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How do you read the Periodic table of the Elements? From a grade 6 learning the table in science class
The periodic table is quite a complex representation of the fundamental building blocks of matter. This is a pretty decent one online. I have no idea what level of knowledge grade 6 kids have on this topic so I'll explain the VERY basic. The 'ATOMIC NUMBER' is the primary determining factor of what an element is. The atomic number is the number of protons within the atomic nucleus of the element's atoms. The periodic table starts with Hydrogen which has 1 proton, then helium which has 2, lithium has 3 etc... The reason why they are not always together, i.e. 4 is not near 5, is because the elements are not only listed in order of atomic number but also they are grouped with elements which share common properties, i.e. 2 Helium, 10 neon, 18 Argon, 36 Krypton are all noble gasses and form the right hand column, 3 lithium, 11 sodium, 19 potassium are alkali metals and form the left hand column... There is plently more information packed into the table, like atomic weight, radius, density, boiling point, which you can learn about more later, but the above should be more then enough to get you started. Vespine04:38, 19 October 2006 (UTC)[reply]
I think you can simplify it all you want to "the sixth grade level" but it is an oversimplification. There is a lot more stuff going on than it would be liked to have said, because then it gets too complicated, and most sixth-grade teachers wouldn't know it. Here's my try:
In it's true form it looks like this.The periodic table of elements is an arrangement to put a general order to the elements. It follows a system of listing the elements in order of atomic number (how many protons are in the nucleus of the atom) going from left to right, and up to down. Each horizontal row is called a period, and each vertical column is called a group. There are seven periods, and eighteen groups, because the pink part that is usually under it is not counted (they don't fit in very well). The periodic table also has many trends that it follows. Obviously, inherently, the lower and farther to the right you go in the table, the bigger the atomic number. For instance, electronegativity (how reactive the element is) generally increases to higher or farther right the element is. The atomic radius (how fat the atom is) generally increases the lower and farther to the left the element is on the table.
I don't pretend to be fluent in Simple ("Layman's terms? I don't know any layman's terms."), but someone needs to do something about that article. It confuses the atomic number and period number, and then says that the least metallic elements are found at the bottom-right (where the metalloids, the usual dividers, of course suggest the top-right instead). --Tardis16:21, 20 October 2006 (UTC)[reply]
I'm wondering how its possible to prepare a compound with manganese in a 3 oxidation state. I've read on the internet of 1. either acidifying manganese sulfate in solution with sulfuric acid and then adding permaganate or 2. adding base to a manganese sulfate solution and waiting while air oxides the precipitate formed (or then one can add sulfuric acid to dissolve the precipitate, leaving the ion in solution). I also read that the first method produces a orangey colour solution and the second produces a dark, brown/black solid. Does anyone know how well these methods work? And how stable would my product be, i.e. could the resultant solution or solid be put into a bottle for use later in other reactions? Thanks in Advance. and P.S. Just in Case, no this is not a homework question, nor does it particularly, I think, sound like say a textbook question. 72.56.169.20500:46, 19 October 2006 (UTC)[reply]
i would worry you would get manganese dioxide (a Mn(IV) compound) from permanganate, but it might be controllable i guess. typically Mn(III)-salen complexes are made by bubbling air through a solution of the corresponding Mn(II)-salen complex in ethanol. i imagine this procedure would be broadly applicable (it's also used to prepare some Co(III) complexes in the same way). an example is in J. Org. Chem. (2006) 1449-57; there are many more out there. Xcomradex01:08, 19 October 2006 (UTC)[reply]
I don't understand what a 1G acceleration means in the time dilation article. Here it is: Indeed, a constant 1G acceleration would permit humans to circumnavigate the known universe (with a radius of some 13.7 billion light years) in one human lifetime.
