May 28 Information

If you throw something at it... Like the sun, and throw it hard. Like 99.99% the speed of light, what will it do to the black hole? Will it affect the black hole's velocity? 65.41.92.123 ( talk) 00:25, 28 May 2008 (UTC) reply

Yes, it would work the same way as any other object, assuming the other object would be able to absorb a star. In fact, absorbing any object would make it move, just more slowly. See conservation of momentum. *Max* ( talk) 00:45, 28 May 2008 (UTC) reply

A gravitational tractor would likely be the most reliable way to move a black hole. (Although given the likely masses involved, it would take a long time and a lot of energy.) -- 128.104.112.147 ( talk) 01:25, 28 May 2008 (UTC) reply

In dipole, two equasions for the magnetic field are given; one describes it as a scalar, the other as a vector. They are

${\displaystyle B(\mathbf {r} ,\lambda )={\frac {\mu _{0}}{4\pi }}{\frac {\mathbf {m} }{r^{3}}}{\sqrt {1+3\sin ^{2}\lambda }}}$

and

${\displaystyle \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi r^{3}}}\left(3(\mathbf {m} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {m} \right)+{\frac {2\mu _{0}}{3}}\mathbf {m} \delta ^{3}(\mathbf {r} )}$

Setting them equal,

${\displaystyle {\frac {\mu _{0}}{4\pi }}{\frac {\mathbf {m} }{r^{3}}}{\sqrt {1+3\sin ^{2}\lambda }}=|{\frac {\mu _{0}}{4\pi r^{3}}}\left(3(\mathbf {m} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {m} \right)+{\frac {2\mu _{0}}{3}}\mathbf {m} \delta ^{3}(\mathbf {r} )|}$

Dividing by common stuff and since the Dirac delta will gives zero,

${\displaystyle {\mathbf {m} }{\sqrt {1+3\sin ^{2}\lambda }}=|\left(3(\mathbf {m} \cdot {\hat {\mathbf {r} }}){\hat {\mathbf {r} }}-\mathbf {m} \right)|}$

Now, ${\displaystyle \mathbf {m} =(1,0,0)}$ and ${\displaystyle \mathbf {r} =({\frac {\sqrt {3}}{3}},{\frac {\sqrt {3}}{3}},{\frac {\sqrt {3}}{3}})}$, so ${\displaystyle \lambda =\arctan {\sqrt {2}}}$. But,

${\displaystyle 1{\sqrt {1+{\frac {2}{3}}}}=|\left(3({\frac {\sqrt {3}}{3}}){\hat {({\frac {\sqrt {3}}{3}},{\frac {\sqrt {3}}{3}},{\frac {\sqrt {3}}{3}})}}-(1,0,0)\right)|}$

${\displaystyle {\sqrt {\frac {5}{3}}}=|\left((1,1,1)-(1,0,0)\right)}$

${\displaystyle {\sqrt {\frac {5}{3}}}=|(0,1,1)|}$

${\displaystyle {\sqrt {\frac {5}{3}}}={\sqrt {2}}}$

These are not equal. Am I doing something wrong or is one formula wrong? Thanks *Max* ( talk) 00:41, 28 May 2008 (UTC). reply

Isn't ${\displaystyle \lambda }$ a function of r. For your example r , ${\displaystyle \lambda =\arctan(1)}$ or 45°. Graeme Bartlett ( talk) 02:47, 28 May 2008 (UTC) reply

Second equation is the one I remember since undergrad, it is correct. The first one, as you say quite correctly, gives B as a scalar, that is, gives an absolute value of B. Therefore, m should not be boldfaced in that equation. Now, as you know, absolute value of a vector is a square root of sum of squares of its orthogonal components. Let us take z-axis along m and x-axis in (m, r) plane. B has no y-component then. The second equation gives, for ${\displaystyle cos\theta =\mathbf {m} \cdot \mathbf {r} /mr}$ and r > 0, the following value of B components:

${\displaystyle B_{z}={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}\left(3\cos ^{2}\theta -1\right)}$,

${\displaystyle B_{x}={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}\left(3\sin \theta \cos \theta \right)}$.

Root of sum of squares of the two is

${\displaystyle B={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}{\sqrt {1-6\cos ^{2}\theta +9\cos ^{2}\theta (\sin ^{2}\theta +\cos ^{2}\theta )}}}$

${\displaystyle ={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}{\sqrt {1+3\cos ^{2}\theta }}={\frac {\mu _{0}}{4\pi }}{\frac {m}{r^{3}}}{\sqrt {1+3\sin ^{2}\lambda }}}$.

