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May 15

Identify the spider

I'm sure it's a pretty common species, and we probably already have zillions of photos of it on Commons... but still... what kind of spider is it?

Photo was taken in Indianapolis, Indiana in July of last year. ~ ONUnicorn^{(Talk|Contribs)}problem solving 02:29, 15 May 2008 (UTC)[reply]

That appears to be a grass spider, genus Agelenopsis (all I did was look up "common spiders" and found it pretty quickly...) --DrVornado (talk) 02:34, 15 May 2008 (UTC)[reply]

Thanks. It does look like the spiders on that page... though the web it was on sure seemed sticky (see all the stuff stuck to it in the picture?) :) ~ ONUnicorn^{(Talk|Contribs)}problem solving 12:37, 15 May 2008 (UTC)[reply]

Inventions & Inventors

1. Like to know the names of atleast 5 inventors who died testing their invention 2.Kindly provide a brief detail about the invention and how the person died

Thanks 118.92.106.168 (talk) 06:17, 15 May 2008 (UTC)[reply]

William Bullock died after getting caught in his rotary printing press.
Otto Lilienthal died from injuries after a gliding accident
Franz Reichelt died testing his parachute-coat.
Thomas Midgley, Jr. invented a system of strings and pulleys to lift him from bed after he contracted polio. He got entangled in the ropes of this device and died of strangulation.
Alexander Bogdanov experimented with the concept of blood transfusion. He died after blood of a student suffering from malaria and tuberculosis was given to him in a transfusion.
Marie Curie developed the radium isolation process. She later died from aplastic anemia, likely due to exposure to radiation. As our article says, "much of her work had been carried out in a shed with no safety measures. She had carried test tubes containing radioactive isotopes in her pocket and stored them in her desk drawer, remarking on the pretty blue-green light the substances gave off in the dark."

Rockpocket 06:45, 15 May 2008 (UTC)[reply]

Believe it or not we actually have an article on the subject: See List of inventors killed by their own inventions. --Cameron (t|p|c) 08:55, 15 May 2008 (UTC)[reply]

Wikipedia, you never cease to amaze me! DMacks (talk) 16:38, 15 May 2008 (UTC)[reply]

Sinking of London

London is sinking by 1-2 mm per year due to sea-level rise. If London is sinking by 1-2 mm per year, then a calculation can be made that by 2100, London will sink 100 mm and by 2200, it will sink upto 200 mm. My question is that if a city like London sinks 200 mm, then what will the effect? Will it become inhabitable? Otolemur crassicaudatus (talk) 09:02, 15 May 2008 (UTC)[reply]

The last major flood in central London was the 1928 Thames flood. Today central London has excellent flood defences - see Thames barrier. Other parts of England are not so fortunate - see 2007 United Kingdom floods. Gandalf61 (talk) 11:01, 15 May 2008 (UTC)[reply]

This may be completely irrelevant but there was recently a film on TV about the flooding of London. It was cleverly named 'Flood', this is the link if you're interested: Flood (film). --Cameron (t|p|c) 11:03, 15 May 2008 (UTC)[reply]

I am not talking about flood, I am talking about permanent sea level rise due to global warming. What will be the effect if London permanently sinks by 200mm? Otolemur crassicaudatus (talk) 11:30, 15 May 2008 (UTC)[reply]

The only logical assumption I could make is that the water level of the Thames and other rivers will also rise by 200mm? It's clearly not a huge issue for London anyway. The Thames Barrier has more than enough leeway for 2cm of water rise. Either of two things would be needed for a devastating effect: a) a significant flood which the barrier can't handle or b) global warming to continue rising the sea level until the level of the Thames is too high. Regards, CycloneNimrod^Talk? 12:03, 15 May 2008 (UTC)[reply]

The Thames Barrier only protects against floods when it's closed. If sea levels rose, it would have to be closed more often, which could hurt the city's economy a bit, I expect. If they rose enough that the barrier had to be permanently closed, a new solution would have to be found. --Tango (talk) 12:59, 15 May 2008 (UTC)[reply]

I guess not a lot, except they may have to raise flood defences by a similar amount. 20cm isnt that much anyway. -- Q Chris (talk) 12:05, 15 May 2008 (UTC)[reply]

You can't make this calculation without some theoretical model of the future behavior of the London-Thames system. It's looks like you're linearly extrapolating the data without any theoretical motivation, which would make this similar to the prediction that the entire world's population will consist of Elvis impersonators by 2015. -- BenRG (talk) 13:11, 15 May 2008 (UTC)[reply]