I don't really understand either. Let's say you're travelling at 299 792 457 m/s. Since it's impossible to gain more than 1 m/s of speed, what does it mean to accelerate at 1 g? How can you accelerate at a constant 9.8 m/s when doing so would mean travelling faster than light? --Bowlhover03:20, 19 October 2006 (UTC)[reply]
It sounds to me like a weak statment making a lot of assumptions and ifs, i.e. "if it was somehow possible to accelerate infinitely and if there was enough energy to propel you and if you didn't have to worry about relativistic effects and if you could somehow turn at relatively extreme angles without worrying about dying, then it could be possible". freshofftheufoΓΛĿЌ 04:43, 19 October 2006 (UTC)[reply]
If you kept relativistic effects then you only need to scoff at the idea of enough energy and the idea of turning at relativistic speeds. But if you accelerated fast enough, you could travel the length of the universe's circumference several times over before you died. Ƶ§œš¹[aɪm ˈfɻɛ̃ⁿdˡi]05:45, 19 October 2006 (UTC)[reply]
Bowlhover might think of it in terms of Einstein's thought-experiment of an elevator in an infinitly long shaft. A steady acceleration of 1G is what would make it feel as if the elevator was were not moving and you were on the surface of the Earth. Now, here's the difference between Newtonian and Eisteinian physics. In Classical physics, a steady acceleration results in a straight-line growth of your velocity; Relativistic physics has it that as you approach the speed of light, a steady acceleration increases your velocity by smaller and smaller increments, and so you never get to go faster than light. B00P07:28, 19 October 2006 (UTC)[reply]
Doesn't a "steady" acceleration mean that your speed is always increasing at the same rate? Or does it mean that you're always applying the same force to the accelerating object? --Bowlhover16:16, 19 October 2006 (UTC)[reply]
Of course, a 1G acceleration would never get you off the ground. Assuming you start from the surface of the Earth. But even if you start from space, you'd have to navigate such that you always follow a line that lies where the gravity of surrounding objects cancels out (sort of follow a gravitational ridge), because else you'd eventually get caught by some star. DirkvdM08:57, 19 October 2006 (UTC)[reply]
Of course a 1G acceleration would get me off the ground. If I'm moving relative to the Earth, I can get off-planet. If I'm accelerating, that's even better. (Objects only accelerate downloads at 9.8 m/s if they're in free-fall. A rocket firing its engines is not falling.) --Bowlhover16:16, 19 October 2006 (UTC)[reply]
So long as your kinetic energy exceeds your (negative) gravitational potential energy with respect to some object (like your capturing star), you won't be "caught". (See the virial theorem, or just think about reversibility, for why direction doesn't matter and you won't accidentally end up in orbit.) You'll be deflected, certainly, but you're always being deflected. So you can pick any course which (with a margin of safety for the deflections) doesn't intersect something you'd rather not encounter. --Tardis15:07, 19 October 2006 (UTC)[reply]
Also, Space is pretty empty - I'd say you'd be unlucky to have your path intersect a star - if we assume an average density of 0.1 stars per cubic parsec - typical in our neck of the woods, then perhaps we need to be within 2 solar radii to collide with the star (the deflection is minimal near to c - light passing the Sun is deflected only by about 1 arcsecond). So to get on average 1 star collision, we'd need to have a disc with radius of 2 solar radii to sweep out a volume of roughly 10 cubic parsecs - roughly a path length of 1×1015 parsecs. Now that's only rough - but even assuming that we need to be within 1 AU of a star to collide, we're still talking of the order of a path length of 1×1011 parsecs. And to even come within 100 AU of a star, you'd have to travel on average 13 Mpc - and that's assuming constant star density - outside of galaxies, star densities are extremely low. Richard B20:00, 19 October 2006 (UTC)[reply]
To come back to the original question: I suppose the article means that the accelartion is such that the astronauts in the space ship feel a force pulling them towards the stern of their vessel which is the same as they would experience o Earth due to gravitation, i.e. 9.8 Newton per kilogram. It is better to think of it as N/kg than m/s², because we avoid the question whether the spaceship accelerates with 9.8 m/s². And, from the point of view of an observer on the Earth, it indeed does so only at the beginning, until it gets near light speed. Now, you might think that once the ship has reached 99% of light speed not much more will happen. However, note that even though the ship does not get much faster any longer from the Earth observer's point of view, it still gets more and more massive and time on it gets more and more dilated. The latter is the reason why the astronauts can cross the whole (visible) universe: In the Earth observer's reference frame, the time on the space ship passes so slowly, that within the astronauts' livetime of a few decades, several billion year have passed on Earth: enough time to cross the visible universe if one is travelling with nearly the speed of light. Simon A.22:16, 19 October 2006 (UTC)[reply]
Thanks That was an Enormus help!
