February 14

In science class, we were discussing the upcoming lunar eclipse and an interesting question came up. If one were on the moon during a lunar eclipse, what would happen? Would the moon be in a shadow? Or would it be brighter? Admittedly, this was assigned as an extra credit question, but I have tried searching online and have no idea. I had the idea that it would be darker on the moon, but I could find no information online to corroborate that. Oh, and please include sources, if possible! Thanks, FruitMart07 (talk) 01:44, 14 February 2008 (UTC)[reply]

You're thinking along the right lines, and if you continue you'll appreciate that it's going to look something like a solar eclipse seen from the earth. Both are eclipses of the sun. The difference is that the earth has a bigger diameter than the moon, and so one would not see a full corona, but might witness Baily's beads. And yes, it will be dark on the moon, in the same way as it is dark on the parts of the earth across which the solar eclipse passes. --Tagishsimon (talk) 02:04, 14 February 2008 (UTC)[reply]

But there's another big difference. Light passing through an atmosphere gets refracted and scattered. (The Earth's atmosphere scatters blue light more, so when you look through a great amount of it, like at the setting sun, you see red because that's what didn't scatter so much.) Now, in a solar eclipse, the light does not enter the Earth's atmosphere until it has gotten nearly all the way to Earth, so it doesn't have time to scatter much. The result is that in the umbra the light is almost totally blocked and it's almost like nighttime.

In a lunar eclipse, on the other hand, the light hits the Earth's atmosphere first and starts spreading out due to scattering, with the red light spreading least. It then travels roughly a quarter-million miles, still spreading out. The result is that some light scatters enough to fill the umbra, which is why during a lunar eclipse you can still see the Moon and it looks reddish. If the Earth isn't blocking all the light because some gets through the atmosphere, what would it look like if you looked at the Earth when it was in front of the Sun? --Anonymous, 02:53 UTC, February 14, 2008.

If you were on the Moon, watching the Earth eclipse the Sun, during totality, the Earth would probably appear as a black disk with a thin reddish ring around it. Other things to look for are stars and city lights. I don't know about the relative brightnesses of the ring, the lights, and the stars, so one or more might be too dim to be seen. --Carnildo (talk) 21:50, 14 February 2008 (UTC)[reply]

Thank you guys so much!! Everyone says this class is really hard, so I'm trying to rake up a lot of extra credit points... I concluded that there would be less light on the moon because the earth would be blocking much of the sunlight. It's a straight-forward conclusion, really, but I just needed to think it through. Thanks for all of the help! (Oops! Forgot to log in!!) FruitMart07 (talk) 00:07, 15 February 2008 (UTC)[reply]

OK, I've spent the past 2 hours reading and rereading everything on Mercury's orbit around the sun, and how it relates to relativity. So, I think I get it, but I would appreciate a nice laymen's explanation that details the major equations. Also, any links to failed theories that attempted to explain Mercury away would be appreciated. Thanks! Zidel333 (talk) 04:09, 14 February 2008 (UTC)[reply]

I don't know if this is what you're looking for, but I think the GR perihelion shift can be correctly understood in terms of the spatial curvature of the Schwarzschild metric, specifically the fact that the circumference of a circle centered at the sun is less than 2π times its "effective" radius. When you've completed what would be one Newtonian orbit, you've actually gone a bit farther than a full circle, so the orbit precesses forward. Let me work this out numerically and see what happens. We're ignoring t, and also θ since everything is in a plane, so we're left with ${\displaystyle ds^{2}={\frac {r}{r-2m}}dr^{2}+r^{2}d\phi ^{2}}$. As you probably know, this can be embedded in ${\displaystyle \mathbb {R} ^{3}}$ with cylindrical coordinates ${\displaystyle (r,\phi ,z)}$ by taking ${\displaystyle r=r,\phi =\phi ,z={\sqrt {2m(r-2m)}}}$. Supposing Mercury's orbit is roughly circular with a radius of R (which is not even close to true, but never mind), the only part of this embedded geometry that matters is the annular region at radius R. To first order this looks like an annular region of a cone with a slope given by ${\displaystyle \left.{\frac {dz}{dr}}\right|_{r=R}={\sqrt {\frac {2m}{R-2m}}}}$. If you slit open a cone along a line segment from the apex to the base, you can flatten it into a disc with a slice missing. The angle of the missing slice, as a fraction of a whole circle, is our predicted perihelion shift per orbit. For a cone slope of n, that fraction is ${\displaystyle 1-(1+n^{2})^{-{\frac {1}{2}}}\approx {\frac {1}{2}}n^{2}}$, or in this case about ${\displaystyle {\frac {m}{R}}}$. Plugging in 1.5 km for m and Mercury's semi-major axis for R, that's about one part in 40,000,000, which is pretty close to the right answer of one part in 12,500,000. There might be a missing factor of two in there from the temporal curvature (the way there is with the bending of starlight) which would bring the prediction to one part in 20,000,000. Of course, Einstein didn't do anything like this; he used some form of linearized general relativity and worked out equations of motion, but I've forgotten the details and it's not as interesting.