BenRG is right about it not being linear. Firstly there are two processes combining: relative to a fixed point (eg. the centre of the Earth), there is an isostatic subsidence of the land due to the effects of ice during the last ice age, and the global rise in eustatic sea level due to the various effects of global warming. The subsidence may be approximately linear, although it's probably slowing down slightly. The rate of global sea level rise is controversial, but likely to be increasing. There's a more detailed explanation of the figures here, and some interesting stuff on the Thames barrier here. Eve Hall (talk) 17:56, 15 May 2008 (UTC)[reply]

As an aside, London is currently inhabitiable. That is to say, it is able to be inhabited. The question is whether or not London would be made uninhabitable due to its sinking. Ahh, English! — gogobera (talk) 17:21, 15 May 2008 (UTC)[reply]

Hi. Well, the 10 cm of sinking in 100 years might not be a problem, but sea level rise in 100 years might be, as it could be anywhere between 200 - 2000 mm. If it's closer to the latter, then more than likely the Thames Barrier would have to be permanently closed, or else 3 million Londoners would need to find new homes. Also, the sea level rise might speed up the sinking, by compressing the soil and eroding it. Venice is also sinking, and sea level rise would make it worse, but make coastal city-dwellers could learn from the Venicians and survive by paddling through the streets. Thanks. ~A H 1^(TCU) 17:42, 15 May 2008 (UTC)[reply]

Permanently closing the barrier isn't an option - where would all the rainwater from upriver go? If sea levels rise such that the barrier can't at least be opened at every low tide, some other solution would have to be found. What is the typical tidal variation at the barrier, anyone know? --Tango (talk) 13:33, 16 May 2008 (UTC)[reply]

Just to clarify, but if I'm reading the introduction to the article correctly, the "sinking" is not due to sea level rise, but combining with it to produce a "double whammy". That is, any sea level rise is a separate issue from the landmass sinking. I didn't have time to read the article, but shouldn't the islands still be rising from post-glacial rebound? Matt Deres (talk) 18:31, 15 May 2008 (UTC)[reply]

London isn't too bad off compared to other parts of the North Sea. Dragons flight (talk) 18:56, 15 May 2008 (UTC)[reply]

Cold Cathode Fluorescent Lamps used as Backlights in LCD computer monitors and televisions

With reference to the Wikipedia article on Cold Cathode:

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

I should be very grateful if either a contributer to the article or a user knowledgeable about the construction and technology of CCFLs used as LCD backlights in televisions and computer monitors could clarify whether these CCFLs actually do contain a radioactive material that emits beta particles to start the ionization of the gas. If this is the case, then would it be correct to assume that facing the back of the computer monitor where the backlight can be seen shining through the grill that one is being exposed to ionizing radiation - since the fast moving beta particles would penetrate the thin glass of the backlight? 78.147.162.37 (talk) 12:13, 15 May 2008 (UTC)[reply]

Julia Howard —Preceding unsigned comment added by 78.147.162.37 (talk) 12:01, 15 May 2008 (UTC)[reply]

I'm no expert, but I should think any possible risk is very, very low here. Please consider self-powered lighting. There, a much more powerful source of beta radiation (that is, a source of high-energy electrons) is used to directly cause the phosphor coating of the tube to glow. Yet the enclosing tube is still glass (albeit, possibly leaded glass) and there seem to be few worries about radiation exposure unless the glass tube is ruptured. In a CCFL, any possible radiation source is there just to facilitate initial ionization, much like in many neon lamps (where ionization sources are definitely used). The magnitude of the radiation from this source must be many orders of magnitude lower than with the self-powered-lighting.

Atlant (talk) 12:47, 15 May 2008 (UTC)[reply]

Heel on wooden floor

When we move on a wooden floor with heel shoe why marks or scratches are left on the floor? —Preceding unsigned comment added by MAQMAQ (talk • contribs) 12:08, 15 May 2008 (UTC)[reply]

I'm not sure I understand the question so let me ask a clarifying question: Are we talking about shoe heels in general or a stiletto heel? In the first case, marks can be left because the rubber material of the heel can rub-off and transfer to the floor. Alternatively, small abrasive particles such as sand that become embedded in the heel can abrade the floor. In the second case, the concentration of weight on the very small area of the stiletto heel can exceed the strength of the floor, causing crushing of the material of the floor.

Atlant (talk) 12:57, 15 May 2008 (UTC)[reply]

Discharging a capacitor on a metal sheet

Hi all,

Let's say I charge a capacitor. The capacitor now has one plate with more negative charges than positive, and one plate with more positive charges than negative. If I then touch one of the plates to a hunk of metal, will I discharge half the capacitor? I'm guessing yes, because the charges between the plate and the metal will balance out, and since the tiny capacitor won't make much of an effect on the hunk of metal, both the plate and the metal will end up with (almost) no net charge. Is that right?

Second question: What will happen if I touch the negative side of a battery to a hunk of metal? Anything at all? Thanks!

