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February 17

Carbonyl reduced to Hydroxyl

I understand that reduction is a gain of electrons and oxidation is a loss of electrons. But I don't understand how when a carbonyl is turned into a hydroxyl it was reduced. To me the number of electrons stayed the same in the functional group before and after. There were no gains or losses of electrons. Could someone explain this to me? 128.163.224.198 (talk) 00:29, 17 February 2008 (UTC)[reply]

It's a reference to the oxidation state of carbon, specificially. Whereas previously the oxygen-bound carbon (in the case of a ketone) was considered to have an oxidation state of +2, it now has an oxidation state of 0. Someguy1221 (talk) 00:34, 17 February 2008 (UTC)[reply]

If you're dealing with biochemical (aqueous) systems, here's another way to think about it: To start with, you have the carbonyl and an available pool of H⁺ ions from the water. In order to go from the carbonyl to the hydroxyl, you have to add two protons and two bonds (one each to both the carbon and the oxygen). To make the bonds, you need four electrons (two for each bond). Two electrons come from the C=O double bond that you "break" into a single bond, so you need to add in an additional two electrons for the second bond. -- 128.104.112.47 (talk) 19:13, 19 February 2008 (UTC)[reply]

Sleeping standing up

Are there humans who sleep standing up? I think that leaning on a corner it would be feasible at least.217.168.1.250 (talk) 00:58, 17 February 2008 (UTC)[reply]

It's not possible without an adequate means of support.The Updater would like to talk to you! 01:03, 17 February 2008 (UTC) —Preceding unsigned comment added by WikiHaquinator (talk • contribs) [reply]

If I can sleep sitting up, or sitting at my desk at work, I can probably sleep leaning against something. But unlike cattle, I do not believe humans can lock their knees and sleep while standing unsupported. Micronaps would occur if a person were sleep deprived long enough, but they would jerk as they started to fall. Edison (talk) 02:33, 17 February 2008 (UTC)[reply]

I could see it happening if someone had some sort of back problem that prevented him/her from laying down, with supports of course. Clarityfiend (talk) 04:56, 17 February 2008 (UTC)[reply]

Astronauts of course can sleep in any orientation, but in orbit there is no differential apparent force on a person distinct from that on the space craft, so there is no up/down. That said, astronauts do have preferences for which way in their sleeping bags they lie (head/feet up/down) but this is just preferences to which way the tethers and zips are orientated to them. Still it does look somewhat bizarre seeing one astronaut giving a video linkup tour of their living quarters, to then see one of there colleagues unzip from a sleeping bag in an opposite orientation. David Ruben ^Talk 13:28, 17 February 2008 (UTC)[reply]

Well, we would not stand unsupported, so we don't have to lock our knees. Imagine that you are on a corner facing it. You can lean your knees and lean your head on the corner. I'll try this at home and come back to you. 217.168.3.246 (talk) 20:35, 17 February 2008 (UTC)[reply]

(OR warning). It's possible to fall asleep standing up, but you tend to wake up fairly soon! A kind of semi-doze can be maintained for some time while clinging to one of those handle-strap things on a bus or train. And I've seen children soundly asleep clinging to a parent's leg or similar. And, of course, there are somnambulists who manage the whole upright thing quite well. Gwinva (talk) 22:01, 17 February 2008 (UTC)[reply]

Electron microscope staining

hey,

Is it possible to stain cells for observation under electron microscope? The Updater would like to talk to you! 01:02, 17 February 2008 (UTC)[reply]

It is not necessary to stain a cell for veiwing with an electron microscope. The cell must be coated with a thin layer of metal (Usually gold I think) to reflect the electrons so an image can be created. It may also be helpful to see electron microscope. --Zrs 12 (talk) 01:27, 17 February 2008 (UTC)[reply]

