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Phantom LOLbooth posted:The equation for calculating the effects of time dilation is actually really simple: So at 50% of light speed you would go 10 light years in 20 years to 'stationary' observers (I know, they aren't actually stationary), but would it seem like 15 or 7.5 years to you?
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# ? Oct 3, 2010 11:23 |
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# ? Apr 18, 2024 00:21 |
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Can someone physics me up a formula that lets me put in the height of the fall in metres, and it spits out how much g-shock a hard drive experienced on impact when dropped? I assume there would be a variable for the surface it landed on, concrete being far worse at slowing deceleration than a thick carpet. Maybe a spot where I plug in how many millimetres of give there is in my floor-covering of choice? My physics knowledge is piss-poor, but I always thought I'd need to work out how fast it was travelling at the point of impact (v^2 = u^2 + 2at ?) and then work out the deceleration speed. However, I've no idea what timescale that would be over, or what distance (would concrete be almost zero? What's a reasonable "almost" here?) Anyway, I think I'm missing stuff that my vague recollections of physics aren't helping me through. An answer would be greatly appreciated.
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# ? Oct 3, 2010 12:52 |
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Gromit posted:Can someone physics me up a formula that lets me put in the height of the fall in metres, and it spits out how much g-shock a hard drive experienced on impact when dropped? There's a wild variation in how the hard drive is affected based on how it lands. For example, if it lands flat, it will probably take more damage than if it lands edge-on, because the most vulnerable part, the discs themselves, are likely to bend more on a flat landing. The only really half-way reliable way to figure this out is to experiment by dropping hard drives or hard drive simulacra onto force plates and look at the peak force. The simple part that you are trying to remember is that the speed just before impact is equal to v = sqrt(2gh), which comes from the formula you wrote incorrectly, vf2 = vi2 + 2 a x, where the initial speed is zero (dropping from rest), the acceleration is g, and the distance is the height h.
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# ? Oct 3, 2010 17:28 |
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Gromit posted:Can someone physics me up a formula that lets me put in the height of the fall in metres, and it spits out how much g-shock a hard drive experienced on impact when dropped? I'm just in high school, but to find the velocity at impact you would use Vf^2 = Vi^2 (which is probably 0 assuming you didn't help it down)+ 2(9.8 assuming you didn't help it down, again)(height from which it was dropped) I think. I don't know how to find g-shock or the give of a given surface, though. I just wanted to post/help/be corrected.
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# ? Oct 3, 2010 20:27 |
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That also depends on the elasticity of the surface. Remember that according to newton's third law, for every action there's an equal an opposite reaction. That "reaction" depends on the surface. It's why it's much nicer to fall onto a pillow than to concrete. They both "respond" to your force, the difference is that on a softer surface more of your kinetic energy is transferred (think of the way a spring compresses) instead of a blunt collision like concrete. What you'd have to find out is how much of the kinetic energy of a falling hard drive is conserved during impact, and that depends on the surface and it's elasticity.
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# ? Oct 3, 2010 22:01 |
Quick Scoop posted:So at 50% of light speed you would go 10 light years in 20 years to 'stationary' observers (I know, they aren't actually stationary), but would it seem like 15 or 7.5 years to you? The distance traveled is correct (although the total apparent distance traveled at the end of the journey would be the same to both groups of observers, only the amount of time it took to cover that distance would seem different), but the time dilation effect is not a 1:1 ratio between percentage of C and time elapsed. So, traveling at 50% of C does not cause a 50% change between your frame of reference and stationary, or "real," time. The effects of time dilation are heavily weighted toward higher velocities. At .9C, ~2.3 years pass in "real" time for every year on board the ship. At .99C, ~7.1 years pass in real time for every shipboard year, at .999C, ~22.4 years pass in real time for every year observed by the people in motion. At .9999999999999C, approximately 2,235,720 years as measured by stationary observers will pass for every year observed by the people on the ship. At .99999999999999C, the number of real years elapsed increases to 7,073,895 for every ship year. Although you only add .00000000000009C to your speed, the dilation effect more than triples. As you can see, the effect is not linear. So, ten years (ship time) at C*.5: 10 / sqrt(1 - (.5^2/1^2)) This tells us that at 50% the speed of light, when ten years have passed on the ship, approximately 11.55 will have passed in the real frame. Ten years (real time) at C*.5: 10 / (1 / sqrt(1 - (.5^2/1^2))) When ten years have passed in real time, approximately 8.66 years will have passed for the people on the ship. As you can see, it's not a 50% reduction, despite the ship traveling at 50% the speed of light. The "1/sqrt(1 - (v^2/c^2))" part that appears in both equations* is called the Lorentz factor, and it really just gives you a ratio between tmoving and treal. Once you have that number, it's just a matter of plugging in the proper values to find the elapsed time in either frame. * Actually, I dropped it from the first equation to maintain consistency in the division at the beginning of the equation. The first equation, with the complete Lorentz factor calculation in it, would look like this: 10 * (1 / sqrt(1 - (.5^2/1^2))) Alternatively, you could write the second equation to use multiplication instead of division: 10 * sqrt(1 - (.5^2/1^2))