And another thing, I don't see how if you're accelerating at 9.8 m/s away from the center of the earth, how you'd get off the ground. I would have thought it's like the bouncing ball paradox, where you drop a rubber ball and it bounces back to 0.6 times the original height, falls again, the 0.6 times that height, and so on, but it never hits 0 because 0.6^infinity is still a number, no matter how infinitely small, but for all pratical purposes, the ball does stop bouncing. So something accelerating at 9.8 m/s might go up a distance of 1/infinity, then the opposing force (gravity) would bring it back down, and the process would repeat, causing (in practicallity) the object to not move. Eh it's been awhile since I took physics. I probably am missing something as simple as -9.8 m/s 9.8 m/s = 0.0 m/s which made my whole paragraph nonsense =P --Wirbelwind02:21, 20 October 2006 (UTC)[reply]
If you're really accelerating at 1g away from the center of the earth (as measured by someone on the earth), you're obviously getting off the ground. If you weren't, your acceleration would be 0 instead (witness all the rocks sitting around, not accelerating anywhere). Now, it's true that if we hang a rocket from a string and light it, only to discover its "acceleration" (that is, its thrust divided by its mass) is 1g, we will find that it simply releases all the tension in the string (since the string isn't having to support its weight) and doesn't actually go anywhere. (Of course, presumably as the rocket's mass decreased its "acceleration" would increase and it would eventually start climbing.) This is the confusion: the "acceleration capacity" of an engine must be added to any other forces on it (the string, gravity, wind, etc.) in order to discover the actual acceleration of the object. Further confusion is the "perceived gravity" within an object: for that you have to consider the actual acceleration of the object and any real gravitational fields in the area. It has nothing whatsoever to do with infinitessimals or geometric series. (But do recall that since, in the normal real number system, "infinity" is not a number, neither is . And things that "never happen" can still be said to be the limiting state of a system!) Finally, since this was originally a relativistic question (even if only an SR one), I'll note that in GR gravity isn't even a force: instead, we say that whether it's the ground or a pathetic rocket engine that's keeping you from falling, you really are being accelerated (relative to not rest but rather the natural state of motion which is "falling freely toward the earth"). So then we simply count up all the accelerations due to real forces acting on you and that's your "perceived gravity", or acceleration in your own frame. (This is the acceleration that can still be 1g when you're already just a snail's pace below c.) Your acceleration takes on a different value, equivalent in concept to the classical notion (where sitting on the ground is 0, not 1g), if measured by an observer outside any gravitational fields. --Tardis16:03, 20 October 2006 (UTC)[reply]
In a previous post, there was mention of circumnnavigation of the Universe. I suppose the "poster" is assuming the Universe is spherical? Could the Universe in fact be a near-perfect sphere? If so, wouldn`t that make it a CLOSED Universe? I tend to think, that with all the evidence about the Universe so far, that it is in fact an OPEN Universe. Anyone care to comment? Thanks Dave 205.188.116.7402:44, 19 October 2006 (UTC)[reply]
The expression "the known universe" usually (and also here) refers to the observable universe. The part we can observe is spherical – which has nothing to do with a possible curvature of space: it is spherical in the same sense in which a billiard ball is spherical. There is a lot about "the known universe" that we don't know. --LambiamTalk11:40, 19 October 2006 (UTC)[reply]
See [1] and [2], as well as our article on "shape of the universe". Basically, although the CMB data does not rule out the possibility of a near-spherical universe, it does not support it either. The evidence also doesn't strongly support any other model of the Universe. --Bowlhover03:38, 19 October 2006 (UTC)[reply]
If the "true" curvature of space has an average magnitude of, say 1 in 1090 m, no observation ever may establish the difference with global absolute flatness. So even if the universe is finite, we may conceivably never see evidence supporting that. The term "closed" is in my opinion less felicitous, because in the terminology of topology also the infinite flat plane and space are closed. --LambiamTalk11:14, 19 October 2006 (UTC)[reply]
It is fairly well accepted that the Universe is flat (BOOMERANG). This would rule out the possibility of the Universe being spherical. There are however other topologies in which the Universe can be an unbounded closed system.
PHYSICS,Mechanical Advantage&related topics pertaining to Pulleys
In a single Pulley system where the Pulley with the Load(L) is hung by a string whose one end is tied to a fixed rigid support and the other end is being pulled upward.
Now,it is observed that when the Load is pulled vertically up by a distance d,the string is pulled up by the Effort(E) to a distance 2d upward.
Kndly help me understand mathematically how the 2d vertcal shift can be proved as against d vertical shift of the Load L.