Variable Speed of Light

I was recently reading a scientific paper written by one of my teacher's uncles on his theory about light having a varying instead of constant speed. First, he set some units: $9.8m/s^{2}=1\,\!$ , by Newton's equation $F=ma\,\!$ and defining $9.8m/s^{2}\,\!$ as the acceleration $m=F/g\,\!$ . Also, defining gravity is $L/s^{2}\,\!$ in which $L=9.75m/s^{2}\,\!$ . He then says therefore, $m={\frac {F}{1}}$ so $m=F\,\!$ . Then he says Length is L, mass is F, t is time, and so $c={\frac {L}{t}}$ . He says work is mass multiplied by Length and energy is the rate of performing work so Newtonian energy is $E={\frac {mL}{t}}$ . He then says using the relativity equation $E=mc^{2}\,\!$ we need to define mass. Since Newtonian mass is ${\frac {Et}{L}}$ the relativity equation is as follows: $E=\left({\frac {Et}{L}}\right)\left({\frac {L}{t}}\right)^{2}$ . He says when simplified this equation shows time is Length. My questions are:

How did he get $c={\frac {L}{t}}$ (this appears to be just an assumption to me)
Is there anything wrong with his logic? and,
When I simplified the equation $E=\left({\frac {Et}{L}}\right)\left({\frac {L}{t}}\right)^{2}$ I distinguished between Einstenian and Newtonian energy and reformulated the equation to $E_{E}=\left({\frac {E_{n}t}{L}}\right)\left({\frac {L}{t}}\right)^{2}$ and when it was reduced I got ${\frac {E_{E}}{L}}={\frac {E_{n}}{t}}$ showing (it appears to me) that $L\neq {t}$ or at least $L=t\Leftrightarrow {E_{n}}=E_{E}$ for the fact that Einstenian and Newtonian energy are different. Who is correct? Does $L=t\,\!$ in fact?

Thanks, Zrs 12 (talk) 01:54, 17 February 2008 (UTC)[reply]

Showing ${\frac {E_{E}}{L}}={\frac {E_{n}}{t}}$ : ${\begin{aligned}E_{E}&=\left({\frac {E_{n}t}{L}}\right)\left({\frac {L}{t}}\right)^{2}\\&={\frac {E_{n}tL^{2}}{Lt^{2}}}\\&={\frac {E_{n}L^{2}}{Lt}}\\&={\frac {E_{n}L}{t}}\\\end{aligned}}$ ${\begin{aligned}\textstyle {therefore}\\\end{aligned}}$ ${\begin{aligned}{\frac {E_{E}}{L}}={\frac {E_{n}}{t}}\\\end{aligned}}$

Zrs 12 (talk) 05:14, 17 February 2008 (UTC)[reply]

ATTENTION: I accidentally put (L/t²). Now it's correct as (L/t)². Sorry. Zrs 12 (talk) 03:31, 17 February 2008 (UTC)[reply]

The theory starts out with gibberish, because if

9.8m/s^{2}=1\,\!

, then in dimensional analysis length equals time squared, which is nonsense. Edison (talk) 02:30, 17 February 2008 (UTC)[reply]

Anything else? (Remember, he is trying to prove that the speed of light is variable so maybe length does equal time squared?) Zrs 12 (talk) 02:56, 17 February 2008 (UTC)[reply]

Except that he didn't prove that length equals time squared, he just sort of declared it, and this seems to form the basis of the rest of the proof. Someguy1221 (talk) 03:19, 17 February 2008 (UTC)[reply]

Ok, thanks. What about the part where I showed

L=t\Leftrightarrow {E_{n}}=E_{E}

? Did I do this correctly or did he? In other words, did I reduce correctly? --Zrs 12 (talk) 03:24, 17 February 2008 (UTC)[reply]

The whole thing is basically gibberish. It's the mathematical trappings of high school physics without any of the physics. It's hard to know how to respond to your questions because you might as well be analyzing Star Trek technobabble. Have you correctly followed the procedure for realigning the tachyon compensators? Well, maybe, maybe not; only your teacher's uncle can say. -- BenRG (talk) 20:36, 17 February 2008 (UTC)[reply]

This is too physics heavy even for the likes of me ;) kidding. Seriously though, the only time the speed of light is not constant is when it passes through a medium, any medium that is not a vacuum. At least that's the extent of what I know regarding this. That's really all I can say about the above since it's mostly mathematical vomit. That's not a personal attack. It's just a really convoluted way to ask a question. Wisdom89 _{(_T / ^C)} 20:58, 17 February 2008 (UTC)[reply]

