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March 29

Electroweak theory

Is electroweak theory used in nuclear reactors or is Fermi's theory enough? What about in studies of stellar evolution? 74.14.110.32 (talk) 01:15, 29 March 2010 (UTC)[reply]

Long story short - Fermi's theory gives good results for those "low" energy applications. Dauto (talk) 20:01, 29 March 2010 (UTC)[reply]

Read Electroweak star for a still speculative highly advanced stage of star evolution that requires the full electroweak theoretical machinery. Dauto (talk) 20:43, 29 March 2010 (UTC)[reply]

Are there any "high-energy" applications outside of particle accelerators? 99.237.180.215 (talk) 20:15, 29 March 2010 (UTC)[reply]

Yes, there are applications. For instance, Fermi's theory does not account for neutral currents which allow neutrinos to scatter ellastically off of matter. That fact was skillfuly used at snolab to clinch the proof that solar neutrinos oscillate into different kinds of neutrinos along their path towards the earth. Dauto (talk) 20:52, 29 March 2010 (UTC)[reply]

What muscle is this, and how do I exercise it?

It's on the side of the body, under the armpit. Basically the one that makes it look like Bruce Lee can fly. Here's a picture: http://en.wikipedia.org/wiki/File:The.Way.Of.The.Dragon.1972.Bruce.Lee.flex.front.jpg

What is the muscle, and would I exercise it by doing tricep dips, or wide-grip pullups, or what? Thanks. —Preceding unsigned comment added by Kevin6174 (talk • contribs) 04:46, 29 March 2010 (UTC)[reply]

Does this help? If not, see axilla. I tried. --Ouro (blah blah) 09:34, 29 March 2010 (UTC)[reply]

Indeed, I do think you're asking about the Latissimus dorsi muscle, and the "Training" section of that article should answer the second part of your question. -- Scray (talk) 09:53, 29 March 2010 (UTC)[reply]

Heat producers

This question appears to be a request for medical advice. It is against our guidelines to provide medical advice. You might like to clarify your question. Thank you.

Responses containing prescriptive information or medical advice should be removed and an explanatory note posted on the discussion page. If you feel a response has been removed in error, please discuss it before restoring it.

-- Scray (talk) 09:55, 29 March 2010 (UTC)[reply]

Space Radiation and the Earth's Atmosphere —how are high-frequency light rays affected?

I have a question about how exactly the Earth's atmosphere protects us from space radiation?

As far as I'm aware, high-frequency radiation (gamma rays, x-rays, and most ultraviolet rays) are in some way "filtered" by the atmosphere and thus never gets to the surface. Whereas lower-frequency radiation (some ultraviolet rays, visible light, infrared waves, microwaves, and radio waves) penetrate the atmosphere and are felt down here.

My question is twofold:

(1) How does the atmosphere interact with the higher-frequency radiation, exactly? Does it reflect the gamma rays, etc. back into outer space; decrease the frequency of it so as to render it into visible light, etc; or some other process?

and

(2) Is this done by the radiation of the atmosphere, the chemical composition, or something else?

--Thank You. Pine (talk) 06:38, 29 March 2010 (UTC)[reply]

The rays are absorbed by the atmosphere, which therefore gets slightly warmer. See [1] for an illustration of this. --Phil Holmes (talk) 09:45, 29 March 2010 (UTC)[reply]

Also, the atmosphere does absorb a lot of the lower-frequency radiation as well except at some atmospheric windows. Dauto (talk) 14:13, 29 March 2010 (UTC)[reply]

Very low frequency radio waves are transmitted through the Earth's ionosphere and atmosphere, but are distorted by electromagnetic interactions. The result is a radio whistler, a subject of great scientific study because their frequency patterns can be used to characterize the physics of the Earth's magnetic field and radiation belts. Nimur (talk) 16:41, 29 March 2010 (UTC)[reply]

Nuclear bomb on Jupiter?