The amount of work done is Force x Distance. The amount of energy stored as potential energy by lifting the weight must be the same as the work put in (law of conservation of energy). Therefore, if the weight goes up force F x distance D raised, and the work going in which is only F/2 force then it must move twice as far: F x D = F/2 x (D x 2). RJFJR13:44, 19 October 2006 (UTC)[reply]
Seriously, the biggest data processing shops have an interesting drill: When a lightning storm is approaching, they fire up the auxiliary power (diesel generators, gas turbines, or what-have-you) ahead of time, switch the input side of their UPSs from the utility-purchased mains power to the diesels, and in that way, isolate themselves from most of the possible effects of the lightning. Remember, very-high speed telecommunications circuits are mostly optical fibers anyway, so there's no concern there about lightning-induced transients. And by routinely running the auxiliary power system, they (almost) never get surprised by it failing to start in a real emergency.
The way " I " see it, you`re dead either way. If YOU, the bug, does the accelerating, then, that should be clear:your 'hole' isn`t deep enough. On the other hand, if the rivet does the accelerating, you`re 'safe', but only until the rivet slows-down to non-relavistic speeds, and re-lengthens to it`s 'regular' length...ouch! Dave 205.188.116.7414:06, 19 October 2006 (UTC)[reply]
Yes. These paradoxes are quite popular in introductions to relativity theory, but surprisingly, most text book never spell out the lesson the student is supposed to learn from them, namely: The standard mechanics idealization to assume perfectly rigid bodies breaks down in special relativity. The solution goes as follows: The bug should be worried even if it only considers its own frame of reference. On first sight, the bug might do the following faulty reasoning: At time t=12.9 ps the rivet head hits the wall and the rivet is stopped instantly. Hence it does not have to worry, because the rivet end stops dead before reaching the bottom of the pit, leaving a gap of ??? cm. (I leave out the numbers to leave some work for you to do. ;->) However, there are no rigid bodies. The bottom of the rivet does not notice immediatly that the head has stopped. It will fly on until the signal from the head (as mediated by the electrostatic forces that hold the material of the rivet together) reaches the end. How long does this take? Well, the length of the rivet in the bug' s frame of reference is 0.9 cm / 2.29, and the signal, travelling at light speed, takes ??? ps to reach the end of the rivet. Until the, the end flies on without breaking at its speed of 0.9 c. So, is this enough time to reach and squash the poor bug? Finish the math and you'll know. Simon A.14:15, 19 October 2006 (UTC)[reply]
You`re absolutely correct Simon. I tried to correct but received an 'editing conflict' before I could post. Either way though, the momentum of the bug would surely squash itself. I know, not the point. Very interesting stuff just the same. Thanks Dave 205.188.116.7414:24, 19 October 2006 (UTC)[reply]
Simon got toward this point, but it's possible to ignore frames of reference altogether here. It is impossible (due to causality restrictions) to arrest an object from the back (as in, grabbing the rivet head) before its front has reached a distance of from the back end. In this case, that's . The bug isn't the only thing that's getting squished; the hole itself is going to be tripled in length! (Assuming that its substrate, like most materials, is not in the position to stop a slug of metal going 0.9c...) --Tardis21:00, 19 October 2006 (UTC)[reply]
One can probably postulate some sort of "relativistic rigidity" where objects maintain their shapes to the degree that causality allows. As for the speed of sound, it wasn't actually relevant; I didn't say that you could stop an object at the length I specified, just that you couldn't stop it at any shorter length. --Tardis15:47, 20 October 2006 (UTC)[reply]
On the other hand, see http://math.ucr.edu/~jdp/Relativity/Bug_Rivet.html for a nice interactive Flash animation by John de Pillis that shows the bug is necessarily skewered. In fact, the animation shows that after collision with the plate, the rivet must "stretch" beyond its at-rest length due to the time delay for the collision signal to travel from the impact point at the rivet head on one end to the leading point at the other end.
What is the major pollutants that the epa would like to get rid of in the air?
What is the chemical composition of the pollutants?
Is there any movement to rid the air of these pollutants?