Thus, the first equation is also correct, provided you replace the boldface m by m in it. Hope this helps. -- Dr Dima ( talk) 03:50, 28 May 2008 (UTC) reply

And this is why I shall not be studying physics at anything beyond 'A' Level :| Regards, CycloneNimrod ^Talk? 09:11, 28 May 2008 (UTC) reply

I'll unboldface m in the article. I calculated λ based on r. The opposite side is ${\displaystyle {\sqrt {3+3}}={\sqrt {6}}}$ and the adjcent is ${\displaystyle {\sqrt {3}}}$, so shouldn't λ be ${\displaystyle \arctan {\sqrt {2}}}$? Does the value of B mean that a point dipole fixed here by other forces here will orient itself perpendicular to the first dipole? I was not expecting that. *Max* ( talk) 17:24, 28 May 2008 (UTC) reply

In the example you give, m = (1,0,0) and r = (1,1,1) / 3^1/2. Therefore, one gets ${\displaystyle \sin \lambda \equiv \cos \theta ={\sqrt {(}}1/3)}$. As you know, ${\displaystyle 1+\tan ^{2}(\lambda )\equiv 1/(1-\sin ^{2}(\lambda ))}$, thus, for ${\displaystyle \sin ^{2}(\lambda )=1/3}$ you get ${\displaystyle \tan ^{2}(\lambda )=1/2}$, which means ${\displaystyle \lambda =\arctan \left(1/{\sqrt {2}}\right)}$ . Please check. As for your second question, I did not quite understand what you mean. If you ask whether there are points in space where B vector is perpendicular to m vector, then the answer is certainly yes, for any r > 0. -- Dr Dima ( talk) 02:32, 30 May 2008 (UTC) reply

Is there any way to magnify the scalar forces? So that it works on large objects? Like make a car levitate or anything? 65.41.92.123 ( talk) 02:13, 28 May 2008 (UTC) reply

I don't know much about this subject, but Zero_point_energy#Levitation_and_inertia looks relevant. -- Allen ( talk) 05:35, 28 May 2008 (UTC) reply

There's no such thing as a zero-point-energy force. The Casimir force is just an ordinary electromagnetic attraction between closely spaced neutral conducting plates. It's unusual in that you can compute its magnitude approximately from an argument involving zero-point fluctuations, but you can also compute its magnitude (more accurately) in the same way as any other electromagnetic force. See this paper. That said, you can levitate macroscopic objects electromagnetically—see magnetic levitation—so in effect the answer to your question is yes. -- BenRG ( talk) 13:03, 28 May 2008 (UTC) reply

I find it interesting that the paper above contradicts the implications in the articles Zero_point_energy and Casimir_force that the Casimir force is strong evidence for the existence of zero point energy. Is it possible those two articles need to be rewritten to de-emphasize Casmir forces as being linked to zero point energy?

Also Ben commented above that there is no such thing as a zero point energy force. My understanding is that you can't extract work out of zero point energy, since it is the lowest possible energy state, but that zero point energy itself does exist as a strictly positive amount of minimal energy. So while you presumably can't extract work from zero point energy, the random quantum fluctuations of zero point energy will prevent any system from reaching an absolute zero energy state. Am I understanding all this correctly? 63.95.36.13 ( talk) 20:44, 28 May 2008 (UTC) reply

Yes you are correct, zero-point energy is beleived to be real, I have read many books published by physicists that zero-point energy is the cause of the casimir force, along with multiple webpages including Wikipedia's, the article above is the first I have read that says the opposite. I would think that this one article is wrong instead of multiple on Wikipedia, others on the web, and multiple books.

Can BritaTM Containters Leach Bisphenol A? Are they made of polycarbonate? Are they made of a plastic that can leach BPA? 68.148.164.166 ( talk) 21:40, 21 May 2008 (UTC) reply

Can you find the plastic number marking on the container? 2 and 3 are safe, I think. Imagine Reason ( talk) 00:43, 28 May 2008 (UTC) reply

I could not find the plastic number marking on the container. 68.148.164.166 ( talk) 06:55, 28 May 2008 (UTC) reply

Ask the manufacturer. Coincidentally, my brother forwarded this email from P&G: "PUR dispenser bodies are manufactured from an acrylic-based polymer classified as recycling code #7. PUR dispenser lids are manufactured from polystyrene, code #6. PUR dispenser filters are made from polypropylene, code #5, and also contain no BPA." I still won't use it, however, because it no longer filters flouride or even chlorine. Imagine Reason ( talk) 02:38, 28 May 2008 (UTC) reply

sir i want to know about can alcohol drink related with height increases. if yes how it work and if no so how some body heavy drinke and they become very tall and stong. —Preceding unsigned comment added by Dss sakti ( talk • contribs) 09:40, 28 May 2008 (UTC) for any sugesion u mail me plz {e-mail removed) reply

I'm not aware of any study or claim saying drinking alcohol increases height. As to the second part of your question, if they drink beer they get a lot of calories. If they get enough exercise to burn those calories they'll get strong. If they don't burn the calories they get obese and may develop health problems.