You still have a capacitor, but now it's formed between the one original plate of the capacitor and the other plate acting in combination with the hunk of metal. The charge that was on the plate of the capacitor that is now in contact with the hunk of metal sheet will re-distribute itself among the new, expanded plate so as to minimize the stored potential energy. If you then separate the capacitor plate from the added hunk of metal some of the charge will be left behind on the hunk of metal and the original capacitor will end up holding less differential charge than when you started.

Atlant (talk) 15:59, 15 May 2008 (UTC)[reply]

However, it seems like the electrostatic field between the positive and negative plates of the capacitor would tend to keep the charges on the plates rather than letting them disperse much onto the metal sheet. Attraction between those positive and negative charges and all that. --Prestidigitator (talk) 16:06, 15 May 2008 (UTC)[reply]

Something similar will happen with the battery: a bit of charge will leak off the battery to the hunk of metal and remain there when you remove the battery, but not much. In the case of the capacitor the charge on each side of the dielectric is fixed but the voltage will change slightly when you add the hunk of metal; in the case of the battery the voltage across the terminals is fixed but the net charge on each side will change slightly. -- BenRG (talk) 19:09, 15 May 2008 (UTC)[reply]

No! It depends on the initial potential difference between the metal plate and the capacitor - sure, the charge on the plate and that side of the capacitor will redistribute itself such that the plate and that terminal of the capacitor end up with a potential difference of zero - but whether that discharges the capacitor a bit - or charges it up a bit more depends entirely on the initial charge on the plate. You could discharge a capacitor (very slowly) by alternately touching the plate to the positive and negative terminal of the capacitor - thereby 'ferrying' some electrons from one plate to the other by carrying them on the plate. But equally, if you put a highly positively charged plate against the positive terminal of the capacitor - you could charge it up a teeny tiny bit. 70.116.10.189 (talk) 11:53, 19 May 2008 (UTC)[reply]

That's actually a very good point, and is the operating principle behind the charge pump and the similar "flying capacitor multiplexer"; thank you for raising it!

Atlant (talk) 14:16, 19 May 2008 (UTC)[reply]

Mice and candle wax

Do mice eat candle wax because it is good for them or because there is nothing else and they are hungry? Because my bro said the mice in his house were eating the candle wax from his candles. What nutritional benefit would a mouse get from eating the candle wax from a candle? Terror toad (talk) 13:13, 15 May 2008 (UTC)[reply]

Depending on what the candles are made of, they can certainly be nutritious. I did a bit of survival training as a Scout and one of the things we were told was good to have with you is an old fashioned tallow candle. It can provide light, help light a fire and if you get desperate, you can eat it - it's just animal fat. Beeswax is also edible. I'm not sure paraffin wax would be too nice to eat, though, and that's the most common wax in modern candles according to candle. --Tango (talk) 14:10, 15 May 2008 (UTC)[reply]

I've wondered the same about soap, which mice will regularly consume from my parents' summer cottage. Can they really digest the saponified fatty acids? —Ilmari Karonen (talk) 15:54, 15 May 2008 (UTC)[reply]

spectral lines of helium

What is the formula for the spectral lines of helium? Em3ryguy (talk) 14:16, 15 May 2008 (UTC)[reply]

I don't think "formula" is quite the right term, as there's no f(x)=y sort of thing, but here's helium's spectrum. — Lomn 14:47, 15 May 2008 (UTC)[reply]

If you're talking about singly-ionized helium, you can use a version of the Rydberg formula. -- Coneslayer (talk) 15:36, 15 May 2008 (UTC)[reply]

No. The full spectrum. Em3ryguy (talk) 15:54, 15 May 2008 (UTC)[reply]

Well, you can look up all the spectral lines of hydrogen in the CRC handbook here (warning - large PDF file). Does that help? --Bmk (talk) 17:06, 15 May 2008 (UTC)[reply]

You may also find this page helpful. -- Coneslayer (talk) 16:01, 15 May 2008 (UTC)[reply]

Joules, Newton Meters, and Foot Pounds

I know in metric/SI units that Joules referes to mechanical work and newton meters refers to torque, even though they're bot N x m. But foot pounds are a little more ambigious. Is it acceptable to use foot pounds interchagably between mechanical work and torque? Thanks Deltacom1515 (talk) 16:14, 15 May 2008 (UTC)[reply]

No idea why that's not formatted correctly. Deltacom1515 (talk) 16:14, 15 May 2008 (UTC)[reply]

According to our article on torque and our article on foot-pound, the foot-pound is a standard unit for both torque and work (in the US customary units system). And btw, if you have a space at the beginning of a paragraph, wikipedia puts the whole paragraph in a weird boxy thing (I fixed it for you). --Bmk (talk) 16:19, 15 May 2008 (UTC)[reply]

