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The X elements would be connected to nanowires. The layer of nanowires under the "hydrogen bonds" layer would run left to right in the "X direction", while the layer of nanowires above the "hydrogen bonds" layer would run forwards and aft in the "Y direction". These layers are then stacked one on another up in the "Z direction".
You write a bit by sending an electrical pulse to move the hydrogen atom from one of the X elements to the other one in the same pair. You read the current bit state through either a capacitance difference or by the reaction to a write pulse.
So what is the best choice for the X element? (A single element would be best to keep from introducing a energy difference between the two states that might decay.) Hcobb (talk) 14:20, 2 June 2012 (UTC)[reply]
The best choice is probably going to depend on what you are actually trying to achieve. What is your goal? You describe this as "single atom", but there are two atoms of X, one hydrogen atom and a whole load of nanowires. In what way do you envisage this system being better than the alternatives? Which choice of X is best will be whichever one maximises that particular advantage, I would think. --Tango (talk) 16:18, 2 June 2012 (UTC)[reply]
On a similar note is it possible to do similar storage in ice? Ice consists of O--H ... O linkages between water molecules. Of course, there's a constraint that each O must have two H's, so you can't shift just one proton, but you could twist a ring of them one way or the other. I wonder if you had some very, very, very, very precise method of measuring molecular vibrations in IR/terahertz, with godlike sensitivity, you could use radiation to shift the protons or read where they are only when they appear in certain contexts of surrounding atoms, to the point (in theory) where you could shift or read any one you wanted, and none other. No idea if that's true, but next time the aliens send us some little icy meteorite, just think of all the information that was in there, if only we could have read it. :( Wnt (talk) 17:55, 2 June 2012 (UTC)[reply]
Protons are all indistinguishable from each other, so a ring of protons all moving one place around couldn't be used to store information because you couldn't tell it had happened. --Tango (talk) 19:13, 2 June 2012 (UTC)[reply]
No - I added a picture above. Look at one of the hexagons in hex ice. Notice as you follow around the hexagon, each water has a proton pointing along the dotted line toward the next. Now picture you slide all those protons over, so now all the hydrogen bonds are covalent and all the covalent bonds are hydrogen around the ring. The waters are now pointing in different directions than they were, but each still has the requisite two hydrogens. So you've encoded a bit of data in that hexagon, and you can do the same in any other hexagon that doesn't overlap it. (Trying to figure out how many overlapping bits you can code in hexagons that do overlap makes my head hurt) Wnt (talk) 19:31, 2 June 2012 (UTC)[reply]
Citing the article: "However, in a few special cases (e.g., media designed to amplify the front-most parts of a pulse and then attenuate the back section of the pulse), group velocity can exceed the speed of light in vacuum, while the signal velocity will still be less than or equal to the speed of light in vacuum."
If one would know, how the pulse is "modulated" couldnt it be possible to calculate the information out of the gathered signals? Or is the relation non-linear so it won't be possible exactly? -- 194.95.142.180 (talk) 15:31, 2 June 2012 (UTC)[reply]
The signal velocity is not faster than light in these situations even though it looks like it is. Greg Egan made a nice interactive demonstration (requires Java). You can close a shutter to modulate the signal and watch what happens to the seemingly superluminal wave pulse. -- BenRG (talk) 04:41, 3 June 2012 (UTC)[reply]
I have heard of different kinds of matter,but actually what kind of matter is light?,Do light have mass and what will happen if light get stuck inside a mirror box(inside of box surrounded by mirror).Will the weight of the box increases a little?Can light be changed into any other form.Actually What is Light?Is it a wave or an energy?Please explain all these in detail. — Preceding unsigned comment added by Ganesh Mohan T (talk • contribs) 16:35, 2 June 2012 (UTC)[reply]