We don't respond by e-mail, please don't include your e-mail on this desk. (You never know who might use it.) -- 71.236.23.111 ( talk) 10:14, 28 May 2008 (UTC) reply

Are you referring to children (who haven't stopped growing) being heavy drinkers? I would expect it to reduce their final height, although I have no references for that. Also, drinking while pregnant can reduce the height of the baby - see Fetal alcohol syndrome. Drinking as an adult shouldn't have any effect on height - you've stopped growing by then. -- Tango ( talk) 10:17, 28 May 2008 (UTC) reply

I have two questions about Glow Sticks. 1. On the article, it says that the ester oxidizes. But, I don't quite understand how it oxidizes on a molecular level. Could you please show me the equation. So far, I only know the LEO and GER. 2. How do glow sticks manage to stay bright for long periods of time? In videos/ demonstrations, they only stay bright for a few seconds.

Thanks. 121energy ( talk) 09:46, 28 May 2008 (UTC) reply

Glow stick and Cyalume give the chemical equations going from oxalylic ester to carbondioxide. -- Stone ( talk) 15:20, 28 May 2008 (UTC) reply

(Moved up to join the original question.)

I'm doing some research for a script. I wonder if there is any drug that makes people... well... more horny? Preferably without any elaborate side effects. It should be used as a kind of "love potion", meaning it should help one character persuade another to have sex with him. I realize there's probably no substance that makes you go crazy for sex without any other side effects, but anything even remotely close to that would be enough, it's not a scientific script so I can make things up a bit to make it fit. It would be nice to just have the name of a substance that is remotely close, to make it a touch more realistic. Thanks! 81.236.199.5 ( talk) 11:12, 28 May 2008 (UTC) reply

The word you're looking for is aphrodisiac. There's a good list of alleged aphrodisiacs in the article. -- Coneslayer ( talk) 11:27, 28 May 2008 (UTC) reply

For fiction you might also find the human portion of Pheromones interesting. Or you could spin an idea off this [1]-- 71.236.23.111 ( talk) 14:39, 28 May 2008 (UTC) reply

Some dopaminergic drugs have alot of anecdotal evidence supporting 'horniness'. MDPV, methamphetamine#Sexual_behaviour, yohimbine and GHB are the most common ones I can think from the top of my head. -- Mark PEA ( talk) 19:36, 28 May 2008 (UTC) reply

Testosterone is well-known to increase libido. -- Sean 15:34, 29 May 2008 (UTC) reply

Here is the dog skull that I cleaned using the advice you gave here. What article can we enhance with this image?-- Lenticel ^{( talk)} 12:08, 28 May 2008 (UTC) reply

How about Dog anatomy or Skull? Only thing is there are many pictures there already. Yours is a good pic, though. Fribbler ( talk) 12:26, 28 May 2008 (UTC) reply

Good picture - I'd add it to dog anatomy, there are lots of pictures there, but none of a dog's skull! -- Tango ( talk) 12:55, 28 May 2008 (UTC) reply

I went to see this film a couple of days ago. In it, the hero hides in a metal box which is thrown through the air for several miles by an explosion, before it comes crashing to the ground, tumbles end-over-end, and eventually coming to rest. The door bursts open and our hero rolls out of the box apparently unscathed by his high-speed journey. Assuming the box is strong enough to retain it's shape, would a man actually be able to such a violent journey? Astronaut ( talk) 12:13, 28 May 2008 (UTC) reply

I doubt it. I would expect him to be crushed by the g-forces on both take off and landing. To propel a person several miles would require massive acceleration. According to projectile motion, the maximum range is ${\displaystyle {\frac {v^{2}}{g}}}$, assuming he travelled 5km, that gives a minimal initial velocity of about 224m/s (higher if he wasn't launched at exactly 45 degrees), assuming the explosion provides all of that speed in 1 second, that corresponds to an acceleration of about 22g. He'd probably lose conciousness, but may survive that. However, that same amount of speed needs to be lost when it lands, and that will be done in a fraction of a section (if the box retains it's shape, it means it didn't absorb the impact and would have stopped dead), resulting in much greater accelerations, almost certainly killing to occupant. (Imagine a car crash at 500mph with none of a car's usual safety features - not likely to walk away from that!) NB: I'm ignoring air resistance - for a heavy, compact box, it's negligable. -- Tango ( talk) 12:49, 28 May 2008 (UTC) reply

In the film, the box bounced across the ground several times before finally stopping; and I suppose there could have been some minor deformation of the outside of the box (ie. it didn't change the inside shape of the box, but could have been quite dented on the outside). Could the deceleration have been such that each bounce removed some of the velocity so our hero didn't get crushed on landing? Astronaut ( talk) 13:34, 28 May 2008 (UTC) reply

Reality would have made for a really short movie. -- LarryMac | Talk 13:26, 28 May 2008 (UTC) reply

Heh. Anybody see "Air America" with Mel Gibson, where the C130 (I think?) crashes in the jungle, and begins a long skid for about a mile over a couple of minutes before stopping, with (to me) pretty funny reaction from those on board? (It was a comedy, btw, I'm not being sociopathic). Gzuckier ( talk) 15:11, 28 May 2008 (UTC) reply