Would exploding a hydrogen bomb on the planet Jupiter or any of the other gas giants turn them into a second sun? And if so, would there be any effects on the Earth?80.1.88.25 (talk) 12:33, 29 March 2010 (UTC)[Trevor Loughlin][reply]

The size and heat of even the largest H-bombs ever set off on Earth is still quite small on a planetary scale even if they are large on a human scale. The amount of energy that is released when, say, an asteroid hits a planet is vastly larger (though not always concentrated in the same area). In any case, even if a significant fusion reaction was set off inside the planet (where the material would be of highest density), without there being more mass and gravity, it would not be self-sustaining like reactions in the Sun. (For the same reason, setting off an H-bomb on Earth does not begin a fusion reaction in the nitrogen in the air or the water in the sea, even though in theory both could undergo fusion under the right conditions. But Earth does not have the right conditions for that.) --Mr.98 (talk) 12:42, 29 March 2010 (UTC)[reply]

Bombs cause explosions, which move things outwards. To turn Jupiter into a second sun you would need to compress it to increase the temperature and density to a point where fusion can occur. However, once that fusion did occur the radiation pressure would cause the planet to expand again and the fusion would stop. You would need some way to compress it and keep it compressed, and I don't think there is any practical way of doing that, even assuming futuristic technology. I think any attempt to compress it would end up using more energy than would be generated. --Tango (talk) 13:07, 29 March 2010 (UTC)[reply]

Agree with Mr.98 and Tango. None of the gas giants are large enough to sustain fusion reactions at their core - otherwise they would be stars already. The minimum theoretical mass for a star with a similar composition to our Sun is about 75 times the mass of Jupiter, and the least massive known star, AB Doradus C, is 93 Jupiter masses - see star. Jupiter became a star at the end of 2010: Odyssey Two, but only with the help of advanced alien technology.

To answer the second part of your question, if Jupiter did somehow become a star, but without changing its size, mass or orbit, I think there would be a minimal impact on the Earth. Jupiter is 1/10 of the diameter of the Sun, and even at its closest, it is four times as far away from Earth as the Sun. This means that its apparent area in the sky is at most about 1/1600 that of the Sun. If we assume a Jupiter-star would emit the same amount of radiation per unit surface area as the Sun, then it would increase the solar radiation on Earth by less than a tenth of 1%. This is the same order of magnitude as the natural solar variation that we experience as the Sun's output changes over its 11-year solar cycle. Gandalf61 (talk) 13:10, 29 March 2010 (UTC)[reply]

To the contrary, I think it would be devastating if Jupiter ever became a sun. Much of Earth's night would be gone, which would wreak havoc on nocturnal animals and mess up the circadian rhythms of the diurnal animals. --206.130.23.67 (talk) 14:39, 29 March 2010 (UTC)[reply]

If Jupiter (at the current size) were burning like a star, it would not eradicate nights on Earth. Your statement appears to be based on the idea that it will shed enough light every night to cause problems. First, it is 1/1047 the volume of the sun. It is more than 4 times further away from Earth than the sun. (Please check my math - I am doing this with a crying baby on my lap, so not concentrating well.) Combined, the luminosity will be minuscule compared with the Sun (or even a full moon). It will be very visible, but mainly as a very bright star, not something that lights up the night. Further, that is only if it is in the night sky. Jupiter is not always in the night sky. -- kainaw ™ 14:50, 29 March 2010 (UTC)[reply]

Since we don't know any way of turning Jupiter into a star, it is difficult to guess what luminosity it would have. A key thing to keep in mind is that human light perception is logarithmic. A full moon doesn't seem that much dimmer than the sun - it's plenty to see by. That is despite the Sun being 449,000 times brighter than a full moon. If Gandalf61 is correct (his guess is as good as anyone's) and Jupiter would be only 1000 times dimmer than the Sun, it would be many many times brighter than the full moon. You wouldn't really notice that it was dimmer than the sun at all. --Tango (talk) 15:09, 29 March 2010 (UTC)[reply]

Using the basic luminosity function L=4πR²σT⁴, you can see that the radius of Jupiter is directly related to the luminosity. Since Jupiter is 1/10th the radius of the Sun, it has 1/100 less luminosity (squared produces 100 instead of 10). If we assume T to be the same, Jupiter would be 1/100th the luminosity of the Sun. That would be the apparent luminosity if Jupiter were the same distance from Earth as the Sun. It is more than 4x further away, so it would have an apparent luminosity less than 1/100th that of the sun. -- kainaw ™ 15:34, 29 March 2010 (UTC)[reply]

