sure on cinnamon one way or the other, but I would not be surprised if it turned out that the basic sodium bicarbonate (baking soda) caused the color to change. Such changes are not harmful in any way, if the pH was the source of the color change, then any change is reversable when the pH is lowered again, as it would be in your stomach. --Jayron3200:05, 12 September 2009 (UTC)[reply]
May have found the culperit. Almost ALL of the cinnamon you are going to get in the western world is Cinnamomum aromaticum or Cassia Cinnamon. Cassia Cinnamon contains relatively high amounts of a compound called Coumarin, which turns a light green fluorescent color at around pH 10.5 apparently. It is likely that the baking soda wash you used for the browning caused the coumarin in the cinnamon to take on a green palour. As noted above, however, such changes are harmless, as the coumarin will revert back to its previous form with no ill effects, in the low pH of your stomach. --Jayron3204:22, 12 September 2009 (UTC)[reply]
This new non lethal weapon the vortex ring gun, is planned to be mounted on a hand held system the Gl-6. It produces an explosion that is then channeled into a vortex, which is fired at the target and knocks it down. my question is how can it do this without violating the law of action and reaction. Surely a force strong enough to knock down the target would have just as strong a reaction and knock down the person fireing it?
(after ec): I guess firing this thing does not knock you down for the same reason as punching a punchbag does not knock you down. In both cases there is a reaction force pushing you back, and in both cases you are ready for it and adjust your posture accordingly in advance. --Dr Dima (talk) 21:16, 11 September 2009 (UTC)[reply]
I agree, it probably works because you brace for it. Alternatively, it could be that you feel a weak force for a prolonged period and the target feels a strong force for a short period (with the air acting as a buffer). --Tango (talk) 21:38, 11 September 2009 (UTC)[reply]
It varies. There are different types of intersex people. They can have unusual genetics, such as XXY, or perfectly normal genetics but have something unusual happen during gestation, such as odd hormones. This can result in someone with XY genes, say, ending up with female sexual characteristics. I suggest you read the intersex article, it has a list of various types of intersexuality. If you are interested in Caster Semenya, we don't know the details of her case yet. --Tango (talk) 23:33, 11 September 2009 (UTC)[reply]
As stated above, it really depends on the case. There is no single cause for intersex conditions (by the way, the term "hermaphroditism" is not really used in the mainstream to describe most of these disorders any more). There are 46,XX individuals who have a chromosomal translocation of the SRY gene onto one of the X's and develop as males. Likewise, there are 46,XY individuals with a mutation of SRY who develop as females. One of the more common intersex disorders is androgen insensitivity syndrome, where the body does not respond to androgens during development, leading to a 46,XY female. Another common disorder is congenital adrenal hyperplasia where the adrenal glands produce excessive sex hormones during development leading to virilization of a 46,XX female. I'd echo what Tango said - that details of Caster's case are unknown and we couldn't really speculate about the cause in her case. IMHO the disclosure of this information to the press was a deplorable breach of Caster's privacy. --- Medical geneticist (talk) 01:36, 12 September 2009 (UTC)[reply]
Fossil Fuels and CO2
How does the burning of fossil fuels create carbon dioxide?
Someone was trying to tell me that burning fossil fuels doesnt create carbon dioxide, it creates carbon monoxide. I knew that wasnt true but i couldnt explain how burning fossil fuels creates CO2 —Preceding unsigned comment added by 71.98.64.15 (talk) 23:52, 11 September 2009 (UTC)[reply]
If it created just carbon monoxide then everybody with a gas cooker would be dead by now. The initial burning of fossil fuels (for instance gas, petrol, diesel, etc. - they are just hydrocarbons - compounds of just carbon and hydrogen) creates carbon monoxide (which is more stable when very hot), but at the outer parts of the flame (where it is relatively cooler) the carbon monoxide burns to give carbon dioxide. The other component of burning such fuels is water - which is why car exhaust pipes sometimes drip when just started. Car engines run very hot - hence make carbon monoxide, which (these-days) has to be removed by the catalytic converter. Ronhjones (Talk)23:59, 11 September 2009 (UTC)[reply]
Most fossil fuels consist mostly of carbon and hydrogen. When you burn them, you add oxygen. The hydrogen reacts with the oxygen - gives off heat and produces water (H2O) - and the carbon reacts with the oxygen to produce more heat and to produce a MIXTURE of CO and CO2 (carbon monoxide and carbon dioxide, respectively). Carbon monoxide is a nasty poison and a greenhouse gas - but it can react with more oxygen to make CO2 - Carbon dioxide is pretty inert - but also a greenhouse gas. The precise mixture of CO and CO2 that you get depends on a lot of factors - which fuel you started with - how much oxygen was available - what temperature the reaction happened at...lots of complicated stuff. In a car, for example, burning gasoline produces mostly CO2 with a fairly small amount of CO. The catalytic converter in your exhaust system adds the extra oxygen so that the CO emissions are cut fairly drastically. If you burn propane, however - you get almost no CO at all. That's why you'll see many fork-lift trucks that have to be run inside warehouses using propane as a fuel - it means that the machine doesn't emit poisonous CO in the confined space of a building. So the person who told you this was talking complete nonsense. SteveBaker (talk) 00:38, 12 September 2009 (UTC)[reply]