Bouncing would require some temporary deformation - it works by deforming the object and storing some of the kinetic energy as elastic potential energy and then that elastic energy turns back into kinetic energy and the deformation rebounds. I'm not sure what that would do to the person inside - the total change in velocity is actually greater, since you go from +224m/s to -10m/s, say, which is a change of 234, rather than the change of 224 if you stop dead. That change probably takes place over more time, though, so the acceleration (which is what's important) would be less. -- Tango ( talk) 14:13, 28 May 2008 (UTC) reply

Stranger things have happened. Wily D 13:35, 28 May 2008 (UTC) reply

That's a lot less strange. He would have been falling about four times slower, plus his fall was cushioned by a handy glass roof, rather than being inside an unyielding metal object. Algebraist 13:42, 28 May 2008 (UTC) reply

A metal box is better then? Or sitting comfortably in a chair? Wily D 13:57, 28 May 2008 (UTC) reply

Actually, you make an interesting point - I neglected air resistance, since it's usually negligible for a box, but at those kind of velocities, it probably isn't any more. Terminal velocity for the box would probably be less than 500 mph, would would improve his chances, but only to the chance of surviving a fall from a plane without a parachute and not landing on anything soft, which is still pretty much zero. Falling while still in a plane (which seems to be the case with most of these miracle stories) would give you a much lower terminal velocity - there's far more drag on a large plane (even after breaking up) than on a person. -- Tango ( talk) 14:13, 28 May 2008 (UTC) reply

The plane weighs a lot more, though. Your terminal velocity is just a function of your density and cross-sectional area. At high speeds, drag is going to go like r²v² and gravity will go like ρr³. If you're roughly spherical, your terminal velocity scales like v_t ~ ρ^½r^½ ... density's probably irrelevent (you're 1, a plane with lots of airspace is probably ~1. So person strapped to a chair = person, person in aircraft depends on chunk size, but they're falling faster. Wily D 14:26, 28 May 2008 (UTC) reply

My very rough calculations say that the density of a fully laden 747 is about 200 50 Note to self: Diameter and radius aren't the same thing! times less than that of a person (based on Boeing 747#Specifications). They are designed to be a light as possible, otherwise they wouldn't be able to fly. Even broken up into a few pieces, it's going to be significantly less than a person, and terminal velocity will be significantly reduced. (Unless you're unlucky enough to end up in a nose dive, perhaps.) -- Tango ( talk) 15:01, 28 May 2008 (UTC) reply

Yeah, deceleration by a long series of bumps, rolls, disintegration, etc. looks spectacular, but will keep peak G forces down to survivable levels. Look for the various Youtube videos of the Michael Mcdowell Texas Nascar Crash from a couple of months back. Since the OP's guy in a box is not strapped in, he has to avoid impact with the inside wall of the box, but if it's spinning, centrifugal force could keep him "strapped down" figuratively and prevent him from fracturing his skull. Kids: don't try this at home. Gzuckier ( talk) 15:08, 28 May 2008 (UTC) reply

I doubt it would spin reliably enough while bouncing. If it's small enough, he could wedge himself inside so he doesn't get shaken around. Even with all the bouncing, landing at terminal velocity in a metal box doesn't sound survivable to me without something to absorb the impact. -- Tango ( talk) 15:12, 28 May 2008 (UTC) reply

But if you are Indiana Jones, cinematic rules override kinematic rules! -- Stephan Schulz ( talk) 15:16, 28 May 2008 (UTC) reply

It's also probably worth noting that the amount of lead in a lead refrigerator would not have protected the hero in question from the thing in question that tossed him. Even if he had, improbably, survived he'd have probably gotten quick sick. -- 98.217.8.46 ( talk) 19:34, 28 May 2008 (UTC) reply

I think the original poster is wrong to assume the metal box was thrown "several miles". As I understood the scene, it didn't move particularly far; it's just that the same thing that put it in motion also damaged the surrounding area, so it may have looked different. You don't see it flying through the air or anything. --Anonymous, somewhere in the real world, 00:01 UTC, May 29, 2008.

Actually, it was shown flying through the air. In fact, from what I remember, its first bounce was just in front of the army car which was driving away from the explosion. The car didn't make it, but the fridge carried on bouncing. Jdrewitt ( talk) 08:50, 29 May 2008 (UTC) reply

Oh. Perhaps I remembered seeing what I expected to see, then. --Anon, 23:49 UTC, May 29.

you all seem to be forgetting that each time the box experiences a collision, its passenger is going to experience an internal collision. Our hero would've been broken and bloody after that trip, even if the box stayed closed and intact. Which it wouldn't have. -- Shaggorama ( talk) 03:41, 29 May 2008 (UTC) reply

I think the most ridiculous thing that has been lost in all of this (having just returned from a showing of said movie) is that he stands in the immediate aftermath of (what one can only assume is) a fairly dirty uranium-fission explosion, what looks like less than a mile from ground zero, with no long- or short-term effects on his health. The physics of that whole movie was (were?) shall we say, less than intelligent. - Running On Brains 05:38, 29 May 2008 (UTC) reply