That is consistent with Gandalf's guess (I think you did the same calculation). As I say, being 1/1600 the brightness of the Sun wouldn't be very noticeable to humans (your pupils would dilate a little more, and then everything would seem the same). That formula doesn't really apply, though, since we're assuming something weird is happening to make Jupiter a fusor. --Tango (talk) 15:56, 29 March 2010 (UTC)[reply]

OK - now that I can work out equations and type without little fingers trying to press all the keys on the keyboard... The apparent luminosity (usually referred to as "magnitude") will be approximately 1/1600 that of the Sun if Jupiter has the same temperature as the Sun. But, that will not be a reasonable assumption. Instead, consider a very small star, still larger than Jupiter, but damn small. VB 10 is the best I can find as a reference. It has an absolute luminosity that is 0.0008 that of the Sun. Therefore, the apparent luminosity of Jupiter, if it was as bright as VB 10, would be about 0.00005 that of the sun (1/20000, which is significantly dimmer than the 1/1600 previously quoted). Further, this would not an object the size of the Sun in the sky. It would be a tiny dot that is visibly brighter than the other stars. So, it is not reasonable, in my opinion, to claim that it will disrupt all nightlife on Earth - which is the point I was attempting to make much earlier. The best I can explain it is comparing lighting up a room with a flood light or with a tiny little LED. Even if the tiny little LED is 10 times brighter than the flood light, it won't do a good job of lighting up the room. In our case, the tiny little LED is 20000 times dimmer than the flood light. -- kainaw ™ 17:34, 29 March 2010 (UTC)[reply]

Would it be visible during the day at 1/20,000th? Googlemeister (talk) 18:17, 29 March 2010 (UTC)[reply]

According to apparent magnitude, yes. It is also a given since we know we can see the moon during the day and it is around 450,000 times dimmer than the Sun. So, an object 20,000 times dimmer will be visible. Jupiter would be a tiny little bright dot in the sky. -- kainaw ™ 19:11, 29 March 2010 (UTC)[reply]

The angular diameter is irrelevant, it is brightness (what you are calling apparent luminosity) that matters. How spread out it is will determine whether you can look at Jupiter without blinding yourself, but that's about it. A flood light and an LED of equal brightness will illuminate a room equally well. 1/20,000 times the brightness of the Sun is 22 times brighter than a full moon. Basically, it would like having a full moon much more often than we do now. That may not have too big an impact, but it would have one. --Tango (talk) 19:22, 29 March 2010 (UTC)[reply]

I pity the poor baby having to put up with Kainaw discussing Jupiter sun's when he/she's supposed to be the centre of Kainaw's attention :-P Nil Einne (talk) 06:53, 30 March 2010 (UTC)[reply]

Also, nuking Jupiter will not turn it into a non-star fireball, because to have a fire you need heat (supplied by nuke), fuel (Jupiter has quite a bit of hydrogen), and Oxygen (technically Chlorine might do), which is lacking, so no fire will be maintained. Googlemeister (talk) 13:54, 29 March 2010 (UTC)[reply]

It's also worth pointing out that the fusion matterial in the hydrogen bomb is either a mixture of deuterium and tritium or a lithium deuteride. Hydrogen itself leads to a slow nuclear burn instead of an explosive one. Thanks god, otherwise the sun would have exploded a long time ago :-) Dauto (talk) 14:49, 29 March 2010 (UTC)[reply]

It's part of a conspiracy called the lucifer project, Skeptoid did a good analysis in this episode. Vespine (talk) 00:21, 30 March 2010 (UTC)[reply]

While there are probably good arguments for why it couldn't happen with something as small as Jupiter, I suspect one could imagine a slightly sub-critical mass of hydrogen that might be ignited with a large enough thermonuclear trigger that would cause it to generate enough energy from fusion to keep the reaction going. Sort of like how many substances don't spontaneously burn, but add an initial spark and they will produce enough of their own heat to keep the reaction going. Type Ia supernova are actually a little like this. It is believed that an initial "spark" of carbon fusion generates heat that triggers fusion in surrounding material and the ensuing cascade of fusion rapidly burns through the entire star (causing it to blow apart). It would be interesting to know what would happen if you artificially heated a nearly critical mass of hydrogen. Would it keep producing fusion, or would the extra heat cause it to expand enough that the reaction would snuff itself out. I don't think one can draw an obvious conclusion either way. Dragons flight (talk) 08:56, 31 March 2010 (UTC)[reply]