I suggest you start off by reading Light and Electromagnetic radiation. It depends on exactly how you define the word "matter", but light isn't usually considered to be a type of matter, since it has zero rest mass. It does have energy (and, therefore, mass, due to mass-energy equivalence), so yes, the box will be very slightly heavier. Just as with other forms of energy, it can be transformed into other forms of energy (that's what a solar panel does). The question "what is light?" is one physicists have battled with for centuries, and there still isn't a complete answer to it. In different circumstances, it has the properties of both particles and waves (see wave-particle duality). If you want a lot of detail on the nature of light, you really need to take a university course on electromagnetism (something I've never done, so I can't give you much more detail than I have here). --Tango (talk) 17:31, 2 June 2012 (UTC)[reply]
Is the photon that hits the mirror the same photon that is emited by the mirror in the reflection. SkyMachine(++) 22:11, 2 June 2012 (UTC)[reply]
Particles of the same type are always identical. If you have two photons, one with momentum k1, the other with momentum k2, then there is only one such state. Interchanging the two momenta of the two photons doesn't yield a different state. Count Iblis (talk) 01:40, 3 June 2012 (UTC)[reply]
No. The photon is absorbed and re-emitted.--Srleffler (talk) 17:32, 4 June 2012 (UTC)[reply]
"Same" (and "identity" in general) is a rather fuzzy concept in quantum mechanics. --Carnildo (talk) 02:15, 5 June 2012 (UTC)[reply]
Indeed. And in case of reflection, which is a coherent process, you can consider doing an interference experiment where a photon arives at a screen directly or by bouncing off a few mirrors. Then the mere fact that there will be an interference pattern means that when a photon arrives at the screen, the way the photon arrived at the screen cannot be determined. Both possibilites exist and these then interfere with each other. So, the photon in case it bounced off the mirrors is the same one in case it took the direct path, otherwise there wouldn't be an interference pattern. Count Iblis (talk) 03:06, 5 June 2012 (UTC)[reply]
Why our Earth revolves around the Sun?Is it because of Gravitational force of Sun(because it is huge)that maintains the distance between sun and Earth(I mean not allowing the earth to go away)Is it because of an explosion inside the sun that created planets and these planets scattered far away from sun and is this the maximum distance the planets can get away from sun(because of gravity).And is it because of the effect of explosion that the planets revolve the sun.
Do all planets begin from the stars?Can we consider Planets as a small,small tiny piece of stars?Why Planets revolve around the sun? — Preceding unsigned comment added by Ganesh Mohan T (talk • contribs) 16:43, 2 June 2012 (UTC)[reply]
Orbits, such of those of planets around stars, are a balance between gravity pulling the objects together and angular momentum keeping them apart - they are constantly falling towards the star, but their movement perpendicular to the star means that the direction towards the star keeps changing, so the direction gravity is pulling them in keeps changing and ends up cancelling itself out over time so the don't actually get any closer. The reason that angular momentum exists is because the planets and the star formed from a rotating cloud of gas and dust. When you hear about planets coming from stars, it isn't the star they are orbiting that they came from. They come from the remains of dead stars that went supernova. You can read about the formation of our solar system (which is a fairly typical system) at Formation and evolution of the Solar System. --Tango (talk) 17:38, 2 June 2012 (UTC)[reply]
I agree with Tango, but would add that gas giants (like Jupiter) can theoretically form from hydrogen and helium only. As such, they don't require the heavier elements formed in supernovae, like the terrestrial planets, such as Earth, do. (Hydrogen was formed directly from the Big Bang, and helium is produced in ordinary stars, like ours.) StuRat (talk) 17:50, 2 June 2012 (UTC)[reply]
Are you sure? What would seed the gas giant? My understanding is that gas giants form by rocky protoplanets accumulating hydrogen and helium. Also, most helium in the universe was formed in the big bang (see Big Bang nucleosynthesis). Helium is produced in stars like the sun, but it tends to stay there (and get turned into carbon and oxygen, which ends up as a white dwarf). A supernova is not just responsible for the creation of heavier elements, it is also responsible for their dispersal. (Smaller stars do expel some of their heavier elements - see Planetary_nebula#Galactic_recyclers - and white dwarfs can sometimes explode in Type 1a supernovae.) --Tango (talk) 19:28, 2 June 2012 (UTC)[reply]
I was thinking about novae dispersing helium from smaller stars than those which go supernova. Hydrogen and helium alone must be able to coalesce into stars, or else we never would have had the first generation of stars, when that's all there was (and subsequently, no later stars either). So, are you saying that there's no middle ground, where there is enough hydrogen and helium to coalesce, but not enough to become a star ? If so, I doubt if those two thresholds correspond exactly, so either there were no small stars or failed stars/gas giants in the first generation, or there were both. StuRat (talk) 18:12, 3 June 2012 (UTC)[reply]