I bet his hair was perfectly styled when he stood up too. That always happens to me when I get caught in nuclear explosions :) And I always get clean clothes in the next scene. Franamax ( talk) 08:18, 29 May 2008 (UTC) reply

He was rinsed off afterwards though to remove any radioactive particles on the outside of his body. As long as he didn't ingest any of these particles then he would be ok. The largest fraction of gamma radiation is emitted in the very first milliseconds of a nuclear explosion, during which time he was in the lead lined fridge (why was the fridge lead lined?) which would have reduced the dose. Jdrewitt ( talk) 08:31, 29 May 2008 (UTC) reply

Hmm.. it is only beta particles that are stopped by lead, and it would need to be a concrete fridge to at least slightly stop an A bomb. 86.153.37.241 ( talk) 00:14, 2 June 2008 (UTC) reply

Gamma radiation will get through lead, however it will still be attenuated and so the dose will be reduced, whether this would be enough, probably not but it depends on the thickness of the lead and the exposure time. Jdrewitt ( talk) 10:14, 2 June 2008 (UTC) reply

Sorry - I didn't check responses to my question: How far can snails travel back to their usual habitat if you remove them from this? within the 4 days. Please would anyone who knows the answer repeat it for me? Thank you. Ruth 555 Ruth555 ( talk) 12:38, 28 May 2008 (UTC) reply

The previous question is archived here. No-one knew a distance, they suggested marking the shells before releasing them so you can tell if they come back. -- Tango ( talk) 12:52, 28 May 2008 (UTC) reply

Original Research is frowned upon here, but it is allowed if no one can come up with a reference. Why not find out and tell us? On day one, pull each snail off, mark it with a red X and gently relocate it 100m away. On day 2, if any snails have a red X you know they can go 100m, so use a green marker and try 200m. If none have made it back, try 50m. Anyway, you have to use different colors and symbols so that on day 3 you know whether the returnees were from day 1 or day 2, etc. Once you have run out of colors use a circle, etc. Ensure that the marks are non-toxic because the experiment ends when you get tired of it and eat them. - SandyJax ( talk) 16:01, 28 May 2008 (UTC) reply

Original research is frowned upon only when describing it in articles. Experimentation is highly encouraged, however it needs to be described in a reliable source before it can be included in an article. It would make a great school science project, though. ~ Amatulić ( talk) 18:38, 30 May 2008 (UTC) reply

What is the instrument with the guy with his fingers in those beer glasses? 68.148.164.166 ( talk) 12:39, 28 May 2008 (UTC) reply

It's known as a Glass harp. -- Tango ( talk) 12:54, 28 May 2008 (UTC) reply

It's not clear why it is shown under the heading Plasmaphone though. Pfly ( talk) 21:03, 28 May 2008 (UTC) reply

(edit conflict) That is NOT a glass harp. A glass harp is played by running your finger around the rim of the glass. In that picture, the musician clearly has his fingers in the water. Moreover, the water level is the same in each glass, so there would be no difference in pitch. My guess is the cords connected to the glasses aren't just lighting them up: I'm betting those fluids are charged. I can only speculate how the instrument works, but maybe by sticking his fingers in the liquid, he changes its conductance properties, or cause a current to pass through his body changing the conductance of the system....I can only guess. In any event, it's probably some sort of fancy electronic instrument or user interface -- Shaggorama ( talk) 03:26, 29 May 2008 (UTC) reply

Good point. I'd assumed his fingers were inside the glasses simply because he was getting them wet, and he wasn't actually playing at that instant. I hadn't noticed the water levels were all the same - there must be something more going on (presumably to do with the wires, as you say). -- Tango ( talk) 13:38, 29 May 2008 (UTC) reply

Ok, I think after all this time I still do not really have a good grasp of what current is (and I've read the articles).

We normally rate batteries by their voltage. So a battery might have 1.5 Volts, say. If we connect a resistor across the battery's terminals, a current goes through it. That current can be calculated by Ohm's Law: I = V/R. So if we have a 10 ohm resistor, we have 0.15 Amps coming out of the battery, right?

If we have a resistor which is 0.1 ohms, we have 150 amps going through the resistor. So I guess we can keep going up to nearly infinity if we keep making the resistance smaller, is that right? (Except eventually the battery will run out.)

However, when I look at solar cells, they give both a voltage and a current (or sometimes they just give a current). So I might have a cell that says it provides 1 amp. So if I connect a resistor across it, is the voltage of the cell going to change? The current? I assume that Ohm's law has to be maintained?

Why are solar cells and batteries different? Can batteries produce any current they want, and cells can only produce a fixed current? Or is it a maximum current? What happens if I have a 5V solar cell connected across an 0.001 Ohm resistor? What is the current?