Read My coment above. Hydrogen burns too slowly to explode. Dauto (talk) 19:51, 2 April 2010 (UTC)[reply]

theobromine's effect on pets like dog or cat

what effect does THEOBROMINE have on dogs or cats?/if harmful what ←−−−×÷√m²m³is the amount that creates harm? —Preceding unsigned comment added by 187.131.196.182 (talk) 13:04, 29 March 2010 (UTC)[reply]

See Chocolate#Toxicity in animals, Theobromine poisoning and Theobromine#Non-human animals. Rmhermen (talk) 13:17, 29 March 2010 (UTC)[reply]

The patterns on windows made by frost

I've looked and all I keep getting is frosted glass which is not what I mean.I wanted to know why ,when windows frost up,the pattern looks like ferns or flowers? —Preceding unsigned comment added by 88.96.226.6 (talk) 13:47, 29 March 2010 (UTC)[reply]

Try also searching for ice crystal. Ariel. (talk) 14:22, 29 March 2010 (UTC)[reply]

The article titled (unsurprisingly) Frost contains a section on window frost. Ice forms a a hexagonal crystals. The frost patterns you see result from a propagation of this hexagonal structure across the surface of the glass, much as in a snowflake. The angles involved in the branching are all generally 120^o (the internal angle of a hexagon), which gives the pattern you are familiar with. --Jayron32 14:24, 29 March 2010 (UTC)[reply]

See ice crystals form (video). Cuddlyable3 (talk) 15:06, 29 March 2010 (UTC)[reply]

Also note that these formations may be fractal in nature. StuRat (talk) 22:18, 29 March 2010 (UTC)[reply]

Oh thank you,I was looking in the wrong places.I've wondered about this all my life...hotclaws 19:04, 31 March 2010 (UTC)[reply]

Sinus Infections

Are sinus infections contagious? --Reticuli88 (talk) 17:14, 29 March 2010 (UTC)[reply]

Almost by definition, anything that can create an infection can infect someone else. IOW, how do you think you got it? Matt Deres (talk) 19:28, 29 March 2010 (UTC)[reply]

I had a bad cold and it resulted into sinusitis. Just wondering if it is possible to infect my coworkers now. --Reticuli88 (talk) 19:50, 29 March 2010 (UTC)[reply]

Other than the information given in sinusitis, we can't really give any more information specific to your case because we don't offer medical advice. You should consult a medical professional if you need to know more. —Akrabbim^talk 20:14, 29 March 2010 (UTC)[reply]

Well, Matt asked for details and I gave it. I just wanted to know if there were any cases where someone infected someone else with sinusitis. --Reticuli88 (talk) 20:30, 29 March 2010 (UTC)[reply]

It was a rhetorical question. You got the infection from somewhere, therefore it is contagious. How contagious, as in how worried should you be if you sneeze, is a more complicated question that depends on a lot of variables that we wouldn't be able to evaluate here. —Akrabbim^talk 20:58, 29 March 2010 (UTC)[reply]

Sinusitis and a cold could be the same organism. Or it could be an opportunistic infection with bacteria after a cold (which is a virus). So you can infect other people, but that doesn't mean it will turn into sinusitis for them - it might simply be a cold. A bacteria sinusitis is unlikely to be infectious. If you are contagious you might transmit the germ, but you can't control in which part of the body it will cause illness in the other person. Ariel. (talk) 21:03, 29 March 2010 (UTC)[reply]

(The following is not offered as medical advice) Hell yes, an infectious disease is infectious, by definition!!!! Edison (talk) 02:03, 30 March 2010 (UTC)[reply]

Torque and angular momentum

The textbook I use derives the relationship τ = dL/dt in the following manner: dL/dt = d/dt(r x p) = dr/dt x p + r x dp/dt = v x mv + r x F = r x F = τ. But doesn't this assume that the position vector for the torque and the particle are the same (ie that the force is being applied directly on the particle)? 173.179.59.66 (talk) 17:30, 29 March 2010 (UTC)[reply]

Whatcha talking about? By definition the coordinate of the point of action of force on a point particle (and the torque as well) is the coordinate of the particle itself. Dauto (talk) 17:53, 29 March 2010 (UTC)[reply]

Well, what if we were looking at a massless lever with a large mass somewhere in the middle. The the force can be applied at the end of the lever, while the particle would have a different r than the force. 173.179.59.66 (talk) 21:41, 29 March 2010 (UTC)[reply]