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# ? Oct 3, 2010 22:54 |
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DontMockMySmock posted:There's a wild variation in how the hard drive is affected based on how it lands. For example, if it lands flat, it will probably take more damage than if it lands edge-on, because the most vulnerable part, the discs themselves, are likely to bend more on a flat landing. I don't care to get that accurate at this point. I'm happy to treat the whole thing as a black box. Or rather, a solid lump of aluminium. DontMockMySmock posted:The simple part that you are trying to remember is that the speed just before impact is equal to v = sqrt(2gh), which comes from the formula you wrote incorrectly, vf2 = vi2 + 2 a x, where the initial speed is zero (dropping from rest), the acceleration is g, and the distance is the height h. It t not the normal variable for time, but x is? I seem to remember u being the second velocity variable when I was growing up, but I could be misremembering, easily. I'm still stuck on working out the deceleration, though, as both myself and "the" mention. There should be an easy way to get a ballpark on it, surely?
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# ? Oct 3, 2010 23:32 |
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There's really no way to calculate what you what, Gromit. Let's put it this way: you're looking for the peak acceleration the hard drive experiences. The absolute best you could do, assuming you knew everything there is to know about how the drive and ground collide, is average acceleration. That's not terribly useful for your purposes, because peak acceleration will be much bigger than average acceleration. There's really only two ways to figure out what you need: either drop it on a force plate like DontMockMySmock said, or stick a really expensive accelerometer on a hard drive and drop it on typical surfaces. edit: To figure out average acceleration, you would have to drop hard drives and measure two things: how long the impact lasts and how high the drive bounces. Use the bounce to figure out its velocity after impact, use your initial conditions to find the velocity just before impact. Then (Average Acceleration) = (Final Veloctiy - Initial Velocity)/(Time). Note that measuring how long the impact lasts is rather challenging. filthychimp fucked around with this message at 00:38 on Oct 4, 2010 |
# ? Oct 4, 2010 00:25 |
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Impluse!
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# ? Oct 4, 2010 01:36 |
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Gromit posted:I don't care to get that accurate at this point. I'm happy to treat the whole thing as a black box. Or rather, a solid lump of aluminium. The only easy or realistic method of calculating what you want would be experiment based. If you were to make a model you would need constants taken from experiments. Laptops sometimes have hardware which detects if it's dropped and stops the hd spinning. Maybe you can use that as a basis.
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# ? Oct 4, 2010 01:37 |
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Gromit posted:It t not the normal variable for time, but x is? I seem to remember u being the second velocity variable when I was growing up, but I could be misremembering, easily. x is distance. You neither know nor care how much time it takes to fall. As fithychimp mentions, you can't analytically calculate the peak force on the drive, and as I mention, the stress on the drive is going to vary wildly depending on how it lands and where it lands and with what lateral velocity and a jillion other factors. Only thing to do is to do the experiment. BUT. You obviously want to minimize the damage to the drive. You can still know what you can do to minimize the damage: keep the hard drives as close to the ground as possible, have the ground be as soft as possible, and don't drop your goddamn hard drives.
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# ? Oct 4, 2010 01:38 |
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Okay, it looks like I'm asking for something that can't be done. Fair enough.DontMockMySmock posted:BUT. You obviously want to minimize the damage to the drive. You can still know what you can do to minimize the damage: keep the hard drives as close to the ground as possible, have the ground be as soft as possible, and don't drop your goddamn hard drives. No, I want to know what sort of g-load they are experiencing when they ARE dropped. I don't plan to drop any of my drives, but as the specs for them always talk about what sort of shock they can handle, people are always going to ask what kind of activity puts them in that range. For all I know, a drop from 1 foot onto concrete is 50 g. But it seems I'll never know unless I buy some equipment and do physical tests.