I don't know anything about the history of other attempts to explain the anomalous precession, but I don't think it's hard to come up with modifications of Newtonian gravity which give the right result. What was exciting about the GR calculation was that it produced the right value without any new adjustable parameters. -- BenRG (talk) 21:47, 14 February 2008 (UTC)[reply]

This is probably an anthropology question – if an Australian aboriginal man wears a red head band or a white headband in a ceremony, what particular role or status do these colours signify? Someone told me what the white one meant, but I forget (add: maybe it meant "elder") and I can't find it anywhere on google etc. Julia Rossi (talk) 06:37, 14 February 2008 (UTC)[reply]

Come on Jack, we're all waiting for you on this one, surely you've got some bonzer mates that can put you right. Richard Avery (talk) 13:35, 14 February 2008 (UTC)[reply]

To my shame, I have no indigenous friends. What I would say, though, is that the Australian indigenous peoples, while speaking with one voice on many matters, are ethnologically and culturally as diverse as the Incas, the Lapps and the Masai. It's not true, for example, that all or even the majority of tribes used boomerangs and didgeridoos, despite these having become symbols for indigenous people as a whole. Two tribes chosen at random would be much more likely than not to have spoken mutually unintelligible languages and to have widely different cultural practices. For a continent the size of Australia, it's not hard to see why this would be the case. The wearing of headbands and the significance of the colours is not something I can comment meaningfully on. Sorry. -- JackofOz (talk) 00:36, 15 February 2008 (UTC)[reply]

Sounds like he joined the Bondi Lifesaving Club Myles325a (talk) 02:13, 15 February 2008 (UTC)[reply]

Don't feel bad, Jack, Brendan Nelson hasn't got any indigenous friends either. Glad to see yu and Myles chimed in even though none of us are the wiser. Carmonozicarmoncarmon etc. ; )) Julia Rossi (talk) 08:32, 15 February 2008 (UTC)[reply]

BTW if it helps, it's a Northern Territory tribe like the Arrente (or pr: Arunta), so bit by bit maybe someone will know something. Julia Rossi (talk) 08:35, 15 February 2008 (UTC)[reply]

I can't remember whether an object that absorbs red light is red or an object that absorbs other light, but bounces back red is red. Which is it?KarateKid101 (talk) 09:51, 14 February 2008 (UTC)[reply]

Think about it carefully. If an object absorbs all light equally, it is black, because you don't see anything reflected. If an object reflects most light equally it is white (in white light), because you see all the light reflected from it. If an object absorbs red light you're not going to see red, but the other colours it does not absorb. For an object to be red, it needs to absorb most of the light that is not red hence what you see is the red light that it does not absorb but reflects. Nil Einne (talk) 10:05, 14 February 2008 (UTC)[reply]