Actually the two bodies both orbit about their barycenter, i.e., their mutual center of mass. Since the sun is so much more massive than the earth the barycenter is very near the center of the sun, so in practical terms we can think of the earth orbiting around the sun. For objects whose masses are similar the barycenter can be a point in space between them. Short Brigade Harvester Boris (talk) 21:27, 2 June 2012 (UTC)[reply]
I know that, but it wasn't important so I didn't complicate my response with it. --Tango (talk) 02:28, 3 June 2012 (UTC)[reply]
Where can I get those safety glasses so I can watch the transit of Venus? What store sells them? A Quest For Knowledge (talk) 18:35, 2 June 2012 (UTC)[reply]
I would suggest watching it by projection with binoculars, rather than through glasses, that way you can get some magnification (never look directly at the sun through binoculars, even with safety glasses - they aren't designed for it and you can easily burn a hole in them, and then in your retina). You just need to point the binoculars at the sun (a tripod, or something else to hold them steady, helps) and then hold a piece of white paper behind them. An image of the sun will form on the paper and you'll be able to see Venus as a black dot moving across it. I've watched a transit of Venus that way before - I didn't have a tripod, so the image was a bit shaky, but I could definitely see Venus. Note, the image will be upside-down. --Tango (talk) 19:34, 2 June 2012 (UTC)[reply]
I don't own binoculars, and I imagine that they are way more expensive than the glasses. A Quest For Knowledge (talk) 19:59, 2 June 2012 (UTC)[reply]
You can also do it with just a pinhole instead of binoculars - get a piece of thin card, make a small hole in it with a pin, and hold that up to the sun with the piece of white paper behind it. This will also make an image of the sun on the paper (you may think at first that it is just an image of the hole, but it really is an image of the sun). That's even cheaper than eclipse glasses! (I'm not going to recommend somewhere to buy eclipse glasses because a) I don't know where you are and b) it's very important to make sure you get good ones that are actually safe to use and I don't want to take responsibility for that - projecting the sun is much safer.) --Tango (talk) 20:20, 2 June 2012 (UTC)[reply]
Here, watch this video. Apparently, some company is mass producing these glasses and distributing them all over the world. I'm just trying to figure out where I can get one. A Quest For Knowledge (talk) 20:40, 2 June 2012 (UTC)[reply]
This is a genuine safety issue. Looking at the sun through any type of glasses is regarded as dangerous. Don't do it. Use the projection method described above. HiLo48 (talk) 23:43, 2 June 2012 (UTC)[reply]
If you get good glasses and they are new and you carefully inspect them for damage before you use them, then they are safe. You do need to be very careful about where you get them from (which is why I'm not going to make any recommendations) and they do need to be in good condition. If you project it with a pinhole, you can just grab any old bit of card, pin and piece of paper, and you'll be absolutely fine. (The only risk with the binocular projection method is that you might ignore the advice not to look through them - as long as you don't do that, it's perfectly safe as well.) --Tango (talk) 02:32, 3 June 2012 (UTC)[reply]
There is likely to be a flood of dubious "View Venus" glasses on the Internet, and using them would be a complete gamble: don't buy safety equipment from people who have no motivation to do anything more than maximize their profit. Use the "reflected pinhole" technique shown here. Johnuniq (talk) 01:26, 3 June 2012 (UTC)[reply]
I actually used the MacGyver version of that technique for keeping track of an annular eclipse several years back. The building where I was working at the time had a large skylight over an atrium area; I reflected sunlight off the glass crystal (outer face) of my watch and onto a shadowed part of a nearby wall. No staring at the sun and no going outside required to check how close the event was to totality.... TenOfAllTrades(talk) 05:01, 3 June 2012 (UTC)[reply]
Welder's goggles or welder's glass #14 will protect your eyes when viewing the Sun - any smaller number is not good enough. I think #14 is the darkest one they make, so if you don't know the number, don't trust it. Bubba73You talkin' to me? 05:28, 3 June 2012 (UTC)[reply]