And as foot-pound notes, some people use "pound-foot" as the unit of torque, reserving "foot-pound" for work. -- Coneslayer (talk) 16:21, 15 May 2008 (UTC)[reply]

Yes. Further, that's what I was taught forty years ago, so it's not exactly a new convention :-)

-- Danh, 70.59.79.51 (talk) 21:09, 15 May 2008 (UTC)[reply]

Wow, way to read. I completly missed that. Guess that answered my question. Deltacom1515 (talk) 16:33, 15 May 2008 (UTC)[reply]

Area of cathode in an MFC

It is said that the area of the cathode in an MFC helps improve the efficiency of the MFC. Is this true? Why is it so? —Preceding unsigned comment added by 59.92.47.158 (talk) 16:39, 15 May 2008 (UTC)[reply]

Particle deceleration

note: question moved from Wikipedia talk:WikiProject Physics

I was wondering, since it's possible to accelerate particles to near the speed of light, is it possible to decelerate particles, and if so, how? (nb. I failed physics, so I wouldn't know myself)Avnas Ishtaroth (talk) 12:49, 15 May 2008 (UTC)[reply]

Yes. The easy way is to put a nice thick lead wall in front of your particle beam, then stand well away from the Bremsstrahlung radiation. You can also (in a hand-wavy sort of sense) reverse your particle accelerator to extract work - the physical processes involved work both ways. See laser cooling and Bose-Einstein condensate for some how and why getting a group of particles moving really slowly in some particular reference frame. - Eldereft ~(s)talk~ 16:46, 15 May 2008 (UTC)[reply]

What particles? Kamiokande uses water. In this case you get Cherenkov radiation. --Lisa4edit (talk) 17:04, 15 May 2008 (UTC)[reply]

The question of "decelerating" a particle is really a question of giving your particle a negative acceleration. That is to say, it's velocity should decrease with time. Of course, the smallest velocity a particle can have is zero. But the question remains, zero velocity with respect to whom. For instance, say you have some charged particles traveling and accelerating to the right due to an electric field pointing to the right. If you turned off the field, the particles would simply travel to the right at some constant speed. (They're in a vacuum so there is no friction.) If you turn on an electric field in the opposite direction, they would "decelerate" until they stopped and then begin accelerating to the left.

Physicists generally talk about acceleration in different directions (to the left and right, etc), rather than "deceleration". Heh, even the deceleration article simply redirects to acceleration! — gogobera (talk) 17:40, 15 May 2008 (UTC)[reply]

I don't believe it's possible to decelerate all particles, however. Photons can't go slower than the speed of light and the theoretical tachyon always travels faster than the speed of light. StuRat (talk) 11:21, 16 May 2008 (UTC)[reply]

Well, photons travel at the speed of light in the medium in which they are travelling (see refractive index). For anything other than a vacuum, that is a speed less than what is colloquially called "the speed of light" (c). Again, follow Lisa4edit's advice to see Cherenkov radiation (I'll add it as a see-also to refractive index). DMacks (talk) 14:45, 16 May 2008 (UTC)[reply]

And in that sense light can actually be decelerated a lot! --169.230.94.28 (talk) 02:41, 17 May 2008 (UTC)[reply]

Thanks, this is very interesting. Is it theoretically possible to stop light? If not, what's the slowest it could go and how could this work?Avnas Ishtaroth (talk) 05:32, 19 May 2008 (UTC)[reply]

It is actually possible to drastically slow light (towards the m/sec range), and there is a stop-and-forward scheme. It's late at night where I am and I will have to forward the Nature articles, but I will look them up tomorrow. Or someone else could search nature.com for "slow photons" in the meantime. Franamax (talk) 05:46, 19 May 2008 (UTC)[reply]

Dandelion invasion

Hi. Today when I hid the sun behind an object that just obscures the sun's disk, I saw what looked like a heavy snowstorm, only it was May, and the sky was a clear blue with only a few fractus clouds in the distance. Anyway, and this is not homework, I speculated that they were dandelion seeds, which was confirmed when I saw them close up. However, around the sun, several hundred seed packets could be seen, ranging from probably 1 - 50 metres above ground. The wind was strong and from the north, I live in southern Ontario. Anyway, there must have been one per every cubic metre or two, as the highest ones near the sun were maybe as small as 10 arcmins apart from each other. Where did they all come from? To my north there are about 20 hectares of land capable of producing seeding dandelions. This means there must have been over a million individual seeding dandelions producing seed packets! In early to late June, there may be as many as 500 dandelions per square metre in some places, I estimate, if they haven't been mowed yet. Are they an invasive species in North America? If the climate warms a few degrees, warm conditions become longer throught the year, and rainfall becomes sporadic but interspresed with heavy and brief downpours, will there be more invasive dandelions? If so, what will happen to the local ecosystems? Also, do flowering dandelions close and reopen to become seeding ones, and how long does this take? A look at any random part of the sky or landscape for a few seconds will reveal at least a few of these seed packets, and sometimes several dozen! Do people consider them weeds because they're an invasive species or because they choke other vegetation? Thanks. ~A H 1^(TCU) 17:34, 15 May 2008 (UTC)[reply]