The article on retrograde amnesia is pretty clear, but I wonder about one thing. It says trying to force someone to remember can be too stressful, but what about repeating a past event without telling the person you are. (I asked in the discussion section, but this venue may be more appropriate.) I know TV often exaggerates things, but will repetition of a specific event help more than just surrounding with familiar things? For instance, in The Jeffersons, George repeated his first date with Louise - a comical disaster - to help her get her memory back. Other sitcoms have used the "familar things" approach, like an episode of Diff'rent Strokes (where it came back gradually, like normal) and the finale of Full House. But in the former, it could be aruged it was repeating a familiar event that triggered Mr. Drummond's memory of Arnold and Willis (threatening to ground them if they ran away), and in the latter, the Full House Chronology [[1]] describes in a footnote that it's implied an accidental repeating of an earlier experienced situation helped Michelle recover her memory. (Being in a strange place and looking for older sister Stephanie for comfort, like Michelle's first day of Kindergarten.) Or, would the "Full House" example be more an example of just using "familr things" to jog the memory, anyway, since it does seem to imply Michelle growing more comfortable with Stephanie first?Somebody or his brother (talk) 13:35, 14 February 2008 (UTC)[reply]

On the Mass–energy equivalence article, it says you can use black holes to convert matter to energy. What is the efficiency of this conversion? 64.236.121.129 (talk) 16:50, 14 February 2008 (UTC)[reply]

It's 100% efficient (which is why it's under the "perfect conversion" section!) The only problem is that it takes a while. (EhJJ)^TALK 18:15, 14 February 2008 (UTC)[reply]

Do you know this, or are you just assuming it's 100%? I seem to recall that another book I saw in the book store say that the conversion is 60 or 70% efficient. Can anyone confirm this? Also what does the energy manifest as? Just heat? Any neutrinos? 64.236.121.129 (talk) 18:20, 14 February 2008 (UTC)[reply]

You can toss anything you like into a black hole and the Hawking radiation will consist largely of electromagnetic radiation, but a hot enough black hole will also emit neutrinos, which are effectively lost energy, and possibly heavier particles as well (which could perhaps be tossed in again). And the electromagnetic radiation can't be converted into useful work with 100% efficiency in practice (I love the image of a miniature black hole being used to boil water which drives a steam turbine). So it does seem pretty dubious to claim that this is a 100% efficient conversion. I'm afraid I don't know the ratios of different particles emitted as a function of temperature.

Also, I should point out that theoretical black holes are able to attain such high efficiencies only by violating most conservation laws. The reason you can annihilate matter with antimatter producing "pure energy" (i.e. photons) is that the conserved quantum numbers attached to a particle and its antiparticle are negatives of each other, and the quantum numbers attached to a photon are all zero, so both sides of the interaction add to zero and the conservation laws are satisfied (except for spin and energy-momentum, which have to be considered separately). With a Hawking black hole you can turn ordinary baryonic matter into photons even though this grossly violates the conservation of baryon and lepton number. I'm not sure what would happen if real black holes actually respect these conservation laws, but you might get out pretty much what you put in, which would make the whole thing rather pointless. On the other hand the bill never comes due if the black hole never evaporates, and baryon and lepton number aren't exactly conserved in general even outside black holes, so maybe this is a non-issue. -- BenRG (talk) 19:39, 14 February 2008 (UTC)[reply]

I should note that matter + antimatter doesn't entirely result in photons. As much as 50% results in useless neutrinos. —Preceding unsigned comment added by 64.236.121.129 (talk) 21:04, 14 February 2008 (UTC)[reply]

I'm not a particle physicist but there is something I cannot get my head around. Documentaries I have seen state that it is not possible to plot the position and speed of a sub-atomic particle at the same time. But if you don't know where something is, how can you measure the speed it is travelling at. If I want to know how fast my cat is running up the stairs then surely I need to where the cat is in the first place and where it ends up with the time taken inbetween! —Preceding unsigned comment added by 82.21.54.17 (talk) 17:03, 14 February 2008 (UTC)[reply]

Simple explanation is that to measure the position of an object you have to interact with it in some way, and that interaction changes the object's momentum. Now measuring the speed and position of your cat by, say, bouncing a light ray off it is not going to change its momentum very much - but that is because your cat is a

macroscopic object. No matter how carefully you try to make a measurement, or how sensitive your equipment is, there is an absolute lower limit on the precision with which you can simultaneously measure an object's position and momentum. This precision limit doesn't make a significant difference to your cat, but it is important for sub-atomic particles because of their small size and mass - see uncertainty principle for more details. (Yes, yes, I know this is an enormously simplified explanation, which glosses over several factors, but it will do for now) Gandalf61 (talk) 17:42, 14 February 2008 (UTC)[reply

]