No, a lever is made of many particles, the external force acts on a massless particle at one end of the massless lever and the action gets transmited to the other end by internal forces between the massless particles used to make the lever. Lots of forces and each one of them act exactly at the coordinate of the particle they are acting on. Dauto (talk) 22:16, 29 March 2010 (UTC)[reply]

Agreed, but then how would you calculate the torque on the massive particle? I would imagine that you would have to use τ = r_F x F, and this would equal d/dt(L_m) = d/dt(r_m x p), with r_F not equal to r_m. So what gives? 173.179.59.66 (talk) 01:12, 30 March 2010 (UTC)[reply]

What? What I said is good for massive particles. r_m = r_F. Dauto (talk) 01:32, 30 March 2010 (UTC)[reply]

I guess I'm missing something lol, maybe an example with numbers would help me. Imagine we have a 2 metre rod of negligible mass with a small rock with a mass of 10 kg attached at the middle (the rod will be rotated about one end). A force of 3 Newtons is applied to the end of the rod. What's the angular acceleration? τ = 6Nm, and τ = d/dt(r_mmv) = d/dt(Iω) = (mr_m^2)α = (100kg.m^2)α, so α = 0.06rads/s. Here, r_F didn't equal r_m. I agree that the internal forces of the rod will ensure that a force with a different magnitude than F will act on the particel at r = r_m, but it remains to be shown that dL/dt still equals r_F x F, which I don't think my book adequately proves. If I'm not making sense, let me know. 173.179.59.66 (talk) 02:38, 30 March 2010 (UTC)[reply]

Everything you said above is correct except for "it remains to be shown that dL/dt still equals r_F x F ". Yes, for that example r_F and r_m are different but that's only possible because you are dealing with an extended object. The book's proof is intended for a set of point particles. But since extended objects are made of point particles, at the end of the day the proof also applies for extended objects. The internal forces will be whatever they have to be in order for dL/dt to be equal to r_F x F. Either that happens, or the massless bar will bend and brake. Dauto (talk) 04:49, 30 March 2010 (UTC)[reply]

How do you know the internal forces will arrange themselves so that dL/dt = r_F x F? I would think it would have to be justified on the grounds of mechanical advantage or something, which my book doesn't seem to address. It seems to me that the proof doesn't apply to extended objects for that reason. 173.179.59.66 (talk) 06:41, 30 March 2010 (UTC)[reply]

Yes, there are some missing steps in the proof but they are almost trivial. Denoting the extended object angular momentum by $L=\sum _{i}^{N}L_{i}$ and its external torque by $\tau ^{ext}=\sum _{i}^{N}\tau _{i}^{ext}$ , we want to prove that ${\frac {d}{dt}}L=\tau ^{ext}$ . But we already know that ${\frac {d}{dt}}L={\frac {d}{dt}}\sum _{i}^{N}L_{i}=\sum _{i}^{N}{\frac {d}{dt}}L_{i}=\sum _{i}^{N}\tau _{i}=\sum _{i}^{N}(\tau _{i}^{ext}+\tau _{i}^{int})=\tau ^{ext}+\sum _{i}^{N}\tau _{i}^{int}$ . So the only step missing in the proof is to show that $\sum _{i}^{N}\tau _{i}^{int}=0$ . This comes about because the internal forces come in action-reaction pairs of oposing forces acting along a common axis which leads to pairs of oposing torques that cancel pairwise identically. Dauto (talk) 15:12, 30 March 2010 (UTC)[reply]

Right, thanks a lot. 173.179.59.66 (talk) 18:59, 30 March 2010 (UTC)[reply]

Actually, no. How do you know the internal torques will cancel? Oposing torques will only cancel pairwise if they're central. 173.179.59.66 (talk) 19:01, 30 March 2010 (UTC)[reply]

I don't know what you mean by a central torque. But I know that the reason that they cancel pairwise is that the pair of forces act along the line connecting the particles. May be that's what you mean by central torque? That's true because this is the only way to preserve the axial symmetry. There is only one important exception. Electric charge particles in motion produce magnetic fields that generate forces on othe moving charges. The magnetic force depends on the velocity of the particles as well as on their positions and can generate non-axial forces. Heck, The magnetic forces may not even opose each other, violating Newton's third law. But all is well since in those situations there will be electromagnetic radiation being produced and if the energy, momentum, and angular momentum of the radiation is also included in the calculation than the usual laws (including