Have a look at the dandelion article to answer some of your questions. A weed is not the same as an invasive species which in turn is not the same as an introduced species. If you get dandelions in your lawn you'll consider them a weed, even if they are useful to local wildlife and co-exist with indigenous species. An introduced species can adapt and become part of the local ecosystem, it may cause some temporary imbalances, or displace a local species, though. If it doesn't multiply in excess and is not harmful to local wildlife, ideally finding some animal that finds it palatable, an introduced species is no better or worse than an indigenous species. (Debatable if it replaces one that then becomes extinct.) An invasive species disrupts the local balance and multiplies in excess. (Common definition). Not all invasives are non-native, but it is easier for an introduced species to become one, because it may lack natural enemies. Temporary proliferation of a certain plant doesn't mean it is invasive. It may serve as a much needed food-source for animals. Dandelions are rather undemanding when it comes to growing conditions, so they can grow in a wide range of habitats. They are rather tasty though (even to humans - try them in a salad), so any mass of them will soon find something to nibble off the leaves. Hope this helps. --Lisa4edit (talk) 18:04, 15 May 2008 (UTC)[reply]

Does light attract itself gravitationally?

The other day I saw an interesting episode of The Universe which discussed the end of the universe through heat death, with the final dark age of the cosmos being one in which all black holes have finally evaporated and all that is left are stray random photons.

This got me wondering, though. Since light is a form of energy, and general relativity says that energy as well as matter bends space/time, does it follow that light can attract itself through gravitational force? After all, if a photon of light contains energy than it should, I think, very slightly bend space/time and thus attract energy and matter. Would it follow that two beams of light fired initially in parallel would slightly bend toward each other even in the absence of other energy or matter? And if so, would it be possible that in the universe's final dark age that stray photons could very, very slowly bend towards each other into wide orbits, producing large galactic size pockets of slightly higher than normal electromagnetic energy or heat? Dugwiki (talk) 17:41, 15 May 2008 (UTC)[reply]

does it follow that light can attract itself through gravitational force?
Yes, light has a gravitational effect that can attract other light. Dragons flight (talk) 18:16, 15 May 2008 (UTC)[reply]
Would it follow that two beams of light fired initially in parallel would slightly bend toward each other even in the absence of other energy or matter?
Not exactly. Gravitational fields propogate at the speed of light as well. Parallel photons would not interact because the gravitational field from one photon could never catch up to the other one. However, if you imagine these as beams of light with many photons, then yes they would interact. The gravitational field from early photons in one beam would effect later photons in the other. This effect would be ridiculously small. Dragons flight (talk) 18:16, 15 May 2008 (UTC)[reply]

Gravitational fields propogate at the speed of light as an older physicist this was not established in my day but was expected to be true, if this has changed can you link me the experimental evidence! (very interesting question, this bit in particular, by the way). Two photons bending towards one another is excactly the sort of evidence that was/is required. GameKeeper (talk) 22:23, 16 May 2008 (UTC)[reply]

See Speed of gravity#Experimental measurement?. The evidence seems to support the theory, but apparently not everyone is convinced. --Tango (talk) 22:27, 16 May 2008 (UTC)[reply]