Not to defend the uncle (I neither know nor care about the above), but 1. there are variable speed of light theories put forward by competent physicists, so dismissing that aspect of it a priori is a little presumptive, and 2. the simple manipulation of simple equations to reach unintuitive results has been done successfully before by none less than Einstein (the 1905 papers on special relativity are basically this—taking a few very well-known equations and trying to make them be compatible with each other in ways that hadn't been done before, specifically combining Galilean relativity with the axiom of a fixed speed of light in a vacuum, and the rest basically falls out from there), so that's not a great reason to dismiss it either. The original derivation of E=mc^2 does not use complicated physics at all; special relativity as a whole is extremely simple "high school" math (Einstein's emphasis was on axioms and concepts; he had to learn a lot of math before he could generalize the theory, however). However it is unlikely that one's uncle is going to do better than Einstein in that game, which would for me be a pretty strong a priori source of doubt, though it is no proof. --98.217.18.109 (talk) 21:40, 17 February 2008 (UTC)[reply]

The first warning sign that something dubious is going on is the use of

g

, the local gravitational acceleration at the surface of the earth, in a theory about light. Any theory about light has to account for the fact that light travels throughout space (i.e. we see stars). Why would the behavior of light throughout the universe depend on the local amount of gravity at the surface of the third planet orbiting an unremarkable G-class star, in the Orion arm of the Milky Way Galaxy? JohnAspinall (talk) 00:47, 18 February 2008 (UTC)[reply]

I kind of thought so. He had but 56 semester hours of math in college. Also, he is a chemical engineer, not a physicist, so it doesn't really suprise me that it seems to be "gibberish". Thank you all, Zrs 12 (talk) 00:33, 18 February 2008 (UTC)[reply]

Engineers of various sorts seem to be the ones behind a lot of this kind of speculation. Physics department around the country have "crank files" of people who write in to dispute or put forward various theories; engineers, in my experience in reading these files (which I have done on a few occasions), seem to go for this stuff the most. I imagine it has to do with their type of education—something that makes them feel quite technically competent but they actually lack a lot of the lower-level understanding that is superfluous for their profession. But that might be an exaggeration. --98.217.18.109 (talk) 02:49, 18 February 2008 (UTC)[reply]

Does your uncle happen to have some underlying reason to want the speed of light to be variable? Wanting to arrive at a desired result can certainly lead people to accept some tortured logic. --Sean 14:52, 18 February 2008 (UTC)[reply]

I haven't a clue. (He's my teacher's uncle) Anyway, this very well may be the case. However, in his memo on the subject, there was no indication of this. This could possibly be the case though. Zrs 12 (talk) 23:10, 18 February 2008 (UTC)[reply]

Speculative Science Question

I was watching a Twilight Zone episode,A Kind of a Stopwatch, the other day. In it, a man gains the ability to "stop time." Objects freeze in midair, people don't move, and none of it affects him. My question is this: If such a thing were possible, then wouldn't he die? Would he be able to breathe the "stopped" air molecules? Would he be able to walk through said molecules? —Preceding unsigned comment added by 192.136.22.4 (talk) 02:04, 17 February 2008 (UTC)[reply]

I can't believe I'm answering this......it's a narrative fiction, you're supposed to just accept the premise as a given, and not examine it too closely, lest you ruin the story. Nevertheless, as I recall it, the protagonist is able to move objects, even though the objects don't move under their own force. So he ought to be able to breathe, eat, move, etc. - Nunh-huh 02:11, 17 February 2008 (UTC)[reply]

Actually, if he has reset the velocities of all the air molecules to 0, there would no longer be any air pressure, and he would not be able to breathe. When you breathe in, you are simply expanding the volume of your lungs and permitting the air molecules to move into a new location. He would need to push the air into his mouth. — BRIAN0918 • 2008-02-17 16:15Z

Yeah just accept sci-fi. In Star Wars sound can be heard even though there is no medium through which sound waves can travel. Zrs 12 (talk) 02:15, 17 February 2008 (UTC)[reply]