{\frac {d}{dt}}L=\tau

) will still apply. This situation is clearly outside of the intended scope of the proof presented by your book. Dauto (talk) 19:52, 30 March 2010 (UTC)[reply]

Okay thanks, I get it now, thanks. 173.179.59.66 (talk) 21:06, 30 March 2010 (UTC)[reply]

Defrosting and re-freezing

If a product has a caducity of some days in the fridge (7 C) or several months frozen (-15 C), what is the problem of defrosting and re-freezing it? You buy it frozen, let it defrost 1 day and then re-freeze it and defrost it again in another further day. That just makes 2-3 days spoiling, is that a huge deal? --Quest09 (talk) 17:49, 29 March 2010 (UTC)[reply]

It wouldn't spoil but it might taste bad. Dauto (talk) 17:54, 29 March 2010 (UTC)[reply]

If you are defrosting it in the fridge, you should be fine (although you may damage the food, particularly the texture). If you are defrosting it to room temperature, it could be a problem. --Tango (talk) 19:23, 29 March 2010 (UTC)[reply]

When a product is frozen commercially they (often) use flash freezing, i.e. freeze it so fast that there is no time of ice crystals to form. This preserves the texture of the product. If you refreeze it, you will allow ice crystals to form, and burst the cells of the food, so it will be limp and have a poor appearance. Ariel. (talk) 21:15, 29 March 2010 (UTC)[reply]

This is an issue that often comes up with frost-free freezers, which do, at times, melt the food then refreeze it. This absolutely ruins some types of foods, such as bread, by driving all the moisture to one end. Thus, you get part that's hard as a rock, and part that's slime. Meat can suffer from "freezer burn" for a similar reason. Ice cream also forms large ice crystals when it refreezes, making it no longer smooth and creamy. Some foods do quite well when melted and refrozen, like juices and soups, just as long as they are fully thawed and stirred before you consume them. StuRat (talk) 22:11, 29 March 2010 (UTC)[reply]

Grasping the EPT

Can I jump from a helicopter and safely grasp the electric power transmission line to hang for a few seconds, like birds that commonly sit onto it? 213.154.8.70 (talk) 21:33, 29 March 2010 (UTC)[reply]

(ec) There are at least two cables in a power line, one taking the electricity away, the other bringing it back. If a circuit is made, and you are a part of it, you will be fried to a crisp. With low power cables, they are insulated, but with high power cables, the bare wire is usually exposed, because insulation wouldn't do anything anyways. As long as only one cable is touched, you should be fine. If two uninsulated cables, however, are touched simultaneously, a circuit will be formed, and you will be turned into a crispy critter. --The High Fin Sperm Whale 21:40, 29 March 2010 (UTC)[reply]

This doesn't seem to be a medical question at all. --The High Fin Sperm Whale 21:42, 29 March 2010 (UTC)[reply]

Original poster: Note that the above answer is not to be interpreted as some sort of guarantee. That is, don't jump from a helicopter and say Wikipedia said you could. Gabbe (talk) 21:45, 29 March 2010 (UTC)[reply]

I would advice against the jumping part (It would likely be to high up to survive a fall and to close to the ground to be able to open a parachute), But they DO, on a routine basis, use helicopters as a working platform for work on LIVE extra high voltage power grid lines. (i.e. without turning off the power!). So if what you are yearning for is to touch a live wire (and survive), then that experience is definitely within reach (pun intended!)(Of course Provided that you have sufficient money to bribe the working crew :-)
Unfortunately I am not able to remember what kind of protection, if any, one needs to have on the helicopter for it to fly safely in the (I presume) unusual ammount of static electricity.
--Seren-dipper (talk) 22:15, 29 March 2010 (UTC)[reply]

Corrected my spelling.
--Seren-dipper (talk) 03:31, 30 March 2010 (UTC)[reply]

You missed one. It's "advise" (rhymes with eyes) for the verb - "I advise you", "I would advise" - and "advice" (rhymes with ice) for the noun "I provide good advice", "the advice was wrong". Conveniently advice the noun ends in 'ice', it doesn't sound particularly wrong when you drop 'ice' into a sentence in place of 'advice': "I (the ice maker) provide good ice", "the ice was wrong (too yellow)". The verb doesn't have ice at the end. --Polysylabic Pseudonym (talk) 06:03, 1 April 2010 (UTC)[reply]