Once when I didn't have #14 glass I stacked two #12, which seems to be about the same. Bubba73You talkin' to me? 21:26, 3 June 2012 (UTC)[reply]
I would just point out that Venus is tiny compared to the sun so even if you did get hold of adequate dark glasses you are not going to see Venus at all without some magnification. See Transit of Venus.--Shantavira|feed me 07:51, 3 June 2012 (UTC)[reply]
What you're saying may be correct, but I don't think the article says what you think it does and it hasn't for a few days at least [1]Nil Einne (talk) 09:23, 3 June 2012 (UTC)[reply]
It's true that the article is somewhat misleading as it suggests that looking at the Sun through safety glasses would allow the viewer to see something useful. I suspect that you would not see the transit because the black dot of Venus will have a diameter about 30 times smaller than the diameter of the Sun, and that will be very small. Using the reflected pinhole I mentioned above will magnify the image, but even then it's small (see the link above). Probably the safety glasses are in the article because it may be some generic information borrowed from an article on viewing a solar eclipse, when something dramatic is happening which would be visible through safety glasses. Johnuniq (talk) 09:44, 3 June 2012 (UTC)[reply]
This has been extensively discussed by professional and amateur astronomers at the Transit of Venus group on facebook at http://www.facebook.com/groups/108400462513165/ If you go back through all the postings they tell you where to get the eclipse glasses and tell you about various ways to view the event and all the problems with those methods. For instance, they don't recommend you use a welder's mask as many modern ones aren't safe for viewing the sun. You won't see very much with eclipse glasses, many places have sold out, and I would have thought it's now too late to obtain them by mail anyway. Richerman(talk) 10:16, 3 June 2012 (UTC)[reply]
This well-referenced page (which was sent to me by a health physicist) includes six different ways you can safely view the Transit of Venus. You may not be able to find the eclipse glasses or welding glasses in time, and the transit is near the visual limit of human eyesight, so you probably should consider one of the other options. --Mr.98 (talk) 13:25, 3 June 2012 (UTC)[reply]
I'm trying to find someone at work who owns binoculars. If that falls through, I'll try the paper/pin hole method. Thanks! A Quest For Knowledge (talk) 21:09, 4 June 2012 (UTC)[reply]
Personally I'm to chicken to try viewing the sun directly (also my vision without corrective glasses is very poor so I'd need something compatible with corrective glasses) and in any case I waited too long to try and look for it here (via projection) in Auckland and the clouds had taken over. But for the benefit of future events (not transit of Venus events obviously), I noticed while looking for live feeds astronomical societies and observatories here in NZ sell solar viewing glasses. These seem a more reliable source the some random internet site. Obviously you want to make sure the society or observatory is reputable and not some random fly by night operation just out to make profit. Nil Einne (talk) 03:01, 6 June 2012 (UTC)[reply]
If cats are raised in litters in their infancy for, usually, at least 8 weeks, where do they develop their solitary instincts from later on in life. I could understand it if the majority of them were only children, so to speak, but if they develop around each other and play together, shouldn't they, theoretically, be pack animals? Thanks for answering my question --Thanks, Hadseys(talk) 21:41, 2 June 2012 (UTC)[reply]
There's really no single answer. It's most likely to vary from individual to individual, based on how they developed after being in the litter. Cats separated from others during this time could develop an independent/antisocial streak, but I don't know of any specific research into this. — The Hand That Feeds You:Bite 23:11, 2 June 2012 (UTC)[reply]
Sociability is more nature than nurture -- it is affected by nurture, but only to a limited degree. Looie496 (talk) 23:34, 2 June 2012 (UTC)[reply]
Humans are raised in litters of typically two or three, but varying from one to over a dozen. As adults they tend to live in pairs or as singles. It's just what nature decided. HiLo48 (talk) 23:41, 2 June 2012 (UTC)[reply]
Cats are very social animals throughout their lives. They wouldn't make good pets otherwise. The reason they aren't considered pack animals is because they don't hunt together. They live in groups, they just hunt alone. --Tango (talk) 02:37, 3 June 2012 (UTC)[reply]
But that is only partially true. Some cats, for example tigers or mountain lions hunt alone, but there is plenty of filmed evidence to show that lions hunt as a pack with a clear drive and ambush strategy. In some respects the original question is making incorrect assumptions. Richard Avery (talk) 07:32, 3 June 2012 (UTC)[reply]