Thanks! — Sam 14:48, 28 May 2008 (UTC)

Ah; the catch is (there's always a catch) that batteries and solar cells are not ideal voltage sources, nothing being ideal, the way they are often simplified as. In the real world, of course the battery/solar cell has its own internal resistance; everything does. So, if your 1.5 volt battery has a 1 ohm internal resistance, then even if it is short circuited, it's not going to deliver any more than 1.5 amps. Same for the solar cell. Same for transformers, anything in the real world; there's always an internal resistance which needs to be factored in in some cases. Of course, also there's internal capacitance and inductance which may need to be accounted for under some conditions; that's why electronic engineers who understand real things not just ideal things do a better job. Gzuckier ( talk) 15:01, 28 May 2008 (UTC) reply

(See also internal resistance). TenOfAllTrades( talk) 15:23, 28 May 2008 (UTC) reply

If you'll allow me to expand on Gzuckier's point, we normally design electronics so that our "voltage sources" are operating pretty close to the regime of being an "ideal" voltage source. That is, while a battery might deliver (say) 5 amps into a dead short, it's commonly used in equipment that's drawing far less current, say 0.25 Amps. There are three reasons for this:

1. It makes the circuitry a lot easier to design if you can mostly ignore the internal resistance of the voltage source (battery).
2. It would be wasteful if the battery were internally dissipating a large share of the total power that was being consumed from it. To a first approximation, if you tried to operate our hypothetical "5 amp" battery with a 2.5 Amp load, roughly half the power being produced from the battery's chemical reaction is being dissipated in the load and half the power within the battery itself. The power dissipated within the battery does no useful work.
3. The internal resistance of a battery rises as it ages. If you design your circuit to consume a lot less current than a new battery is capable of, an old, decrepit battery may still be capable of powering the circuit.

Hydraulic analogies are often useful in this case. A 1.5 volt battery might be compared to a water tank 1.5 metres off the ground. That tiny AAA cell represents a tank of 1 litre capacity. That big D cell represents a tank of 20 litres capacity. The bigger tank also has a proportionally-larger shut-off valve on the tank's discharge connection. Now, connect a drainage pipe to the tank. For a very skinny pipe, both tanks will provide adequate water pressure but the AAA-sized tank will run out of water twenty times as fast as the D-size tank. Now connect a bigger discharge tube to the tanks. For the tiny tank, the tank's discharge valve will limit water flow into the big pipe and there'll be very little pressure in the discharge pipe. But for the D-sized tank, its larger discharge valve can still provide full water flow into the discharge pipe and full water pressure. Batteries (and solar cells and most other voltage sources) work just like this.

Atlant ( talk) 17:28, 28 May 2008 (UTC) reply

Ok, that makes a little sense. However, I still don't quite understand why solar panels are rated just by their amperage. Is a "one amp" panel describing the current that would be delivered into a short? — Sam 20:49, 28 May 2008 (UTC) —Preceding unsigned comment added by 63.138.152.238 ( talk)

It's more likely a roughly defined point at which the internal resistance can be considered negligible, and thus at which the cell is suitable for powering a given application. As Atlant notes in the 5-amp (short current) battery example, it wouldn't make sense to use it in a 2.5 amp application, either. — Lomn 21:57, 28 May 2008 (UTC) reply

I don't know about the case of solar cells, but normally when both a voltage and an amperage are quoted the voltage is an estimate while the amperage is a maximum. E.g. a 240V 13A wall socket will maintain a potential difference of about 240 volts and you shouldn't try to draw more than 13 amps from it or something bad will happen (in this case a tripped circuit breaker). -- BenRG ( talk) 23:27, 28 May 2008 (UTC) reply

A battery or solar cell could be characterized by the open circuit (no load voltage), which will be the highest possible output voltage, higher than seen when it is supplying current, and the short circuit current, which is quite high (several amps) even for an AA alkaline cell. But such a high current would limit the useful life to a very short period for a battery, and the voltage would drop dramatically. Another rating system would be so many amps (or milliamps) at a certain output voltage for normal or optimal operation. For example, a 200 Watt Kyocera solar panel [2] specifies that for max power output, the current should be 7.61 amps and the voltage should be 26.3 volts, the product of current times voltage equalling 200.61 Watts,which implies a total external circuit resistance including the load of 3.46 Ohms. but says the short circuit current is 8.21 amps and the open circuit voltage is 32.9 volts. The power under short circuit conditions would be 8.21 amps times zero volts, or zero Watts. The power for open circuit conditions would be 32.9 volts times zero amps, or zero Watts. For conditions close to short circuit or open circuit, the output power would be nonzero but far smaller than the 200 Watt maximum. Edison ( talk) 02:41, 29 May 2008 (UTC) reply

i want get the details of solar power process and its panel types with photo graph & the details of batterys used to store power with maximum capacity and its configration details

if provide.any source or call center to discuss about said details —Preceding unsigned comment added by 124.125.42.50 ( talk) 15:11, 28 May 2008 (UTC) reply