I might have seen the documentary film about this on National Geographic Channel.
--Seren-dipper (talk) 22:22, 29 March 2010 (UTC)[reply]

To the OP: Yes. Watch this video. And these. Ariel. (talk) 22:33, 29 March 2010 (UTC)[reply]

I assume no liability for the following statement. It would be doubtful you could jump from a helicopter and grab even a non-electric rope without falling to your death. Utilities commonly have "live line work" done by personnel carried by helicopters. The personnel wear a metallic garment, and a probe absorbs the arc as the helicopter assumes the potential of the high voltage line (345 kv or 765 kv, for example). Then the worker installs a line separator or whatever, at the line potential, but isolated from ground potential. So for information only, and not proffered as advice, I assume that if I jumped from a helicopter and somehow grasped a high voltage line, without falling to my death, there would be a surge of electric current when I passed near the wire, causing an electric arc which might cause injury or death, which might cause my hands to grasp the wire, or to be unable to grasp the wire. I have seen photos of high voltage line maintenance in China in which the worker shot a nonconductive line across a transmission wire, then pulled herself up to the conductor, then attached a little wheeled cart which she used to travel to the line separator needing service. (By no means should you attempt this). The trick is to be at the line potential, without a path to ground or a different phase. Once there, corona discharge might be painful or disabling, even without arcing to ground or a different phase. The higher the voltage, the more difficult it would be to hand on. Birds land and perch on distribution lines, but not on transmission lines. Edison (talk) 01:59, 30 March 2010 (UTC)[reply]

There was a programme on the BBC nor long ago - Richard Hammond's Invisible Worlds - which showed an engineer sitting on a power line. I don't think it said whether it was a distribution or transmission line, but they used a special camera to show the electric field around him. It has to be said he was wearing a special metal mesh suit as well! --TammyMoet (talk) 08:19, 30 March 2010 (UTC)[reply]

Thanks, but would the power hit me if I touch myself (is it a circuit)? 213.154.0.144 (talk) 15:33, 30 March 2010 (UTC)[reply]

The question is unclear and difficult to answer. If I touch my toes with my fingers, then yes, there is a circuit through the body, and some current could flow, if something is creating a voltage difference in the body. It is not clear why there would be a difference of voltage. Don't try such an experiment with household or utility electricity. There is the likelihood of electrocution due to touching two conductors or a conductor and ground (the wood of a utility pole counts as ground), and arcing can occur across some distance of air, perhaps several feet, depending on the voltage. At high voltages, there is corona discharge even without a connection to ground or a different phase. This is due to the air being ionized near the sharper points connected to the conductor. The shape of a conductor affects the location and amount of corona discharge. Edison (talk) 16:25, 30 March 2010 (UTC)[reply]

But would there be a corona discharge between hands or legs? Is there a way to prevent it (maybe by placing a non-conductive material under me or knowing the upper limit of discharge's improbability)? 213.154.0.237 (talk) 06:08, 31 March 2010 (UTC)[reply]

Probably if you took a chopper beside a wire, allowed yourself to be charged by contacting the helicopter to the wire so you wouldn't get a static shock, then climbed out and hung on the wire, nothing would happen. The two wires close to each other do not carry opposite charges. --Cheminterest (talk) 20:46, 31 March 2010 (UTC)[reply]

What is the psychological term denoting the strong reaction (weeping, sobbing) when divulging a traumatic experience?

Resolved

When a patient for the first time divulges an until then repressed traumatic event, then many people (patients) experiences a (to them) surprisingly, and sometimes even frighteningly strong reaction of crying, sobbing, shaking etc.
What is the specific psychological term for this (strong) reaction?
--Seren-dipper (talk) 21:36, 29 March 2010 (UTC)[reply]

Catharsis? We have an article Repressed memory. (I haven't read it.) Bus stop (talk) 21:38, 29 March 2010 (UTC)[reply]

Yes! Thank you! Catharsis and Abreaction (=psychotherapeutic catharsis) is what I was looking for. :-)
--Seren-dipper (talk) 03:16, 30 March 2010 (UTC)[reply]