I think you have oversimplified what the documentary (should have) said. I have no way of knowing what they actually said. A better phrasing might be: "not possible to plot the position and speed of a sub-atomic particle to an arbitrary degree of precision at the same time." Gandalf61's comment follows from here. JohnAspinall (talk) 19:23, 14 February 2008 (UTC)[reply]

Gandalf, the uncertainty principle is distinct from the observer effect. It is possible to create two particles so finding the velocity of one will give you the velocity of the other and finding the position of one will give you the position of the other (I'm not certain how it works. It might be that they have the same position and velocity, but you make one after the other.) You can then measure the position of one and the velocity of the other. Anon, there are ways of finding the velocity of something without using two measurements of its position, such as the Doppler effect on a photon reflected from it. — Daniel 00:51, 15 February 2008 (UTC)[reply]

As I said, I do know that my explanation is enormously simplified, and glosses over several factors, one of which is the subtle distinction between uncertainty principle and observer effect. Yes, we could get into the whole topic of quantum entanglement. But I gave an explanation that I thought was appropriate for the questioner's level of knowledge. If you think you can give a better answer to the original question, then please have at it. Gandalf61 (talk) 10:34, 15 February 2008 (UTC)[reply]

This is a good question. I don't know how the speeds of quantum objects are measured in practice, actually. But it's not so strange for something to have a well-defined speed without having a well-defined position. Wind does, for example. The uncertainty principle shows up in the theory of classical waves, so there's a closer analogy there. The speed of a quantum particle corresponds to the frequency of a wave, and the location of a particle to, well, the location of a wave (in space or in time). You might know that you can tune one instrument against another by listening for beats between the frequencies. As the frequencies get closer the time between beats gets longer, which lets you know that you're tuning in the right direction. But to be sure that the frequencies were exactly the same you would have to wait forever. The frequency difference and beat time are related by an equation that looks like the uncertainty principle. This isn't a weakness of this tuning technique, it's a universal property of waves. A wave doesn't actually have a well-defined frequency unless it has a large extent in space or time. -- BenRG (talk) 13:43, 16 February 2008 (UTC)[reply]

Assuming that Nuclear Weapons were not used, who would be likely to win in the event of a war between the USA and Europe? —Preceding unsigned comment added by 195.188.208.251 (talk) 17:04, 14 February 2008 (UTC)[reply]

According to Military budget of the United States, the US spends 47% of the world military budget, so dollar for dollar, the US could take on approximately the rest of the world, let alone Europe (taking the premise of a unified Europe). Also, US military technology is generally recognized to be somewhat superior to that of other countries, so I'd speculate that each US dollar spent on defense is more potent then a European dollar. Europeans do outnumber Americans by a bit more than 2:1, but the population of Europe is also older. Overall, I'd bet on the USA. --Bmk (talk) 17:33, 14 February 2008 (UTC)[reply]

The

French, and Swedes have good air forces, and the British subs would be troublesome. But not for long. Funny, isn't it, that throughout history madmen have been dreaming of conquering the world, and now sane men actually could but won't. --Milkbreath (talk) 17:52, 14 February 2008 (UTC)[reply

]

Don't forget the individual units themselves. The

A-10 are all pretty frightening. Let's also recall that the USA has remotely piloted stealth bombers and those fancy artillary units that makes safe zones that troops can find with GPS. Thank God they don't go and attack random countries. Oh wait, never mind. 206.252.74.48 (talk) 17:59, 14 February 2008 (UTC)[reply

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Interesting side note: Although the USA spends the most on its military, it is only third in military spending per capita (after Israel and Singapore). See this source. --Bmk (talk) 18:11, 14 February 2008 (UTC)[reply]

Also, the US has military bases in Germany, Italy, The Netherlands, Portugal, Spain, and Turkey (Well, alot of those are NATO bases, but there's American forces there) which gives them a strategic advantage. NATO bases would cause interesting problems, although I suppose that by the time it came to war NATO will have been dissolved. Also, we don't attack random countries, only countries with alot of oil, get it right. Mad031683 (talk) 18:21, 14 February 2008 (UTC)[reply]

Hi. Oh dear.