Seen this before. There is an old (say 1970) Arthur C Clarke story where a jewel thief is given a device which stops everything more than a couple of metres away from him (or accelerates everything closer). That kind of works a bit. The story is interesting because it begs the question what would you do if you had all the time in the world. Crime? Revenge? Assasination? Sadly once you become a parent the only answer has to be "sleep". --BozMo talk 07:45, 17 February 2008 (UTC)[reply]

Except you'd still suffocate in your sleep since you'd basically be in a 2 metre bubble of air which would run out of oxygen :-P Nil Einne (talk) 18:53, 17 February 2008 (UTC)[reply]

When Isaac Asimov wrote the novelisation of Fantastic Voyage he addressed some of the basic physics which the screenwriters hadn't troubled with - for example (IIRC) when they're running out of air they tunnel into the lungs to get some, but in the book he actually had them shrink the air from the lungs. --ColinFine (talk) 21:31, 17 February 2008 (UTC)[reply]

Reading a more realistic sci-fi story some time ago, I've seen a solution for this kind of problem: the protagonist had to hold his breath while he was using his superpower. This way, the whole story was a bit more balanced, he could use his "power" only for around a minute. The power was to pass through matter and/or become invisible, but this "solution" can also be applied to the power of slowing down time. --V. Szabolcs (talk) 18:50, 18 February 2008 (UTC)[reply]

Yeah, has any one heard the story of how nothing exploded into everything and then life some how started from nothing. That means spontanious generation. Pretty far out huh!

Use of Opiates in the treatment of psychiatric disorders in women!

Question is about medical/chemical applications re wife's treatment. This questioner is reminded: Do not request medical or legal advice. Any such questions may be removed. If you need medical or legal advice, do not ask it here. Ask a doctor, dentist, veterinarian, or lawyer instead. See your doctor. Julia Rossi (talk) 08:49, 17 February 2008 (UTC)[reply]

I'd like to add that if you do not agree with your wife's psychiatrist, you (as a couple) could consult a different psychiatrist. Generally, if you get a few "no, we don't do that" responses from several unrelated physicians, it's probably the best way to go (versus asking a bunch of non-medically trained, anonymous volunteers on the internet). Also, feel free to read our article on hydrocodone if you'd like to be an "educated patient", but use that information only to help you ask your doctor the right questions. (EhJJ)^TALK 21:16, 17 February 2008 (UTC)[reply]

absorption

Intensity of light decreases with the rise in concentration of the liquid through the light passes through: please explain this in detail.Jskirankumar (talk) 09:34, 17 February 2008 (UTC)[reply]

Beer's law. Someguy1221 (talk) 09:44, 17 February 2008 (UTC)[reply]

Photoreceptors

Hi all I am usure as to how a rod/cone cell become hyperpolarised. I know that when a photoreceptor such as a rod is resting K+ ions leak out of the inner segment and Na+ ions diffuse in through the outer segment. When a rod is stimulated the Na+ channels close but the Sodium/Potassium pump continues to pump Na+ ions out and K+ in. I'm not sure how this makes the charge go from -40mv to -70mv. Hope this makes sense and please correct me if anything I have written is wrong! —Preceding unsigned comment added by 172.207.224.229 (talk) 10:49, 17 February 2008 (UTC)[reply]

If I'm not mistaken, the channel lets 3 Na+ go out and then 2 K+ in. So in net total, +1 goes out of the cell, and the inside of the cell becomes more negative..206.240.25.247 (talk) 13:34, 17 February 2008 (UTC)[reply]

When rods are stimulated by photons, the signaling mechanism that is elicited activates cGMP phosphodiesterase which depletes and inacticates cGMP, stopping the depolarization due to sodium ions influx. Since the resting membrane of the cell is closer to the equilibrium potential of potassium, the cell hyperpolarizes. Wisdom89 _{(_T / ^C)} 20:42, 17 February 2008 (UTC)[reply]

Why is there more static electricity in dry weather?