I would think that solar energy would be a good place to start and then, when you have specific questions, ask again. -- kainaw ™ 19:08, 28 May 2008 (UTC) reply

Thank you for responding to my latest enquiry. The idea of nail varnish is a good one. Please be patient with my lack of expertise on this site. I'm aware that this is a response, rather than another question, but am not sure how to slot this comment into the correct place! As a newcomer, I'm enjoying the site immensely! What a variety of different subjects - I've spent hours reading through the topics instead of getting on with my own work - I'm in process of writing a novel. Ruth555 Ruth555 ( talk) 15:21, 28 May 2008 (UTC) reply

I answered the question above. - SandyJax ( talk) 16:13, 28 May 2008 (UTC) reply

Hello Ruth - we're glad you're enjoying yourself. When you want to add to an existing section, just click on the "edit" link on the right side of the line that contains the section title. -- LarryMac | Talk 16:19, 28 May 2008 (UTC) reply

I know that rust is produced, but what gas is created?

I've noticed bubbles continually forming. Is it chlorine? —Preceding unsigned comment added by 24.81.198.165 ( talk) 20:11, 28 May 2008 (UTC) reply

It's got to be hydrogen. Rust is a mix of iron oxides. Since there isn't enough dissolved oxygen in water to rust things very fast, the water must be dissociating into hydrogen and oxygen. The oxygen combines with the iron in steel to form rust, and the leftover hydrogen atoms combine to form hydrogen gas. -- Carnildo ( talk) 20:45, 28 May 2008 (UTC) reply

Hello. Is C₂H₅OH a hydrocarbon? When burning, is the chemical equation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O? Can this reaction possibly be double displacement? Thanks in advance. -- Mayfare ( talk) 21:21, 28 May 2008 (UTC) reply

Have a look at Ethanol, Alcohol and Metathesis reaction (double displacement). J kasd 21:41, 28 May 2008 (UTC) reply

You should also take a look at combustion. bibliomaniac 1 5 04:02, 29 May 2008 (UTC) reply

Hydrocarbons contain hydrogen and carbon only. Ethanol contains oxygen so is not a hydrocarbon. 86.153.37.241 ( talk) 00:16, 2 June 2008 (UTC) reply

This French guy has just recently failed in an attempt to float his helium balloon at 40,000m and then jump out, free-falling for around 15 minutes before opening his parachute. In every article I have read about this, it says at 35,000m he would have been expected to break the sound barrier. Is this possible? I know at that height the air is relatively thinner and there would therefore be less friction, but surely his terminal velocity would not rise from c. 150mph (nearer the ground) to over 750mph? Also, would there be a sonic boom and what effect would this have on him, considering he wouldn't be in the safety of a jet-plane?-- ChokinBako ( talk) 21:26, 28 May 2008 (UTC) reply

I can't comment on the specifics, but one thing people sometimes forget is: the only reason a falling object has a terminal velocity at all is air resistance. So, if the air resistance almost goes away, the resulting speeds could be very high. Friday (talk) 21:30, 28 May 2008 (UTC) reply

Also, the speed of sound varies with temperature. Lower temperatures at high altitude decrease the speed of sound. See sound speed gradient for a discussion of this phenomenon. — Lomn 21:47, 28 May 2008 (UTC) reply

Indeed 150mph is not an accurate terminal velocity near the ground. According to Terminal velocity an experienced skydiver can achieve about 200 mph by optimising body position (of course an experienced skydiver wouldn't want to be at that speed to near the ground, but hopefully people get what I mean). According to [3] the record is 321MPH without any special equipment and Joseph W. Kittinger achieved 619mph Nil Einne ( talk) 23:08, 28 May 2008 (UTC) reply

Who is a TECHIE COOLIE & Why are they so called? 117.197.240.153 ( talk) 21:44, 28 May 2008 (UTC) reply

It could be a racial slur about someone who is a "techie". See techie and coolie. Friday (talk) 21:50, 28 May 2008 (UTC) reply

Alternatively, it may derive from the use of "coolie" to indicate a brute-force laborer. That would still leave it a derogatory remark, though perhaps not an outright slur. In any event, Google shows very little use of the phrase, so there's not likely a definitive answer. — Lomn 21:54, 28 May 2008 (UTC) reply

It could be a derogatory remark about the techie's employer, saying that they treat their workers like coolies. That seems the most likely interpretation to me, but of course I'm guessing. --Anonymous, 00:05 UTC, May 29, 2008.