U) 18:24, 14 February 2008 (UTC)[reply

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Ok, we attack countries that have oil at random. The paradox is the the vehicles used to invade the countries are mostly fueled by oil products - so soon we will fight for oil just to get oil in order to fight...for oil. 206.252.74.48 (talk) 19:28, 14 February 2008 (UTC)[reply]

Is this science? (rhetorical)87.102.114.215 (talk) 18:25, 14 February 2008 (UTC)[reply]

I don't think it is. It's also hypothetical. Not really appropriate.64.236.121.129 (talk) 18:29, 14 February 2008 (UTC)[reply]

Also it answers itself, if the supply is running out, its more worth fighting over it. Mad031683 (talk) 18:42, 14 February 2008 (UTC)[reply]

There are numerous forums where you can have this discussion to your hearts content. To get started go to YouTube and search for videos of the USS Enterprise, or M1A1 Abrahams, or whatever piece of military hardware helps you most. The enter the comments below the video and leave a message such as "USA ROKS, frenxch pussies can kiss my ass" or something similar.. Before long you will have lots of well wishers inviting you to their websites, as well as many heart warming replies from nice people around the world. Best of luck.87.102.114.215 (talk) 18:56, 14 February 2008 (UTC)[reply]

Ho ho - my web search turns up this "LOL i remember russian soldiers are forced to become prostitute by their officers according to UN LOL! " - given this information I think USA will win unless it catches AIDS from mexican woman first. Once again in peace and love to all of you.87.102.114.215 (talk) 19:11, 14 February 2008 (UTC)[reply]

I don't think the Europeans would keep those low military budgets if their existence was seriously challenged by a foreign power such as the USA. Whenever a country faces conflict, the portion of its GDP dedicated to military sky-rockets. Nowadays the US army is far stronger than those of the European Union countries combined, but the EU has a larger population and a slightly larger economy. Furthermore, Russia alone (if we let it join the European team) has a large army, something to keep in mind. To sum up, I guess that if the war began tomorrow, it would be a one-sided win for the US, but if we let the EU have 10 years for preparing itself, I wouldn't be that sure. --Taraborn (talk) 20:26, 14 February 2008 (UTC)[reply]

I don't think russia would make a big diference - once recruit was so badly beaten by his comrades he had to have his penis and testicles amputated.87.102.114.215 (talk) 21:11, 14 February 2008 (UTC)[reply]

Although if war began tomorrow the USA might have more difficulty extricating itself from the Middle East to deploy them elsewhere than various European powers. AlmostReadytoFly (talk) 23:38, 14 February 2008 (UTC)[reply]

ok that's enough isn't it. TRY SOMEWHERE ELSE.87.102.114.215 (talk) 00:35, 15 February 2008 (UTC)[reply]

Well, all the pointers say the US should win, but then Vietnam should have been a cake-walk, and you got your butt tanned big time there, and now you are getting second helpings in Iraq because the first lesson didn't sink in, so it just goes to show that "the bigger they are, the harder they fall". I think Monaco might be a safe bet. Myles325a (talk) 02:18, 15 February 2008 (UTC)[reply]

Draw. The U.S. navy is far too strong for the Europeans to reach North America, but on the other hand, without nuclear weapons, the American army isn't big enough to conquer Europe, never mind the supply problems. After all, it's having trouble handling a single, mid-sized country. Clarityfiend (talk) 06:51, 15 February 2008 (UTC)[reply]

The Europeans kept low military budgets in the 1930s when they were menaced by Nazi Germany, such that the Germans were able to take over Czechoslovakia and other territory without any military oppisition from the powers such as Britain and France which had pledged to support its territorial integrity. What reason is there to believe they would arm themselves now to defend adequately against some threat from the U.S.? But as for invading, conquering and occupying Europe, I just can't see it happening, when after five years of occupation of Iraq the U.S. still finds it necessary to engage in large scale aerial bombing of Iraqi cities and cannot drive down a road without large IEDs being detonated. Edison (talk) 17:06, 15 February 2008 (UTC)[reply]