I wonder. 206.240.25.247 (talk) 13:30, 17 February 2008 (UTC)[reply]

I think the water vapour can bring the static electrictiy away. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 14:09, 17 February 2008 (UTC)[reply]

How does that work exactly? 206.240.25.247 (talk) 14:11, 17 February 2008 (UTC)[reply]

From reading several articles on electrostatic discharge, it seems that humidity tends to limit the electrostatic buildup because it makes surfaces slightly conductive, which provides a path for separated charge to recombine. (See [1] for a list of links on ESD. [2] contains animations of the process of tribocharging, in which surfaces of dissimilar materials become charged after contact, friction/pressure, and then separation. [3] explains the connection between humidity and "return current".) --71.175.22.137 (talk) 15:05, 17 February 2008 (UTC)[reply]

See the section "Simple experiments" in the Static electricity article. The preface notes that a humid atmosphere provides a conducting path for the rapid neutralization of static charge... --hydnjo talk 19:50, 17 February 2008 (UTC)[reply]

absortion1

why does the distance that the light travels through the material also effect the intensity of the light when the concentrator is constant? I shall be thankful for your convincingJskirankumar (talk) 16:45, 17 February 2008 (UTC) explanation[reply]

Assuming that the medium is homogeneous (i.e. uniform), you'd expect light to be attenuated by the same amount every time it travels through a unit distance in the medium. For concreteness, let's say

(intensity of emergent ray) = α × (intensity of the incident ray).

After traveling

L

unit-length layers of the medium, you'd expect the emergent ray to have an intensity of

\alpha ^{L}

× (intensity of the incident ray).

--71.175.22.137 (talk) 17:38, 17 February 2008 (UTC)[reply]

If a helium filled airship wants to gain altitude...

How does it do this? Lift is determined by how much helium gas it has, so how can they make it fly higher, when they are already at a given altitude? Malamockq (talk) 20:21, 17 February 2008 (UTC)[reply]

The density of a gas can be decreased through heat - this would allow the craft to gain altitude. Wisdom89 _{(_T / ^C)} 20:32, 17 February 2008 (UTC)[reply]

So similar mechanism to a hot air balloon then. Thanks. Malamockq (talk) 20:44, 17 February 2008 (UTC)[reply]

Yup, you got it. Wisdom89 _{(_T / ^C)} 20:54, 17 February 2008 (UTC)[reply]

No, that's wrong. They couldn't burn anything inside the helium bag because anything they piped into the bag to burn, as well as its reaction products, would be heavier than helium and would weigh the ship down. Heating the outside of the bag is theoretically possible but it would be slow to act since the heat won't transfer efficiently to the mass of helium. And the remaining choice, an electric heater inside the bag, would require a heavy power source.

There are actually three solutions that were used in the heyday of airships (whether hydrogen- or helium-filled, doesn't matter).

The first and simplest is that when you take off, you carry something heavy on board as ballast, and when you want to make the airship lighter, you drop it. If you've seen pictures of hydrogen or helium passenger-carrying balloons, you'll note that the passenger gondola is typically ringed with sandbags. (You drop the sand, not the whole bag!) With an airship the preferred ballast is water for several reasons: it's cheap and widely available; it can be loaded with a pump and a hose; nobody will mind much if some of it lands on them; and it can be used to supply the onboard washbasins and recaptured (into a second set of ballast tanks) from the drains. The Graf Zeppelin, for example, carried about 12 tons of water ballast. (If they started to run low on ballast, this would be an emergency situation the same as running low on fuel, and they would land as soon as possible to replenish it.)

The second way is the same way that an airplane does it: by pointing the nose higher or by moving control surfaces in the tail. Of course, this only works when the engines are exerting enough force to keep the vehicle moving forward at sufficient speed.

I slipped up here. The use of the tail control surfaces, as in an airplane, would be for the purpose of pointing the nose higher, not to provide increased lift directly (since it would be off-center). Pumping ballast toward the rear of the airship would be a second way to lift the nose. --Anon, 06:02 UTC, Feb. 19.

The third way is to burn some fuel, assuming that something like ordinary gasoline or diesel fuel is used. Of course, this is slow to act, but you have the burn the fuel anyway to keep moving. Now most airships have used a single type of fuel and no way to burn it except in the engines, so they had little control over this way of affecting their weight, but the Graf Zeppelin was an exception. It carried two types: Blau gas, which was a propane-based gas mixture about the same weight as air, and gasoline. If it was desired to make the airship gradually lighter, they would burn gasoline in the engines; if not, they would burn Blau gas.