I live in the cochella valley in California. We have many cockroaches (they are called palm beetles, but a cockroach by any name smells the same). I have noted what seems to me a very peculiar chracteristic: We find many dead ones in our garages and sidewalks and around the periphery of buildings. They are always on their back. I have not been able to see one in the throws od death so I'm not sure how or why they get on their backs. Does any one know or have a theory. Please, serious answers only. Thanks, wsc —Preceding unsigned comment added by 98.149.228.69 (talk) 00:09, 12 September 2009 (UTC)[reply]
Not being "right" does not mean it isn't in widespread usage. King Cnut had a better chance of turning back the tides than one has of changing an "incorrect usage" of a term like this. Consider the confused state of a word like berry, for example. Its so bad that MOST of the most commonly eaten fruits called "berries" are not really "berries" (botanically speaking) and yet no one is changing their name. SO we are probably stuck with names like "palm beetle" and "palmetto bug" even though cockroaches are neither beetles or true bugs. --Jayron3204:10, 12 September 2009 (UTC)[reply]
Perhaps being on their back (maybe after an encounter with another bug, or falling from a wall) is the cause of death, because they find they are too weak to right themselves.--Shantavira|feed me07:54, 12 September 2009 (UTC)[reply]
My wife and I are thinking of having kids, but we think it would be prudent to have an environmental impact assessment done first. Whom should we consult? 99.225.250.31 (talk) 03:54, 12 September 2009 (UTC)[reply]
I would imagine that would be extremely hard to do, bordering on nearly impossible. There are an absurd amount of factors, namely how you raise your child (to be the type of person who drives a Hummer or a bicycle, for example), how long they live, and what they do with their life. You could look up average impacts for people living in your area, although I'd suggest trying to raise the child to simply be what you and your wife consider to be best. You can also only have the one child, thereby reducing the population of the world (once you die); even better, you two can adopt! ~ Amory (user • talk • contribs)04:21, 12 September 2009 (UTC)[reply]
Interesting point. Adoption would seem to be essentially impact-neutral, as the child already exists and adoption in general isn't driven by supply & demand theory. You may be increasing the average costs & impact of that child by bringing it into a wealthier situation, but compared to creating one from scratch you're orders of magnitude ahead!
As all man made detrimental effects on the environment are made by humans it would be logical not only not to have children yourself but also to discourage everyone else of having children. I know this is highly cynical but that's not my fault. I think the sheer question as such is deeply flawed but in a way where the flaw can't be easily pointed out. And I think modern society needs to think about this type of questions. As of whom you should consult, this depends on what kind of answer you want. 95.112.165.86 (talk) 08:39, 12 September 2009 (UTC)[reply]
the organic chemistry of carbon dioxide fixation (i.e. mechanism of RuBisCO)
Yet another carbonyl question. Does the RuBisCo enzyme use the idea of nucleophilic attack on the central carbon of carbon dioxide? I tried looking at some articles but they use really elaborate physical chemistry, wavefunctions and ... no electron arrow mechanisms. Yes, I realise electron arrow mechanisms are sort of simplified but can I please see one? (Mechanisms for simpler evolutionary predecessors to RuBisCO are also welcome). John Riemann Soong (talk) 07:04, 12 September 2009 (UTC)[reply]
It depends on which part of the mechanism you are looking for, the enzyme action or the overall reaction between the CO2 and the ribulose biphosphate. Probably the RuBisCO operates on the "Lock and Key" model; that is there are binding sites for both the CO2 and the ribulose biphosphate which are brought together to form the 3-carbon sugars which enter the Krebs cycle. Our article on RuBisCO does not detail where on the molecule those binding cites are, but given that the molecular weight of RuBisCO is about 540,000, such detail is most likely out of the scope of a Wikipedia article. The exact way the electrons are "pushed" to bring about the 3-carbon sugars is also not detailed in our article, but the six-carbon intermediate is apparently 3-keto-2-carboxyarabinitol 1,5-bisphosphate, which means that if you can work out the structre of THAT; you can work out the electron pushing diagram since you have two relatively simple reactants (the ribulose biphosphate and the CO2) and a relatively simple product. --Jayron3212:01, 12 September 2009 (UTC)[reply]
This [34] is a far better visualisation of the critical step, showing the role of Mg2+, and the mechanism too. (Don't even ask me how they worked that out - I haven't got a clue - probably isotope labling experiments or something).
(Note that the enol based attack isn't as simple as usual since there is a OH next to the keto - making the intermediate structure approximately a 1,2 dihydroxy alkene... see that technically the reaction forms a bond between two carbonyl carbon atoms!!)
In this particular example, the wriggly line is the symbol that I don't know about. There surely is an article on wikipedia, but I can't locate it. Can somebody please supply a link? Thanks! --TrogWoolley (talk) 10:24, 5 September 2009 (UTC)
Resolved
.jpg){kind=spacer}
Please do your own homework.