Ok, but the Iraqis didn't do anything to us. The average American wouldn't have known an Iraqi if he was waterboarding him back in 1980. Whereas the Europeans have been going out of their way to get on our last freakin' nerve ever since WWII, especially the French. I mean, look at it. The French hairdressers are on strike! How do you think that makes a red-blooded son of Uncle Sam feel? And that on top of pulling out of NATO. You just have to piss America off enough to swing public opinion in favor of the required total war mentality, and the population imbalance can be redressed pronto. Sorry, did I say all that out loud? --Milkbreath (talk) 17:17, 15 February 2008 (UTC)[reply]

You mean, all across France, with the hairdressers on strike, even as we speak, people are forced to tolerate ungroomed armpits? Ah, the humanity. - Zotz (talk) 00:44, 16 February 2008 (UTC)[reply]

Why have so many red-blooded sons of Uncle Sam forgotten that the USA owes its independence to France? Isn't the American Revolution taught in history classes in the US anymore? Pfly (talk) 07:05, 16 February 2008 (UTC)[reply]

Isn't the chronological order of the American and French Revolution taught elsewhere? It's not that Americans aren't cognizant of the help France provided by using America as a proxy in their conflict with England, it's that they've also learned the lessons in gratitude taught them by the French nation in the years since the last World War. - Zotz (talk) 07:15, 16 February 2008 (UTC)[reply]

For the record, no, the American Revolution is not really taught in any great detail. There is a wonderful mythological version which is quickly glossed over (OMG they taxed our TEA WTF!! and then we got the Constitution!), but nothing very serious and most high school students forget the salient points almost immediately after being tested on them. Most high school seniors today could not locate France on a map. Many probably could not locate Europe on a map. It is a pretty sad state of affairs. --98.217.18.109 (talk) 21:59, 17 February 2008 (UTC)[reply]

Of course you can't find Europe on a map, it's in orbit around Jupiter. 206.252.74.48 (talk) 17:39, 19 February 2008 (UTC)[reply]

If you were hit by free electrons (lightning bolt or something), while holding a rifle like an M-16. Would the gunpowder in the bullet shells ignite? 64.236.121.129 (talk) 18:28, 14 February 2008 (UTC)[reply]

Easily possible -- a lightning strike could certainly ignite gunpowder.

.22 cartridge as a fuse replacement and got it to fire (don't recall the current/voltage used, but far less energy than a lightning strike). It's hard to say anything definitively, though, as lightning does weird stuff. Would current pass through the rifle? Even if so, would the rifle's construction serve as a sufficient lightning rod? It'll likely vary from strike to strike. I'd put money on a strike on the rifle magazine causing the bullet to fire, though. — Lomn 18:40, 14 February 2008 (UTC)[reply

]

Of note: M-16 shells do not use gunpowder. Also, the butt, handle, and barrel guards are plastic, not metal. The "handle" on top is not supposed to be used a handle. Still, it is possible for electricity to travel along the metal and, for some unknown reason, want to travel into the bolt mechanism and then, for some unknown reason, want to travel to the shell and ignite the propellant. It would make more sense for the electrons to travel along the outside of the rifle to the human and then to ground. --

™ 20:15, 14 February 2008 (UTC)[reply

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M-16 shells use smokeless powder I believe, which I mentioned in my topic. 64.236.121.129 (talk) 20:50, 14 February 2008 (UTC)[reply]

It appears that M-16s use WC844 ball powder, which is discussed as being distinct from smokeless powder in ways I don't really understand. I'd imagine it's a similar outcome in any event. — Lomn 21:25, 14 February 2008 (UTC)[reply]

What? I think it's just one of the (many) varieties of smokeless powder. Where do you see a suggestion otherwise? Friday (talk) 21:30, 14 February 2008 (UTC)[reply]

I was using this as a reference, but like I said, I didn't grok the details. It may well be a simple distinction in subtypes of smokeless powder. My confusion probably stems from reading all this with a "gunpowder=smokeless powder" mental filter, thus seeing Kainaw's comment as "M-16s don't use smokeless powder", which led to me trying to find out what they did use. Short version: my bad. — Lomn 22:01, 14 February 2008 (UTC)[reply]

It would be pretty surprising if any explosive used to propel bullets could not be ignited from an electrical spark. But the cartridge generally provides a metal shell completely surrounding the smokeless powder, gunpowder, etc. Certainly a full-fledged lightning discharge could ignite the rifle, the rifleman, and the cartridge, just from the high temperature. I have seen fairly large pieces of metal vaporized by lightning. Edison (talk) 16:59, 15 February 2008 (UTC)[reply]

When possible health effects of microwave radiation is considered they usually look at the average energy concentration involved. What bothers me is that there is usually no mention that artificial sources of microwaves typically have lots of energy in a tiny band (relative to the frequency, not absolute). And microwave wavelengths are typically around the order of the size of humans, human organs or a human scull.