would it be possible that in the universe's final dark age that stray photons could very, very slowly bend towards each other into wide orbits, producing large galactic size pockets of slightly higher than normal electromagnetic energy or heat?
I think the answer is no, but I'm not entirely sure. No matter how large you imagine the universe to grow, I don't think you could every get a large enough density of photons (given initial conditions such as exist today) to create bound orbits for other photons. I think the energy density of light is too low and the velocity too high to ever allow for gravitationally bound conglomerations of photons in the future universe. Dragons flight (talk) 18:16, 15 May 2008 (UTC)[reply]
That sounds right. For photons to orbit, you need something close to a black hole (I think something 8/27 times the density of a black hole would work - a very large neutron star, maybe). Radiation density on the scale of neutron stars seems highly improbably, if even possible. --Tango (talk) 18:43, 15 May 2008 (UTC)[reply]
Remember that black holes don't have a single density (even an average one): the Schwarzschild radius scales with the mass, which in turn scales with the radius cubed at constant (average) density. So any substance whatsoever can form a black hole if it fills a large enough spherical region. The cosmic background, say, with a density of 4.64×10⁻³⁴ g/(cm³), would form a black hole as a ball with a radius of two trillion light years. (It might be closer to 1.4 trillion; I'm not sure that I shouldn't give it a gravitational density of ${\frac {2u}{c^{2}}}$ ; see my oft-repeated link on the subject.) Using instead the critical density of 9.2×10⁻³⁰ g/(cm³) the radius is 14 billion light years; that this is approximately the age of the universe times the speed of light is not a coincidence. --Tardis (talk) 17:57, 16 May 2008 (UTC)[reply]
Excellent point, however even with such a large universe the CMB wouldn't allow for photon orbits - they are usually calculated for a black hole in otherwise empty space, if the universe has roughly uniform density, all the gravitational forces roughly cancel out and you won't get orbits. It's not the density that needs to be on those scales but rather the variation of density (or some function of it, I don't have an envelope to hand). --Tango (talk) 18:18, 16 May 2008 (UTC)[reply]
Not true, Tango. The presence or absence of a spherically symmetric mass distribution outside of a given spherical surface has no effect on the gravitational field inside that spherical surface. This is the Shell theorem, which follows from Gauss's law for gravity. (This is why classical models of a static universe are gravitationally unstable.) Tardis's argument is valid. --169.230.94.28 (talk) 02:36, 17 May 2008 (UTC)[reply]
I can't see how that can work when you're talking about the universe as a whole. By symmetry, it has to cancel out, otherwise the net pull would be in a certain direction, destroying isotropy. --Tango (talk) 12:05, 17 May 2008 (UTC)[reply]

(Edit conflict) While photons don't have rest mass, they do have relativistic mass. However, it certainly isn't much mass (for example, I believe a 400 nm photon should have a mass of about 6×10⁻³⁶ kg). The wave nature of photons causes beam divergence (see diffraction), and as far as I know any gravitational curving of the path due to other photons is negligible compared to that. --Prestidigitator (talk) 18:25, 15 May 2008 (UTC)[reply]

Suspensions

I am looking for examples of situations where there is a need to keep solid particles suspended in water. One example I know of is oil drilling - as the drill progresses the little bits of broken rock need to be carried away in a stream of water - certain additives are added to the water to help keep the rock bits in suspension. Can you think of any other situations? I am more interested in industrial or agricultural applications, less in medical or food applications. Thanks! ike9898 (talk) 17:42, 15 May 2008 (UTC)[reply]

Magnetorheological fluids, though I think those generally use oil, presumably for its higher density and viscosity. One of the ways of making artificial diamonds uses a slurry of diamond grit slowly deposited by a controlled temperature differential. Generally, a gritty liquid like you describe is called a slurry. - Eldereft ~(s)talk~ 20:55, 15 May 2008 (UTC)[reply]

Detergents have non-surfacant ingredient. This keeps the dirt from settling back on the clothes or other surfaces, so it can be flushed through the drain. --Lisa4edit (talk) 06:44, 16 May 2008 (UTC)[reply]

greenhouse gases

how much methane and CO2 do 6 billion people breate out or fart out each day? Paul kahlich (talk) 17:49, 15 May 2008 (UTC)[reply]

We emit 1.5-3.5 kg of CO2 per person per day (1-2 lbs of carbon per day per person). But that doesn't matter because all the carbon we emit comes, directly or indirectly, from plants which gathered their carbon out of the atmosphere in the first place. Hence respiration has no impact on the global carbon balance. Dragons flight (talk) 18:01, 15 May 2008 (UTC)[reply]

If that were true - why would we be concerned with emissions from cows? The problem is that while you're correct in saying that the CO₂ production is balanced by the carbon we consume in eating plants - the methane we produce is not. Methane is a much worse greenhouse gas than CO₂ (75 times worse!) and it's not reabsorbed by the next generation of plants that we grow to replace the ones we ate. Eventually, methane does decompose into CO₂ - but it has a half-life of something like 100 years and since human populations are growing, we're inherently increasing the greenhouse problem. However - as bad as this is, it's nothing like as bad as with animals like cows who ruminate on their food - producing vastly more methane pound-for-pound than humans. 70.116.10.189 (talk) 04:29, 16 May 2008 (UTC)[reply]

Correction. See the methane article, second paragraph. The half life in the atmospere is 7 years, not 100. And because of the decay, the effect gets ever smaller, so one can never state the global warming potential (the effect compared to CO2) without specifying over which period. Most quoted is a number of 23 (stronger than CO2), which apparently is for a period of just over 100 years. Note that 105 years is 15 half-life periods (15 x 7), so by then the amount of methane is reduced to 1/(2^15) = 0,00003 times the original quantity. In other words, it's almost all gone and any global warming effect will be by the resultant CO2. Amrad (talk) 10:41, 16 May 2008 (UTC)[reply]

But temperatures have gone down in spite of rising CO2 levels... isn't global warming attributed to water vapor or sunspots?--WaltCip (talk) 12:16, 16 May 2008 (UTC)[reply]