Perhaps he works next door to the Code monkey. Edison ( talk) 02:22, 29 May 2008 (UTC) reply

RE: http://www.youtube.com/v/kS9inNW4KS8 -- Why is so much energy released when water is added to a grease fire? I'm wondering whether the combustion of the fuel in an internal combustion engine could be similarly augmented, by injecting some water into the cylinder, either with the fuel or separately, before or immediately after ignition. -- Nonlocal ( talk) 21:58, 28 May 2008 (UTC)Nonlocal reply

All that's happening is that the water is (very quickly) brought to a boil by the hot oil. So, it bubbles and expands and the hot grease goes everywhere, and the fire spreads. There is such a thing as water injection in automotive technology (see Water injection (engines), but it's about cooling, not getting energy out of the water. Friday (talk) 22:24, 28 May 2008 (UTC) reply

Man, what is it with the British and their very disturbing PSAs! Yikes! -- 98.217.8.46 ( talk) 01:23, 29 May 2008 (UTC) reply

Disturbing = unlikely to be forgotten Theresa Knott | The otter sank 06:35, 29 May 2008 (UTC) reply

I think my main confusion is that the British seem to think their PSAs should actually be useful rather than just hollow ad campaigns used to write-off a tax deduction or something along those lines. ;-) American PSAs are almost always totally toothless, laughable, poorly designed and poorly executed. -- 98.217.8.46 ( talk) 15:27, 29 May 2008 (UTC) reply

That's a pretty awesome demonstration in that video. I think the fire grows so rapidly because the explosive boiling of the water conitnually throws up grease because it's contained in the pot (and there's also probably an unusual amount of grease in the pot). Normally, grease fires don't result from pots of grease like that; usually, the flaming grease is in a pan and there is less of it. In that scenario, adding water spreads the fire around instead of making it grow like that. But in any event, don't put water on a grease fire! -- Shaggorama ( talk) 03:13, 29 May 2008 (UTC) reply

A kilogram of TNT releases less enery than a candle of 1kg, but it looks different, becaus the candle takes hours and the TNT far less than a second for the reaction.-- Stone ( talk) 05:25, 29 May 2008 (UTC) reply

The paper, On the interaction of a liquid droplet with a pool of hot cooking oil, Short communications, by S.L. Manzello, J.C. Yang and T.G. Cleary (Fire Safety J 38 7 (2003), pp. 651–659), notes that such occurrences of rapid boiling of water "have been termed vapor explosions, explosive boiling, or rapid vapor explosions." -- Nonlocal ( talk) 16:39, 29 May 2008 (UTC) reply

As I understand the modern conception of physics, conservation laws arise because of symmetries in the laws of physics ( Noether's theorem, I think). Conservation of momentum is the result of the universe being symmetric under spatial translation, conservation of angular momentum is due to spatial rotational symmetry, and the conservation of energy is due to time translational symmetry. However, we also have the concept of space-time, where there is no distinction between space and time (after accounting for a few -1's, i's, and c's). So if we have conservation of angular momentum due to rotational symmetry in the x-y plane (a rotation that occurs between two spatial axises), is there a conservation law for rotation in the x-t plane? (That is, a rotation that occurs between a spatial axis and the time axis.) - How about for rotations and translations involving the other 6+ dimensions proposed by string theory? -- 128.104.112.147 ( talk) 22:32, 28 May 2008 (UTC) reply

The best way to think about it is to consider the metric. It will be a function of your coordinates, (t,x,y,z), say. If it doesn't depend on one of them, you have a conserved quantity (momentum in that direction, roughly speaking - see Killing field for more details). If you use different co-ordinates, for example, spherical coordinates, you may find that the metric doesn't depend on one of the angles, that gives you conservation of momentum in the direction of that angle - angular momentum, in other words. If you choose a coordinate system in which one of the coordinates is a rotation in the x-t plane, I imagine you would find that the metric does depend on that coordinate (those -1's, i's and c's do need to be accounted for!), so you don't get a conserved quantity. I'd have to find pen and paper to work it out for sure. The extra dimensions of string theory would definitely allow for extra conserved quantities, but if you ever actually have Killing fields in those directions, I don't know. -- Tango ( talk) 23:01, 28 May 2008 (UTC) reply

The conserved quantity associated with rotation around the origin in the xy plane is x p_y − y p_x, i.e. the angular momentum around the origin in the xy plane. This holds for any pair of perpendicular axes, even if one of them is timelike. So the conserved quantity associated with rotation in the xt plane is x E − t p_x. I'll let you work out what that represents. Note that it depends explicitly on t, unlike the conserved quantities people normally talk about, but that's not a deep difference—you just plug in different values of t at different times.

But the laws of physics are not symmetric with respect to ordinary rotation in the xt plane; you need a Lorentz transformation instead. I'll let someone else figure out what the corresponding conserved quantity is, though. — Ilmari Karonen ( talk) 22:16, 29 May 2008 (UTC) reply

The Lorentz symmetries give rise to conservation of four-momentum, surely? Algebraist 22:26, 29 May 2008 (UTC) reply

I don't know whether the Calabi-Yau manifolds that describe the extra six dimensions in string theory have any symmetries that lead to conserved quantities. In Kaluza-Klein theory I think that the conserved quantity associated with translation in the fifth dimension is the electric charge. -- BenRG ( talk) 00:19, 29 May 2008 (UTC) reply