Source: The Golden Age of the Great Passenger Airships: Graf Zeppelin & Hindenburg by Harold G. Dick with Douglas H. Robinson, 1985, Smithsonian Institution, ISBN 0-87474-364-8.

--Anonymous, 23:35 UTC, February 17, 2008.

Interesting, I had a feeling this was the case. For example, if the airship was filled with hydrogen, it wouldn't be very smart to try and heat it, considering it is flammable. Malamockq (talk) 02:37, 18 February 2008 (UTC)[reply]

It is possible to heat helium without igniting it. One doesn't have to use a direct flame or combustion. Just saying. Wisdom89 _{(_T / ^C)} 09:38, 18 February 2008 (UTC)[reply]

I know that, but in airships they can also use hydrogen. Heating hydrogen to make it gain altitude prolly isn't a good idea, nor how they do it. Malamockq (talk) 16:39, 18 February 2008 (UTC)[reply]

Question : Why is it not possible to change the buoyancy of one of the gasbags by pumping the gas into and out of rigid tanks? APL (talk) 01:14, 19 February 2008 (UTC)[reply]

I think this has been discussed before, in an extreme form. If you could really do that, you wouldn't need any gas -- you'd just evacuate the tanks and off you go floating into the wild blue yonder. But no one knows how to make taks that rigid, without adding more weight than the weight of the gas. Your suggestion seems like a compromise somewhere in the middle -- my guess is it won't work either, for the same reason, but I haven't tried to run the numbers. --Trovatore (talk) 01:30, 19 February 2008 (UTC)[reply]

On second thought maybe I misread your question. I thought you wanted to make the gasbags themselves rigid. Maybe you're saying, when you want to lose altitude, pump some of the helium into a tank so that the bag partially deflates. I suppose that works in principle. Would probably use a lot of energy though, and it might be awkward to have the size of the bag change in flight. --Trovatore (talk) 02:47, 19 February 2008 (UTC)[reply]

Yes, that is what I was asking. I was not asking about the mythical Vacuum Blimp. I was asking about slightly deflating the gasbags without completely venting the hydrogen. Seems like depending on the type of airship, it wouldn't be that awkward to have the bag change in flight. Even if you completely deflated a gasbag on a Zepplin the envelope would presumably stay the same shape. APL (talk) 04:18, 19 February 2008 (UTC)[reply]

Well, I'm no airship engineer; I can't say whether there's some way the idea could be made to work (better than the alternatives). But I would note that there's still an uncomfortable tradeoff: If you make the tanks small, then in addition to having to make them very strong, you'll have to supply a lot of energy to compress the gas into them (integral of pressure wrt volume). Whereas, if you make them big, they'll be heavy, and you have

to lift them with the gas you're planning to put into them. --Trovatore (talk) 04:54, 19 February 2008 (UTC)[reply]

Sanity check: Either A) the gasbag is rigid (doesn't change size), in which it has a constant volume, and moving mass (with weight) between the gasbag and a similar rigid pressure tank doesn't change the total volume of the system, or total mass either, so there is no effect to overall bouyancy. Or, B) the gasbag is NOT rigid, meaning that as you pump gas into it, it expands to keep the same pressure as the outside air. Which means that as gasbag total mass increases, the total volume increases, too. Again, you don't gain anything.

The only way to gain something is to either change total airship volume, while keeping mass constant, or change total airship mass, while keeping volume constant. Note that hot air balloons do the latter - they heat the air, which expands out the bottom, making the whole balloon mass less while keeping the volume the same => it will rise. To descend, just let it cool off. The air will contract, and more are will come in the bottom, for more mass at the same volume.