Temperatures have gone down? Are you joking or do you read taboids of the worst kind? Maybe you read that at some localities temperatures have dropped, but globally, the rise in temperature is so immense that even after a few decades it's a certainty even for the scientific community (as opposed to individual scientists), which is usually extremely careful with its claims. And concerning the solar activity, yes, that plays a role, but it's negligible compared to the effects of the rise in CO2 levels. Btw, sun spots are just another effect of increased solar activity, certainly not a cause for warming. Amrad (talk) 13:46, 16 May 2008 (UTC)[reply]

density of paraffin wax

I'm looking for the density of paraffin wax in liquid phase. The wax we're using melts at 85°C (used in a car thermostat just in case you're wondering) but I'm not familiar with any good sites or databanks? Thanks in advance —Preceding unsigned comment added by 193.190.253.149 (talk) 18:58, 15 May 2008 (UTC)[reply]

~~Well, a database referenced in our paraffin article, found here, lists "paraffin oil" as 0.8 g/cm^3.~~ Nevermind - turns out paraffin oil is another word for kerosene (as I found out in our paraffin oil. --Bmk (talk) 20:28, 15 May 2008 (UTC)[reply]

It seems that mineral oil is at least close to being liquid paraffin (the alkane lengths seem somewhat smaller than in paraffin wax, but that probably doesn't matter too much). (Its density is actually given only in the latter article and not its own; should fix that.) --Tardis (talk) 16:15, 16 May 2008 (UTC)[reply]

Trees vs Lamp posts

What are the relative strengths of lamp posts and trees (say, of equal width?) when struck, such as by a speeding car? Has there been experiments done on this?--Fangz (talk) 19:30, 15 May 2008 (UTC)[reply]

Well, many lamp posts and street signs are designed to collapse when hit by a vehicle (rather than kill the occupant), so that would definitely need to be taken into account. I saw a bus hit one once and it just toppled over without any resistance whatsoever. --98.217.8.46 (talk) 19:39, 15 May 2008 (UTC)[reply]

A lamp post that's solidly planted in the ground will be much stronger than an equivalently-sized tree. However, every lamp post I've seen has had a breakaway joint at the bottom that shears off easily when struck by a vehicle, but I've never seen a tree with a breakaway joint. --Carnildo (talk) 22:16, 15 May 2008 (UTC)[reply]

Apropos safety, shouldn't those falling lamp posts have a speaker that goes "timbeeeeeer", to warn pedestrians? Amrad (talk) 10:28, 16 May 2008 (UTC)[reply]

A few questions about neurotransmitters

1. Is it possible for the brain/nervous system to run out of transmitter chemicals? 2. If so, how and from what is the supply replenished? 3. If neurotransmitters are re-used, where are they stored when not in use? —Preceding unsigned comment added by 69.224.182.55 (talk) 20:37, 15 May 2008 (UTC)[reply]

Is the infomation in neurotransmitter not suffient for your needs? DMacks (talk) 20:57, 15 May 2008 (UTC)[reply]

[EC] Certainly not under normal circumstances - neutotransmitters are manufactured within neurons. I would recommend reading our article on the chemical synapse - I think it will clear things up. According to that article, most neurotransmitters are taken into either the presynaptic or postsynaptic cell for some degree of decomposition, and then either recycled or disposed of. --Bmk (talk) 20:59, 15 May 2008 (UTC)[reply]

Squirrel feces

Why don't you ever see Squirrel feces on the ground? You see Bird / duck feces, but never other large animals like raccoons. LLOTAAMI (talk) 20:56, 15 May 2008 (UTC)[reply]

I see squirrel feces all too often. Just put a bunch of birdseed on the ground, and wait. -- JSBillings 22:40, 15 May 2008 (UTC)[reply]

That works for raccoon doo-doos too. They really liked corn on the cob we had dangled from the gutter! (Don't try this if you have a cat! The raccoon may get hungry after it's done with the bird seed.)--70.91.165.182 (talk) 00:49, 16 May 2008 (UTC)[reply]

I have a science project that I am allowed to get help for some information

Hi, my name is Lindsay, a fifth grade student. I need help with this topic: LIST WATER'S FREEZING POINT AND MELTING POINT IN BOTH DEGREES, CELSIUS AND FAHRENHEIT.I really need help on it and I couldn't find any imformation on it. I'm still doing my rough draft that is due May 20,2008.PLEASE HELP ME!!!! —Preceding unsigned comment added by 71.243.52.218 (talk) 21:56, 15 May 2008 (UTC)[reply]