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# ? Oct 4, 2010 08:56 |
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Gromit posted:No, I want to know what sort of g-load they are experiencing when they ARE dropped. I don't plan to drop any of my drives, but as the specs for them always talk about what sort of shock they can handle, people are always going to ask what kind of activity puts them in that range. You're pretty famous here for what you do, so I think I know what you're wanting. HTML Excel Someone here can check my maths, but it seems accurate. Now, I have no idea what the deceleration of a hard drive is on concrete, so I made assumptions there. It should give you a general idea though.
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# ? Oct 7, 2010 00:39 |
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Great, thanks heaps for that. But as you say, knowing what the deceleration values are for standard surfaces is hard to picture.
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# ? Oct 7, 2010 01:16 |
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Gromit posted:Great, thanks heaps for that. But as you say, knowing what the deceleration values are for standard surfaces is hard to picture. Get a superball. Now, if the surface were perfectly elastic, if you dropped the ball from a height of 1 meter, it should bounce to a height of 1 meter. This would mean that the surface it bounces on returns all the energy. However, we know that the surface is not like that from our real world experience. Drop the ball from a height of 1 meter. Measure how high it bounces back up. Use the difference between the expected value of a perfectly elastic surface and how much it actually bounces to find out how much energy is absorbed by the surface. Use this to find out the elasticity of the surface. There's your answer. What we're trying to tell you is that there's really no other way to find out what you want to know aside from experimentation. It's like if you asked "how fast does a planet rotate?" and you were upset when we asked "Well that depends on the size of the planet, what other planets there are, how far it is from the star it orbits, if it has any other satellites, etc." the fucked around with this message at 05:59 on Oct 7, 2010 |
# ? Oct 7, 2010 05:56 |
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This is a rather silly question, but since you've already linked some webcomics in here I don't feel terrible asking it. Why is this impossible?:
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# ? Oct 8, 2010 01:11 |
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helopticor posted:This is a rather silly question, but since you've already linked some webcomics in here I don't feel terrible asking it. First of all, you have great taste in webcomics. Second of all, that's far from a silly question - it's actually pretty complicated. The strong nuclear force has two manifestations: the force that holds protons/neutrons together, and the force that holds quarks together to form protons/neutrons/other hadrons. These are analogous to electric dipole forces and electric monopole forces. The analogue to electric charge is, in this case, "color charge," which I've talked about before in this thread. All hadrons are color-neutral, but there are color "dipoles" that let them interact with other hadrons, analogous to how a water molecule is charge-neutral but its electric dipole still gives it a So bear with me for a moment while we come at it from a completely different angle. You've probably heard the concept that in quantum field theory, forces between particles are accomplished by the exchange of virtual or real force carrying particles. The electromagnetic force is the exchange of virtual or real photons, the weak nuclear force is the exchange of virtual or real W and Z bosons, gravity is the exchange of virtual or real gravitons (probably), and the strong nuclear force is the exchange of virtual or real gluons. Photons are massless and have no electric charge. Gluons are massless, but they have color charge - a gluon is both the carrier and a source of the strong force. In principle, a massless force carrier should give rise to a force that has infinite range and has a multipole expansion where the monopole goes as 1/r2, the dipole goes as 1/r3, the quadrupole goes as 1/r4, etc. (actually it's not so simple since there are three color charges instead of one, but it's analogous). But since gluons are their own source, the force actually gets stronger as you pull things further apart, which fucks up everything and leads to a situation where there are no free particles with nonzero color charge (I talked about this some earlier, too). In any event, it ends up that the equivalent of dipole interactions between nucleons is carried not by gluons, but by pions, spin-1 mesons that are composed of the lightest quark-antiquark pairs. Here's the thing: pions aren't massless. A massless particle can't carry the force forever. That's why the weak nuclear force is so weak and short-range - its force carriers are massive. Consequently, the range of the strong nuclear force ends up being about a femtometer, the size of a nucleus. In fact, you can get a good approximation of that size just by taking the pion mass, about 140 MeV/c2, and converting it into a distance using hbar = c = 1. If you do the same thing with zero mass, you get infinity, which correctly predicts that photons have infinite range. So, in summary, the strong nuclear force can never ever ever be relevant on scales as large as an atom or bigger.