When during exposure a person does not move relative to the source of radiation such as when speaking on a cellphone held tightly against the ear, won't resonances occur (analogous to what happens when you sing in the shower) that push energy concentration in certain spots way above the average? If yes, is this taken into account with most studies and can we place a ceiling on the extent to which this effect can cause potentially dangerous hot spots?

Secondly, and perhaps this is still a controversial subject, I read about a possible HIV treatment that uses the resonant frequencies of the virus as a way to "shake them to death". I hope it's OK that I'm not mentioning the source because I really can't remember. They say the frequency is calculated at around 60GHz and that the proposed treatment won't affect normal cells which have a much lower resonant frequency.

Does anyone know in what sense this "resonance" is meant, because in order for 60GHz to resonate at that small scale I assume that there has to be some sort of wave that propagates much slower than the speed of light. And then, using this logic, roughly at what frequency DOES normal cells resonate and what would happen if they were exposed to EM radiation at this frequency? (What I'm asking is slightly different from the proposed treatment mentioned which instead proposes using a laser pulsing at that frequency).

41.241.187.40 (talk) 20:08, 14 February 2008 (UTC) Eon Zuurmond[reply]

• Yes, some spots of your hand/head will absorb more than others when you use a mobile phone. There's lots of people doing research and modelling in this and other areas, MTHR in the UK for example. I don't think they've found anything to worry about yet.
• I think the HIV destruction was by ultrasound not microwaves, so yes they will travel much slower. You'd have a job getting an HIV virus to absorb much microwave energy at 60 GHz anyway, you'd probably fry the surrounding cells first and end up killing the virus that way.
• I seem to remember that there were initially some worries about the TETRA system having a pulse repetition frequency of around 17 Hz which was close to the resonant frequency of a particular cell or reaction between cells. There's probably something about this in the Stewart Report if you want to look it up. JMiall₰ 20:21, 14 February 2008 (UTC)[reply]

and get high off of it —Preceding

I would think not. The large intestine usually only absorbs liquids, so you can get drunk in this manner (one

Darwin award winner died in this fashion), but not high off solid material. THC is not soluble in water either. Lastly, I highly discourage taking this path to get high, or any path. 206.252.74.48 (talk) 19:42, 14 February 2008 (UTC)[reply

]

I believe the THC in marijuana undergoes a small chemical change when it is heated (decarboxylation?) which converts it to its psychoactive form. Supposedly injesting uncooked marijuana will not get you high. (This is according to a book by Ed Rosenthal). ike9898 (talk) 20:09, 14 February 2008 (UTC)[reply]

you reduce your credibility by saying "or any path". —Preceding unsigned comment added by 79.122.19.82 (talk) 20:56, 14 February 2008 (UTC)[reply]

What? I just vehemently oppose psychoactive drugs, and alcohol. My life has been the better for it. 206.252.74.48 (talk) 21:01, 14 February 2008 (UTC)[reply]

Someone who tells you marijuana can be consumed anally might just be blowing smoke up your ass. Edison (talk) 20:29, 15 February 2008 (UTC)[reply]

actually come to think of it, if you can make brownies, you can put them in your rectum, i think you could absorb it that way similar to a cotton rag with alcohol like puertorican youth do or a crystal meth enema like gay men

talk) 05:32, 20 February 2008 (UTC)[reply

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"In the U.S. illicit methamphetamine comes in a variety of forms, at an average price of $150 per gram for pure substance" from Methamphetamine#Production_and_distribution, and ounce is ~28g87.102.114.215 (talk) 21:38, 14 February 2008 (UTC)[reply]

Note that an ounce is a LOT. I would think (based on friends I used to have) that it would take a regular user maybe a week or two to go through just a gram. --140.247.11.3 (talk) 01:04, 15 February 2008 (UTC)[reply]