The "right way" would certainly be to have rigid gasbags that we could pump down to a vacuum while they kept their size/shape, but that means a lot of structural strength, which, ahem, would make them too heavy to lift. -SandyJax (talk) 17:29, 19 February 2008 (UTC)[reply]

No, you missed a point in alternative (B): It's true that the weight of bag-plus-gas is increasing, but the weight of storage-tanks-plus-gas is decreasing by the same amount (not counting relativistic effects :-). So you do in fact gain buoyancy when you move gas from the rigid tanks into the inflatable bag, and decrease buoyancy when you move it the other way. Whether this is a good, or even feasible, design, is another question, but it does work in principle. --Trovatore (talk) 20:29, 19 February 2008 (UTC)[reply]

SandyJax, I think you misunderstood. I was suggesting changing the total displaced volume of the airship by inflating and deflating the gasbags while recapturing the gas in pressurized, rigid holding tanks for reuse, keeping mass constant. (total displaced volume should not be confused with total enclosed volume when talking about rigid frame ships like Zeppelins.) However, Trovatore has satisfied my curiosity by convincing me that my idea would require some pretty heavy equipment that would probably take up a pretty large percentage of the airship's carrying capacity. APL (talk) 05:38, 20 February 2008 (UTC)[reply]

Absolute frame of reference

According to the special theory of relativity there is no absolute frame of reference. But something bothers me. The kinetic energy of an object is 1/2 m v^2

For an object of 2 kg.

To get from 0 m/s to 1 m/s requires 1 joule of energy.

to get from 100 m/s to 101 m/s requires 201 joules of energy.

So why can you measure the amount of energy required to increase the velocity of the object by 1 m/s and thus calculate how fast the object is travelling on the absolute frame of reference (which special relative claims that it does not exists).

202.168.50.40 (talk) 21:39, 17 February 2008 (UTC)[reply]

To increase the momentum of an object in one direction, you must also increase the momentum of some other object in the other direction (conservation of momentum). You cannot directly observe the energy given to either one; you can only observe the energy given to both. And this energy doesn't vary by reference frame. This is true in classical mechanics too, so for simplicity I will not deal with relativity, and give an example in one-dimensional classical mechanics:

Suppose you have two masses

m_{1},m_{2}

stuck together, moving at speed

v_{0}

. Then you have an explosion that imparts a momentum

p

to both of them in opposite directions (let's say it accelerates

m_{2}

in the positive direction and

m_{1}

in the negative direction). Then

m_{2}

will be traveling at speed

v_{0}+{\frac {p}{m_{2}}}

and

m_{1}

will be traveling at speed

v_{0}-{\frac {p}{m_{1}}}

. Now let's compute the difference in total kinetic energy of the objects before and after:

{\frac {1}{2}}m_{1}\left(v_{0}-{\frac {p}{m_{1}}}\right)^{2}+{\frac {1}{2}}m_{2}\left(v_{0}+{\frac {p}{m_{2}}}\right)^{2}-{\frac {1}{2}}m_{1}v_{0}^{2}-{\frac {1}{2}}m_{2}v_{0}^{2}

={\frac {1}{2}}\left(m_{1}v_{0}^{2}-2pv_{0}+{\frac {p^{2}}{m_{1}}}+m_{2}v_{0}^{2}+2pv_{0}+{\frac {p^{2}}{m_{2}}}-m_{1}v_{0}^{2}-m_{2}v_{0}^{2}\right)

={\frac {1}{2}}\left({\frac {p^{2}}{m_{1}}}+{\frac {p^{2}}{m_{2}}}\right)

Notice that this increase in total kinetic energy is not dependent on the initial velocity

v_{0}

. --Spoon! (talk) 23:38, 17 February 2008 (UTC)[reply]

Pitch Drop experiment

The most famous version of the experiment was started in 1927 by Professor Thomas Parnell of the University of Queensland in Brisbane, Australia, to demonstrate to students that some substances that appear to be solid are in fact very-high-viscosity fluids. Parnell poured a sample of pitch into a sealed funnel and allowed it to settle for three years.

Considering how long it takes for pitch to flow. Exactly how long did it take for Prefessor Thomas Parnell to pour the pitch sample into the funnel? 202.168.50.40 (talk) 22:56, 17 February 2008 (UTC)[reply]

According to this, he heated the pitch before pouring it in - so I assume it didn't take that long. --Kurt Shaped Box (talk) 23:15, 17 February 2008 (UTC)[reply]

How clever! —Keenan Pepper 00:30, 18 February 2008 (UTC)[reply]

You've seen our article about the Pitch drop experiment, right?