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# ? Oct 8, 2010 02:02 |
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What's the difference between the uncertainty principle and the observer effect? I had a lecturer today say that you can't measure something, say an electron, without hitting it with a photon and changing it's path, hence the uncertainty principle, which I'm guessing is wrong. But is there a link between the two?
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# ? Oct 8, 2010 06:19 |
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4 Day Weekend posted:What's the difference between the uncertainty principle and the observer effect? I had a lecturer today say that you can't measure something, say an electron, without hitting it with a photon and changing it's path, hence the uncertainty principle, which I'm guessing is wrong. But is there a link between the two? Your lecturer is spouting an often-believed misconception. In reality, the uncertainty principle is more fundamental than that, and really has nothing to do with the observer effect. To prove the uncertainty principle, you need only speculate that there exist two observable quantities, you do not need to specify whether they are observed nor how. And you do some complex math (that is to say, math involving complex numbers) and you come up with the uncertainty effect: the product of the squared standard deviations of two observables is greater than or equal to the square of the quantity of the commutator of the operators associated with those observables divided by twice the imaginary unit. A very general and very powerful idea, but one that has nothing to do with the observer effect. The observer effect (the idea that observing a system changes that system) is important in quantum mechanics, though. In the Copenhagen interpretation, an observation collapses the wavefunction. It's responsible for the quantum Zeno effect, which says that observing a system makes it evolve slower (because, in the Copenhagen interpretation, you're collapsing the wavefunction back to its original state rather than letting it evolve).
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# ? Oct 8, 2010 06:50 |
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What's the maximum math I need for astrophysics work? I know that QM uses linear algebra. What about the others?
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# ? Oct 8, 2010 07:22 |
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the posted:What's the maximum math I need for astrophysics work? I know that QM uses linear algebra. What about the others? The answer is "whatever a physics degree requires of you" which is probably something like linear algebra, multivariable calculus, PDEs, complex analysis, and assorted other minor junk. With the exception of general relativity, astro doesn't use any math that isn't used in every field of physics, and if you do general relativity, your general relativity class should teach you the relevant math.
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# ? Oct 8, 2010 08:00 |
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So, I have kind of a dumb question. I was wondering about the apparent and actual positions of stars. I understand that gravity can bend light through Einstein's general theory of relativity, which is why we have apparent and actual positions. What I'm having trouble understanding is why a solar eclipse changes the apparent positions of stars in the sky. Why does the brightness of the sun change the positions of the stars in the sky even though the mass of the sun is the same? The mass of the sun creates the gravitational pull, not the amount of light it produces. Gaaaahh, I'm having a lot of trouble trying to express what I want to ask. This is all so way over my head it's not even funny.
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# ? Oct 8, 2010 21:12 |
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This makes me sad, but I don't think that our species will ever really explore space. Not in the way that popular science fiction portrays, at least. FTL violates causality, which seems to be something that the universe is utterly bent on maintaining. Nonlocal correlations between quantum particles are what I base this hypothesis on. I feel that, like the Heisenberg uncertainty principle, this phenomenon reveals something fundamental about how the universe functions, not any kind of deficit in the equipment we're using to get our data. This means that we're going to always be subject to time dilation at relativistic velocities, and all of the other implications of relativity. I wish wormholes were possible, but that just doesn't seem to be the case.
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# ? Oct 8, 2010 22:36 |
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Incendiary Pig posted:Why does the brightness of the sun change the positions of the stars in the sky even though the mass of the sun is the same? The mass of the sun creates the gravitational pull, not the amount of light it produces. The eclipse and change in brightness isn't what causes the shift in positions due to gravity, it merely allows you to see this positional shift which would otherwise be whited out due to the sun's brilliance. That positional shift is ever-present just like you're intuiting, your problem is in conceptualizing the point of the eclipse in observing this.