The Reference Desk has a firm policy that we will not do your homework for you. However, since this is a project about water, I suggest you start at our water article. The top of that article notes that for chemical and physical properties (such as freezing point), you'll want to refine that to water (molecule) (I'm listing both to illustrate the research process). If necessary, Google can be used to do the Celsius/Fahrenheit conversion: search "98.6 fahrenheit in celsius", for example, and Google will return 37 degrees celsius, which is the average human body temperature. The relevant formulae can be found at both Celsius and Fahrenheit, incidentally. — Lomn 22:02, 15 May 2008 (UTC)[reply]

If you don't already know it, the freezing and melting point of a substance are both approximately the same temperature. Also note that "the freezing/melting point of water" can mean different things. 100% pure water has one freezing/melting point, while the salt-water found in the oceans has a different one. If the teacher didn't specify, they probably mean 100% pure water. The pressure also affects the freezing/melting point, but I'd assume they mean normal air pressure at sea level, called "one atmosphere", unless otherwise specified. You should say all this in your results, though: "The freezing/melting point of 100% pure water at one atmosphere pressure is ...". There are two other temperature scales, named the Rankine scale and Kelvin scale, so including the freezing/melting point of water on those scales might impress the teacher. You might also want to include the freezing/melting point of ocean water on all 4 temperature scales. That should be enough to get an A, I would think. StuRat (talk) 11:04, 16 May 2008 (UTC)[reply]

Benzene resonance

Are benzene and other aromatic compounds are often represented as a resonance of 2 structures with double bonds. Is benzene really a superposition of 2 cyclohexatrienes, i. e. if you somehow detect an electron between carbon 1 and carbon 2 and then do a measurement (very quickly after the first measurement) is there a higher chance of finding an electron between carbon 3 and carbon 4 or carbon 5 and carbon 6 than in the other gaps between the carbon atoms? Thanks. 201.66.22.129 (talk) 22:31, 15 May 2008 (UTC)[reply]

Resonance is a single hybrid structure that is an average of the individual resonance structures, not a set of discrete structures that rapidly change one to another. If you could freeze the pi electrons in-place and observe where these electron particles are at a certain time there may be more between some carbons than others, but a series of such measurements would not have them jumping one position to the next around the ring at alternate times, and that's a poor description of how electrons actually are anyway. So the "real" picture is that there is some pi bond at all positions at all times, but not equivalent to a normal "double bond" at any position ever. DMacks (talk) 22:45, 15 May 2008 (UTC)[reply]

Thanks, I didn't say "very quickly" because I thought of alternating states but because I thought the measurement could disturb the molecule. I am more interested in knowing if the electrons are really entangled in such a way that a measurement would cause them to fall into one or the other configuration for a short time. 201.66.22.129 (talk) 23:32, 15 May 2008 (UTC)[reply]

No. The actual molecular orbital diagram for a conjugated system is completely non-classical: looks very little like "double bonds in various places" at all. If one tried to figure out where each pair was (either by freezing in place, or by somehow observing them), one wouldn't see any pair that was anything like an alkene. Actually, the electron pairs aren't even all equivalent to each other, so the whole concept of three alkene-like pairs chasing each other around (or merging into a single cloud) is pretty wrong. But rather than the electrons themselves being entangled or the "cause" of this mess, it's more that all the p orbitals are entangled (LCAO is a reasonable approximation) and then pi electron pairs simply (fill the lowest molecular orbitals). Each molecular orbital looks almost nothing like an alkene. WP really needs a MO diagram for an aromatic system but doesn't seem to have one. Here is what I mean: notice the shapes of the colored regions for each pair. DMacks (talk) 01:39, 16 May 2008 (UTC)[reply]

I'm surprised that

\pi _{2}

through

\pi _{4}

aren't degenerate sums of three MOs, that might be split by an impressed magnetic field. (I.e., shouldn't the geometric symmetry in the molecule under rotation by 60 degrees in the plane result in additional orbitals at these energy levels?) -- Fuzzyeric (talk) 00:41, 18 May 2008 (UTC)[reply]

Need to have orthogonal combinations--if the MOs aren't independent, the electrons in them don't have unique quantum numbers (even two energetically degenerate MOs need to be two different MOs). So if you have a basis of "three MOs", you can only get three independent sums. OTOH, if you start with "three pi bonds", then you have six MOs (3 π + 3 π*), which makes sense because they came from 6 p AOs. If you try to work out some linear combinations of those 6 "double-bond orbitals", you wind up with something similar to the original picture from my previous URL.

Taken another way, the number of nodes in the MO is a good representation of how (un)stable the MO is. A non-node is a good overlap between adjacent p AOs and a node is an antibonding combinations of adjacent p components; more nodes->less stable. So now you just need to figure out how many orthogonal ways you can arrange a certain number of nodes on the benzene coordinate system. Again, the same result results. DMacks (talk) 21:56, 19 May 2008 (UTC)[reply]