The US Environmental Protection Agency, under control by Republicans, has had it's original mission (to protect the environment) subverted. Their new purpose is to protect industry from environmentalists. For example, they are working to prevent California from enforcing tough environmental legislation. Some common pollutants are ozone (O3), carbon monoxide (CO), carbon dioxide (CO2), and particulates (various compositions). Many of these also occur naturally, but in smaller quantities (CO2) or at higher altitudes (O3). StuRat14:34, 19 October 2006 (UTC)[reply]
I generally hate the EPA because it's history makes me angry. e.g. DDT and recycling. It is too much to say everything to say what I think and why. I hate that last sentance. But too much. The major pollutants they hate would have to be CO, SO2, CFCs, and nitrogen oxides. O3 and CO2 I would have to say are not in the actionable effective brunt of many initiatives, and much further down on the list, because their acuteness is much lower. Something tells me I'm going to get a disagreement on this. :) Oh well, all for the better anyway. — X[Mac Davis] (SUPERDESK|Help me improve)05:07, 20 October 2006 (UTC)[reply]
Do you have any idea how much CO2 is released into the atmosphere. The ones you mentioned are certainly a problem, but the amounts of CO2 are easier to take on first. Unfortunately, the worst polluter (United States) didn't sign the Kyoto Protocol... - Mgm|(talk)10:08, 20 October 2006 (UTC)[reply]
The bug/rivet paradox reminded me of this...A photon travels at the speed of light, always. If it had a wristwatch on, the watch would never tick a single second. Since no time elapses, wouldn`t that place said photon EVERYWHERE in the Universe at once? If so, wouldn`t that make the Universe infinitely bright? Since the Universe is NOT infinitely bright, can one conclude that the Universe is OPEN? Am I missing something obvious here? Thanks, Dave 205.188.116.7414:40, 19 October 2006 (UTC)[reply]
Sorry for delay. I got 'blocked'. AOL. If photon is speeding-away from my frame of reference, its 'watch' wouldn`t tic. If watch isn`t 'ticking', wouldn`t the photon go an infinite distance, in ZERO time? Dave 172.130.33.10215:01, 19 October 2006 (UTC)[reply]
Yeah, you are. It's true that in the "photon's rest frame" (a physically dubious concept, but historically interesting and useful for some thought experiments) that no time passes. However, in everyone else's reference frame, it is moving at the speed of light, a large but finite speed, so from your perspective the photon is not everywhere in the universe at once. I'd also like to note that it's not that photons give off light: they are light. In order to be seen, they have to be absorbed by your eye. Photon is now a featured article, so you can learn a lot more there. -- SCZenz15:53, 19 October 2006 (UTC)[reply]
Its true; as the photon travels at the speed of light, in one year it travels precisely one light year from the obeservers perspective. Laïka18:37, 19 October 2006 (UTC)[reply]
Very interesting, since a few years ago, I took a physics course which dealt with the fundamentals of it, where we had to calculate relative time differences. The example was probably something like, two watches were synchronized on earth, and one was place inside a space shuttle while the other was left on earth. The shuttle went to space, traveling at some great speed (nowhere near the speed of light). The shuttle returns to space and the clocks put side by side, and the one from the shuttle is a minute behind the one on earth. This is more background info in case someone doesn't know what the topic meant. But from what I understand, time still elapses for photons, just at a much slower rate, else it would create the paradox that one light photon is everywhere in space at once, but that's obviously not true, else you can never have a dark room. So in that sense, you can't prove with photons and this theory that the universe has no boundries. --Wirbelwind01:55, 20 October 2006 (UTC)[reply]
If the photon were wearing a watch, it would see time passing at the same rate as you or I see our watches (while they are still attached to us). However, if it were possible for an observer to see the photons watch he/she would not see it ticking. What the photon would see is everything that ever happened happen at exactly the same time (and at exactly the same place). It would not perceive time to pass more slowley for itself, only that time for everything else will speed up. The Universe being bright in every direction is, however, and interesting problem. It is known as Olbers Paradox. If the article is there it should explain how it is resolved.
I think your thought experiment breaks down at the very beginning. I believe all the Lorentz style equations require a rest mass or a rest frame. I don't think it makes sense to talk about the photon frame of reference with respect to other frames that are related by invariant mass. Also wouldn't the uncertainty pinciple take over and your error would go to infinity if you tried to narrow down the frame of reference? In other words, at inifintely bright (a point) it's infinitely large in which case it's meaningless.
Hi, all! Two questions: a) WHAT is the nature of TV static? (Is the 'fuzz' a natural behavior for a TV, or are external electromagnetic radiation factors more important?) b) More importantly, WHY is there no article that talks about it in Wikipedia? If there is, then by all means do say so! And if there is not... well, I don't think I have to say more, now do I? Kreachure17:45, 19 October 2006 (UTC)[reply]
That would be a cosmic ray, which isn't exactly the same as a wave. But it's true cosmic rays can cause some noise (of almost-negligible effect, though).