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# ? Oct 8, 2010 23:21 |
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Incendiary Pig posted:So, I have kind of a dumb question. I was wondering about the apparent and actual positions of stars. I understand that gravity can bend light through Einstein's general theory of relativity, which is why we have apparent and actual positions. What I'm having trouble understanding is why a solar eclipse changes the apparent positions of stars in the sky. Why does the brightness of the sun change the positions of the stars in the sky even though the mass of the sun is the same? The mass of the sun creates the gravitational pull, not the amount of light it produces. Imagine if you will a group of stars visible at nighttime during the winter. We record the positions of those stars relative to each other during the winter. Then, summer rolls around. Now, the sun is between us and those stars. If we look at those stars, the sun will bend their light and cause us to record different positions. That's all there is to it. The reason eclipses are involved is because usually you can't SEE those stars during the summer, because the daytime sky is too bright to make them out. During a solar eclipse, it becomes easy to see those stars, and thus this measurement can be made. Reality Is Digital posted:This makes me sad, but I don't think that our species will ever really explore space. There's plenty of science fiction out there where relativity is obeyed and you still have humans exploring the stars. It just takes longer. We're still a breakthrough or two away from being able to build a ship capable of going at relativistic velocities, and even at relativistic velocities it will take years to get there, but that doesn't mean we'll never explore the stars. I have entirely different pessimistic reasons to believe that we'll never explore the stars, though - but then, I read a lot of cold-war-era science fiction.
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# ? Oct 8, 2010 23:25 |
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One of the most difficult things to overcome about particle physics for me is understanding that subatomic particles are not little physical balls. How would you describe subatomic particles and their differences to the layman ? Not just electron and protons, how about gluons for example ?
Fozaldo fucked around with this message at 00:44 on Oct 9, 2010 |
# ? Oct 9, 2010 00:42 |
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Fozaldo posted:One of the most difficult things to overcome about particle physics for me is understanding that subatomic particles are not little physical balls. How would you describe subatomic particles and their differences to the layman ? Not just electron and protons, how about gluons for example ? I don't think there's anything wrong with thinking of them as little balls, with the caveat that no matter how tiny you think they are, they're tinier than that. Generally, fundamental particles (that is, particles with no substructure; I'm pretty sure I listed them all earlier in this thread) are thought of as "pointlike," having a position and momentum but no physical size. That's really how you can imagine any of them: electrons, quarks, gluons, photons, etc. Composite particles are made up of whatever they are made up of. For example, a proton does have a size, it's (probably) a little tangled ball of quarks, antiquarks*, and gluons* about a femtometer wide. *You may have heard that a proton is made up of three quarks, two up quarks and a down quark, but that's a simplified picture and the reality is more complicated. Those three quarks are there, and are referred to as "valence" quarks, but there are other virtual quark-antiquark pairs mixed in with the gluons all the time. When people say a proton is made of three quarks, they mean "valence quarks." Usually this distinction doesn't matter, but if you want to visualize a proton accurately, you have to know about it.
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# ? Oct 9, 2010 01:17 |
Smock, what you do you feel about this. No doubt you've seen it before https://www.youtube.com/watch?v=f94j9WIWPQQ.
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# ? Oct 9, 2010 02:22 |
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the posted:Get a superball. Oh, if your suggestion of doing some experiments just requires me to drop a ball onto a surface I'm fine with that. What I don't want to have to do, and how it was made to sound to me, was that I needed to drop a hard drive onto that surface in order to get a rough estimate. Dropping a ball is something anyone can do, and I can build up a list of different surface values quite easily. But to make things clearer for me, how do I use these height difference values in the formula to give me a more accurate result? Please also bear in mind that at no point did I get "upset" at anyone's responses to my queries.
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# ? Oct 9, 2010 02:38 |
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Gromit posted:Oh, if your suggestion of doing some experiments just requires me to drop a ball onto a surface I'm fine with that. What I don't want to have to do, and how it was made to sound to me, was that I needed to drop a hard drive onto that surface in order to get a rough estimate. If only it were that easy. No, you need to do experiments for each pair of colliding objects. Further compounding the issue, the has the wrong take on this problem. What the is describing is how to find the energy dissipated in the collision. I'll bet you right now that hard drives don't bounce very well, and the collision is mostly inelastic. Dropping a hard drive on a carpet will be about as inelastic as dropping a hard drive on concrete, but obviously there are much larger g-forces on the concrete. The only way to get decent acceleration values is to slap an accelerometer on a hard drive and drop it on the floor. If you were to treat the hard drive as a solid lump of metal, you could maybe try and apply contact mechanics, but that's really hard, and almost certainly inaccurate in something as internally complicated as a hard drive.