Atlant (talk) 01:52, 18 February 2008 (UTC)[reply]

Sudden tics when drowsy?

Sometimes, especially in class, if I am leaning back and attempt to sleep, parts of my body will suddenly tense. This is really uncomfortable, and I'd very much like to sleep without this happening. Is there a good article on this muscle behaviour? 81.93.102.185 (talk) 23:18, 17 February 2008 (UTC)[reply]

Perhaps you might consider attempting to sleep somewhere else than in class? You probably want to read hypnic jerk and myoclonus, but if you want more than that, please see a doctor, because medical advice is not allowed here. --Anonymous, 23:47 UTC, February 17, 2008.

I doubt that this is a medical issue so I'll provide my insight and two cents. What you are experiencing are most likely hypnic jerks upon entering into stage one of non-REM sleep. Also, strange sensations when falling asleep might be a sign of sleep paralysis, which is completely benign. Wisdom89 _{(_T / ^C)} 09:36, 18 February 2008 (UTC)[reply]

Who would win?

In a fight between a Predator and a Terminator, which would be more likely to win? --81.79.122.2 (talk) 23:40, 17 February 2008 (UTC)[reply]

Not only does this ask for speculation, it is not a science question. --Anon, 23:48 UTC, Feb. 17.

Sounds like a great idea for a movie proposal. And you get to decide! Someguy1221 (talk) 00:23, 18 February 2008 (UTC)[reply]

Predator by a mile - even though this isn't a legitimate science question. One blends into the environment and preys on a victim, while the other loafs around blindly. If you want to liken it to something in nature, how about a preying mantis verses a spider? Wisdom89 _{(_T / ^C)} 00:27, 18 February 2008 (UTC)[reply]

But human Arnold beat a Predator so therefore clearly a robot Arnold must destroy a Predator. —Preceding unsigned comment added by Shniken1 (talk • contribs) 01:21, 18 February 2008 (UTC)[reply]

You do have to remember that a Predator, while probably of approx. equal strength and speed to a Terminator, is only slightly less vulnerable to gunfire/edged weapons/fists and feet than your average human. A Terminator can absorb significantly more damage and still keep coming. If the Predator cloaked, the Terminator would probably be able to calmly track its vague outline against the surroundings (and get a target lock) just fine. --Kurt Shaped Box (talk) 01:34, 18 February 2008 (UTC)[reply]

In the AVP movies, Preds usually wear bullet proof armor on their bodies. Does a pretty good job of deflecting bullets. Terminators are still more durable despite that though. Not nearly as agile though. Malamockq (talk) 02:42, 18 February 2008 (UTC)[reply]

The Predator might come out on top if he figured out what the humans obviously seem unable to when fighting Terminators - namely that plugging away at a cyborg's heavily-armoured head and torso is *very* ineffective and instead concentrated on shooting its spindly legs out from under it. Once crippled, the Pred could take his sweet time and concentrate plasmacaster fire on the Terminator from a distance, until he managed to damage something vital. --Kurt Shaped Box (talk) 16:35, 18 February 2008 (UTC)[reply]

See answer #4 here, and substitute the strings "Predator" and "Terminator" as appropriate. —Steve Summit (talk) 01:48, 18 February 2008 (UTC)[reply]

They are evenly matched. It's all up to the writers. That's what makes "vs." stories interesting. Pitting two evenly matched opponents against each other. Having Predator vs. Galactus, wouldn't make much sense for example. Malamockq (talk) 02:40, 18 February 2008 (UTC)[reply]

Despite the speculation comment, the opinions have come thick and fast :) In the latest Requiem AVP movie, the predator takes off his armour to experience an honourable fight to the death with his opponent, much like you see in the martial arts movies where the guy who has just disarmed the other guy, throws his sword away as well. So I suppose in the predator's mind, his opponent was a respected and worthy one. Sandman30s (talk) 12:26, 18 February 2008 (UTC)[reply]

Ten bucks on Terminator. Humanity shall prevail. --Ouro (blah blah) 13:02, 18 February 2008 (UTC)[reply]

Just ask Google [4] :P --V. Szabolcs (talk) 15:35, 18 February 2008 (UTC)[reply]