That's peachy! Question a) answered. Now, how come none of the articles you mention even have the word television in them? Upon further examination, there doesn't seem to be any explanation of how the many sources of radiation from the universe affect TV reception, and thus create the famous TV static... are we supposed to conclude the fuzz effect from the random stimulus of all signal frequencies of vibrationon our own? Do any of those articles explain why is it black and white fuzz dancing around? I think not, and if a simple question like this isn't answered in Wikipedia, then (I guess I did need to say it) there has to be an article on that, or at least a mention on any article related to this subject. So, I recommend to anyone who is familiar with this subject to Be bold and create TV static (or whatever name you want to put it)! I'm sure people who are more interested in data compressionstochastic processes or telecommunications in general than me, would be very thankful to have an encyclopedia that talks about this common subject. (Otherwise, why do you answer questions like this for? So that it reflects here only, and not in the improvement of Wikipedia?) Kreachure22:08, 19 October 2006 (UTC)[reply]
Can anybody get their head around what it would be like to exist in a universe with more than one dimension of time. What would it be like ?mmatt19:08, 19 October 2006 (UTC)[reply]
Same as a second dimension of space would be. You would have two dimensions of time to travel through. So it would be more like a plane of time. Maybe you could 'go around' a point in time same way as you could go around a point in space if you have two dimensions? What would it mean to be able to see and even travel in two directions in time as you can do in two dimensions of space? I dont know what the implications are or is it even possible to comprehend - Thats why I asked. Maybe the answer is that its gobbledygook. Is that what you think ? mmatt20:33, 19 October 2006 (UTC)[reply]
Objects would still navigate linearly through the timeplane (how boring, nothing exciting) however, it is possible to fit an infinitely long line of time into a finite are of time plane. Be we do not how objects would navigate the timplane, wether it would random, or controllable, or set. PhilcTECI20:50, 19 October 2006 (UTC)[reply]
I think taking the "dimension" metaphor of time too seriously is misleading here — it is not a "road" or a "channel" or a "plane" that one can just imagine having multiples ones of. Our time and fourth dimension articles have a good description of the two major philosophical views of time, neither of which take the "time is something one walks on" metaphor too seriously. Time as a dimension governs the activities of the spatial dimensions (or is governed by them, depending on how you look at it). "Another dimension" of time might simply mean that under certain conditions these spatial dimensions would exhibit different effects (the increase of entropy outlined in the second law of thermodynamics, for example, might happen faster or slower than it would in the reference frame governed by the other dimension of time, or the second law of thermodynamics might not apply at all, for example). I think getting too wrapped up in spatial metaphors is a sure-fire way to coming up with bad science-fiction approximations, if not gobbledygook. --Fastfission21:40, 19 October 2006 (UTC)[reply]
Inside a black hole, t becomes a spacelike dimension and the exterior space dimensions become timelike, but that has more to do with what kind of paths you can travel - outside a black hole, you can sit somewhere so that your position doesn't change but your time co-ordinate does, but inside the black hole you've got to travel to the center. The other problem is that it's because you're trying to measure with co-ordinates based on space outside the black hole, so you wouldn't be able to observe it yourself - the time co-ordinate you take with you would stay timelike. Confusing Manifestation00:32, 20 October 2006 (UTC)[reply]
The following quote from the article F-theory is of interest: "Max Tegmark has written a paper arguing that life cannot exist in a universe with more than one macroscopic temporal dimension, because differential equations would not be hyperbolic in such a universe, rendering prediction of "future" states impossible". For the same reason, it is not possible to have a notion of "what it is like" to exist in such a universe. --LambiamTalk13:52, 20 October 2006 (UTC)[reply]
Isn't this related to the notion of parallel universes? At every point in time you can take different directions, splitting reality up in many copies that are ever so slightly different, but then diverge ever more (or possibly come back together again?). We might actually live in such a reality, but we can only see one timeline. Just like a flatlander can see something moving in three dimensions instantly diappear. Except that that happens at the same moment they see it appear - so they can't really see it. Does this make sense? DirkvdM08:40, 22 October 2006 (UTC)[reply]
If i compared electricity in a wire to water in a tube, and voltage was the amount of water in the tube, what would current/amperage be? would it be something like pressure, or am I just not understanding this. Ilikefood21:15, 19 October 2006 (UTC)[reply]
Ilikefood...It might be interesting to note that other analogies can be made, such as: a bucket with a hole in it, capacitor; a check-valve, diode/rectifier; a neoprene 'balloon', somewhere along a pipe, might be akin to a coil, etc...Dave 172.130.33.10222:52, 19 October 2006 (UTC)[reply]
Since voltage is also known as potential, it's similar to the water pressure that can force water through a pipe. Current is often compared to the flow of water in a pipe, in many many electrical engineering books, and I've seen a few of them. --Wirbelwind01:38, 20 October 2006 (UTC)[reply]