Among non-domesticated cats, African lions (Pantera leo) are the only truly social species. All others spend most of their adult lives as singles, only briefly joining up in pairs to mate, and females raise their ofspring without the help of others. Sub-adult cheetahs may stay together as a sibling group after leaving their mother but by full maturity they would be solitary. Roger (talk) 09:44, 3 June 2012 (UTC)[reply]
"Cat", without qualification, means "domestic cat". That's what I was talking about. There are other members of the cat family that have different social structures, but that isn't relevant to the question at hand. --Tango (talk) 13:43, 3 June 2012 (UTC)[reply]
Domestic feral cats often form colonies with well-defined hierarchies. There are a few studies of them, I'll have a go at finding some later. Personal OR, in a multi-cat household they will live together in a degree of harmony, with a top cat (usually female) and other subservient cats. Entire males do not fit into this hierarchy and will often wander off or be excluded. This mirrors the situation with prides of lions. Domestic cats need to be "socialised" to fit into human society from a very early age, say 5 weeks: if this does not happen they will often be more individualistic and make less good pets. --TammyMoet (talk) 08:04, 3 June 2012 (UTC) I found this longitudinal study in the UK. --TammyMoet (talk) 08:11, 3 June 2012 (UTC)[reply]
Interestingly Felis silvestris, the species from which domestic cats are derived are definitely solitary. Roger (talk) 09:50, 3 June 2012 (UTC)[reply]
It's possible that the domestic cat has more than one ancestor: I'm thinking about the Egyptian Mau cat and other breeds derived from Africa. --TammyMoet (talk) 11:56, 3 June 2012 (UTC)[reply]
The Egyptian Mau is simply a breed of domestic cat - not a different species. The domestic cat is definitely a derivative of F. silvestris and is in fact usually designated as a subspecies F. silvestris catus. Roger (talk) 17:01, 3 June 2012 (UTC)[reply]
I doubt that "more than one ancestor" was meant to suggest that domestic cats might be a hybrid of different species. Sean.hoyland - talk 17:31, 3 June 2012 (UTC)[reply]
From the article: "The exact origin of the Egyptian Mau is not recorded and therefore cannot be known for certain.[5] The Egyptian Mau is often said to be descended from African wild cats". So there is dispute at least about your statement, Roger. --TammyMoet (talk) 17:33, 3 June 2012 (UTC) And again a bit further down: "Egyptian Maus are typically slender and muscular and they are thought to be one of the progenitor breeds of the modern domestic cat." All the statements I've quoted are referenced. --TammyMoet (talk) 17:36, 3 June 2012 (UTC)[reply]
What it does not say is that "African wild cats" as well as "wild cats" from as far apart as Britain and China are all subspecies of Felis silvestris<- Read this article, it is written with much better scientific rigor than the domestic cat article.Roger (talk) 17:45, 3 June 2012 (UTC)[reply]
I once read – somewhere or other – the hypothesis that Fluffie is descended from spontaneous crosses of F.sylvestris and F.chaus kept together in Egyptian temples. —Tamfang (talk) 17:42, 3 June 2012 (UTC)[reply]
That was probably written before modern genomic analysis. —Tamfang (talk) 01:58, 5 June 2012 (UTC)[reply]
I would call house cats semi-social. That is, they can live in groups or alone, depending on how they are raised and the temperment of the individual. Lions aren't that much different. They can live in prides, but don't really get along very well. Fully social animals, like herd animals, seem to get along better (and also do worse by themselves). StuRat (talk) 18:00, 3 June 2012 (UTC)[reply]
"All others spend most of their adult lives as singles, only briefly joining up in pairs to mate, and females raise their ofspring without the help of others." I am not sure this is always true. I saw an outdoors cat bring her three newborn kittens into a box where another (unrelated) cat was already raising her three kittens. I was afraid that they would be aggressive to each other('s kittens) and tried to seperate them but the cat #1 kept bringing her kittens back until I let them be. Each cat took care of the other cat's kittens along with her own (breastfeeding, cleaning, etc). They even groomed each other. I cannot explain it to this day. Surtsicna (talk) 23:21, 3 June 2012 (UTC)[reply]
Then again, I also saw (and photographed) a cat breastfeed an adult tomcat larger than her along with her kittens - so much about sociability. Perhaps cats around here are simply kinky. Surtsicna (talk) 23:24, 3 June 2012 (UTC)[reply]