The typical price of an ounce might be less than implied by the stated price per gram, since larger amounts may be sold wholesale, to allow a markup for the risk and bother of selling it to all the crack heads and risking arrest. Law enforcement people love to show some huge shipment of confiscated drugs and announce its huge "street value." A far smaller sum would likely have changed hands had it reached the drug lord it was intended for. Edison (talk) 16:50, 15 February 2008 (UTC)[reply]

Why are dinosaur names written with a capital letter and in italics? I've seen this in both Spanish and English. --Taraborn (talk) 20:07, 14 February 2008 (UTC)[reply]

See binomial nomenclature. ike9898 (talk) 20:10, 14 February 2008 (UTC)[reply]

It's standard formatting and nomenclature - kinda like for bacteria, which is done in italics. Wisdom89 _{(T / ^C)} 20:11, 14 February 2008 (UTC)[reply]

Oh, kinda obvious. Thanks :) --Taraborn (talk) 20:27, 14 February 2008 (UTC)[reply]

I'm working on a science project for an earthquake unit. I need to know of an earthquake that was measrured in the mercalli scale, where and when it happened. I need the answer before 5 'oclock eastern time. Please help me. QUICK! =0 —Preceding unsigned comment added by 72.221.113.46 (talk) 20:24, 14 February 2008 (UTC)[reply]

From what I can find, the Modified

Mercalli Intensity Scale is still in use, this page shows the ratings for some areas of California after the 1989 Loma Prieta event. I haven't found any specific quakes using the original 1902 10-point version, but knowing it was in developed in 1902 might lead in the right direction. --LarryMac | Talk 21:11, 14 February 2008 (UTC)[reply

]

Given the 1902 date, the 1906 San Francisco earthquake is an excellent contemporary example. — Lomn 21:22, 14 February 2008 (UTC)[reply]

Are we supposed to answer or honor obvious homework questions? Wisdom89 _{(T / ^C)} 21:39, 14 February 2008 (UTC)[reply]

We are allowed to help.. as much as we want to.87.102.114.215 (talk) 21:54, 14 February 2008 (UTC)[reply]

Yes and no. We're explicitly encouraged not to do someone's homework for them, but also encouraged to be helpful in assisting. I figure that "name a measured earthquake" is a pretty trivial homework question if that's the whole thing -- likely any major earthquake postdating the Mercalli system has been measured by it. Do you think I've crossed that line via the post above? — Lomn 21:57, 14 February 2008 (UTC)[reply]

The poster still has there report/essay to write - so no I don't think you've overstepped the line - you were very helpful in the case of an honest and straightforward request for info. Well done.87.102.114.215 (talk) 22:16, 14 February 2008 (UTC)[reply]

What cleaning chemical is produced when you mix salt, bi-carb of soda and vinager, I saw it used on the TV for cleaning drains 121.200.39.12 (talk) 22:12, 14 February 2008 (UTC)[reply]

The salt isn't part of the main reaction, but bicarb of soda (aka sodium bicarbonate) and plain vinegar (aka acetic acid) react to form sodium acetate, carbon dioxide gas and water. It's the reaction itself, in producing small bubbles of carbon dioxide, that does a lot of the cleaning work (it's like a gaseous exfoliation), although the individual chemicals have cleaning properties of their own, which is why it's such a multi-purpose cleaning method. Confusing Manifestation(Say hi!) 22:27, 14 February 2008 (UTC)[reply]

I suppose the salt introduces an abrasive providing it's not dissolved. Julia Rossi (talk) 23:45, 14 February 2008 (UTC)[reply]

The

aqueous portion of the vinegar which is only 3 % acetic acid. So it will probably not be abrasive. —Preceding unsigned comment added by 79.76.141.105 (talk) 00:36, 15 February 2008 (UTC)[reply

]

That all depends on how much vinegar there is compared to the solids. Salt isn't infintely soluble, and it and the bicarb compete with each other in terms of solubility (common ion effect for sodium ions). Any excess would remain as a solid abrasive. Then later it could be rinsed away with excess water, so there's no solid residue to scrape out of ther drain. DMacks (talk) 20:25, 15 February 2008 (UTC)[reply]