One possibility here is that the neoteny we have bred into them (making them all act like kittens rather than adults) backfired in this case. StuRat (talk) 23:39, 3 June 2012 (UTC)[reply]
Roger, who you quoted, was talking about non-domesticated cats (tigers, leopards, etc.). Domestic cats, whether living with humans or feral, are very social and it isn't uncommon for them to look after each other's kittens. --Tango (talk) 00:03, 4 June 2012 (UTC)[reply]
I wouldn't say "very social", as I'd reserve that for animals which can only exist as a group, like bees, ants, meerkats and most people (living on our own, without the tools, clothes, food, housing, and knowledge provided by other people, is almost impossible for us). StuRat (talk) 00:09, 4 June 2012 (UTC)[reply]
In this article, it says that he will reach 700MPH and nearly break the sound barrier (exceed the speed of sound). But a human body has a terminal velocity of about 125MPH in the lower atmosphere, so in the region where he is going 700MPH the atmosphere would be much thinner, making the speed of sound considerably higher than near sea level, right? Bubba73You talkin' to me? 22:06, 2 June 2012 (UTC)[reply]
For an ideal gas (to which Earth's atmosphere below about 80 km is a very close approximation), sound speed depends on temperature and not on density or pressure. The sound speed will be different way up there because the air is warmer or colder, not because it's thinner. Short Brigade Harvester Boris (talk) 00:59, 3 June 2012 (UTC)[reply]
In Speed of sound, the first equation under "Basic equation formula", the speed of sound depends on the square root of the density. Bubba73You talkin' to me? 02:02, 3 June 2012 (UTC)[reply]
That's where the "ideal gas" bit comes in. Notice the quantity under the square root sign is the ratio of pressure to density. For an ideal gas, by definition, p = (ρ Rs T) where p = pressure, ρ = density, Rs = the specific gas constant for the gas in question (Rs=R/M), and T is temperature. So in the ratio (p/ρ) under the square root, substitute for p from the ideal gas law to get (ρ R T / ρ). The ρ (i.e., density) in the numerator and denominator cancel, leaving only temperature and the gas constant. Short Brigade Harvester Boris (talk) 03:01, 3 June 2012 (UTC)[reply]
Once again, the "ideal gas" qualifier comes in. Read the bulk modulus article that you linked. Under "thermodynamic relation" you will see that the bulk modulus for an ideal gas is just the pressure times Cp/Cv (for which the article uses the symbol γ). Short Brigade Harvester Boris (talk) 15:15, 3 June 2012 (UTC)[reply]
I'm not a professional in that area but I used to have a booklet that gave me those equations when I needed them until I've lost it in the wave of moving around. Does someone maybe know a site where such formulas are provided plain and simple, just for a layperson like me? So we maybe even have an article that would give me that info in simple terms? Greatly appreciated, TMCk (talk) 22:53, 2 June 2012 (UTC)[reply]
If you know the voltage, P = VI (power = voltage * current); in units, 1W = 1V x 1A. So for domestic power in the UK (V = 240V), a 5A fuse will power up to 5x240 = 1200W. Don't try plugging a 2kW kettle into a 5A fuse. --ColinFine (talk) 11:06, 3 June 2012 (UTC)[reply]
By "convert" I meant "calculate" but anyhow, the answer I got is exactly what I was looking for. Thanks a bunch, TMCk (talk) 14:06, 3 June 2012 (UTC)[reply]
I recently measured the amps.volts, voltamps and watts from a refrigerator powered by 120 volt alternating current, and the power factor was 0.62, so that the actual watts was 0.62 times the volts times the amperes. For different types of AC appliances,the conversion from voltamps to watts might vary a bit. It is certainly simpler with direct current appliances: then the watts is simply the volts times the amperes. AC appliances also consume "vars" or reactive power. Edison (talk) 04:06, 4 June 2012 (UTC)[reply]
The formula for power is this: Power = Voltage × Current. Power is measured in watts, voltage in volts, current in amperes (or amps), and resistance in ohms. Ohm's Law can help you with the power equation: Resistance = Voltage ÷ Current. There are two converses of Ohm's Law that may help you too: Voltage = Current × Resistance and Current = Voltage ÷ Resistance. Watts are units of power and amps are units of current, so the two cannot be converted, but if you know the voltage of the circuit, multiply the current (in amps) by the voltage (in volts) to get the power (in watts). If, however, you know the power but not the current, divide watts by volts to get amps. If you need any more help on an electricity-related topic, contact me on my talk page and I'll help you as much as I can (I'm offering more help because I just took a science class on electricity and magnetism, so everything is fresh in my mind). ChromaNebula(talk) 15:57, 5 June 2012 (UTC)[reply]