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# ? Oct 9, 2010 03:55 |
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Yeah, I tried working it out and it got hairy. Oh well
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# ? Oct 9, 2010 05:12 |
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With regards to the two slit experiment that shows how observation changes the outcome of the problem wherein having photon detectors that can tell if a photon has passed through it at the slits changes the probability that a photon will be detected on the other side of the wall that the slits are in: How do those photon detectors work? I once heard how standard photon detectors work when a photon hits a charged plate this causes a cascade of photons to another plate, which goes on until the effect is amplified to an audible click, but these ones absorb the photons, not letting them pass.
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# ? Oct 9, 2010 07:12 |
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Speaking of light: what is the difference between interference and diffraction? I have a general grasp of it on a "physics 101" level, but it ain't as strong as I want it to be.
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# ? Oct 9, 2010 07:14 |
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oiseaux morts 1994 posted:Smock, what you do you feel about this. No doubt you've seen it before What am I supposed to "feel" about it? Dude is pretty depressed about being instrumental in the deaths of thousands of innocents and the possible future apocalypse. I probably would be too. Kenneth Bainbridge, upon witnessing the Trinity nuclear test, commented "now we are all sons of bitches." I like that quote better, even if it is less poetic. HRKR posted:With regards to the two slit experiment that shows how observation changes the outcome of the problem wherein having photon detectors that can tell if a photon has passed through it at the slits changes the probability that a photon will be detected on the other side of the wall that the slits are in: As far as I know, in real life, this experiment isn't actually done with photons, but instead with electrons, which are easier to detect without absorbing. I'm not even sure how to google around for the answer to this question, actually, and I don't feel like spending too much time looking for it now. If you find it, let us know. We can always just imagine that we can detect photons without disturbing them, though. antwizzle posted:Speaking of light: what is the difference between interference and diffraction? I have a general grasp of it on a "physics 101" level, but it ain't as strong as I want it to be. Diffraction is a type of interference. Interference is simply a general term for the phenomena that occur when two waves occupy the same space. Diffraction is a subset of interference phenomena where a pattern of constructive and destructive interference fringes is produced.
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# ? Oct 9, 2010 10:03 |
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DontMockMySmock posted:Imagine if you will a group of stars visible at nighttime during the winter. This is all so simple that I feel more stupid now than when I asked the question. I can't believe I didn't figure this out. I guess it's a good thing I'm gonna be a cop and not a physicist. Drunk people are a lot easier to handle than the mysteries of the universe.
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# ? Oct 9, 2010 18:15 |
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HRKR posted:With regards to the two slit experiment that shows how observation changes the outcome of the problem wherein having photon detectors that can tell if a photon has passed through it at the slits changes the probability that a photon will be detected on the other side of the wall that the slits are in: Any detector necessarily absorbs the photon. There are a couple of ways to detect single photons, but the most common ones work in pretty much the same way. In particular, you probably heard about a photomultiplier tube. A PMT works by letting the photon hit a piece of metal, which causes an electron to be emitted by the photoelectric effect. A second piece of metal, at a different voltage, attracts the electron. The electron accelerates, gains energy, and when it hits the second piece of metal it emits several electrons itself. This is repeated a few times, with all the free electrons liberating even more electrons. When there are enough, they are collected in the last electrode, where they can be detected as an electrical current. If anyone has described a two-slit experiment where the photons were detected first, then they're wrong. At best you can detect a photon, but then immediately emit a very similar photon. There's a semi-famous example of a weird two-slit experiment known as the Afshar experiment, but there the photons are detected after having interfered with each other, not before.
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# ? Oct 9, 2010 23:45 |
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Dilb posted:If only it were that easy. No, you need to do experiments for each pair of colliding objects. I am amazed that there isn't an easy way of working this out for a solid lump of aluminium (which is how I'm willing to treat a hard drive for the accuracy I care for), but will defer to all your knowledge which is far greater than mine in this. Thanks for the info, all. I'll happily use the XLS that oRenj9 supplied in lieu of anything else, knowing full well that it is quite inaccurate. I just wanted something that was close to an order of magnitude or two. "Tens of gees", "hundreds of gees" - that sort of thing would have been fine. Anyway, consider the topic done unless someone has some startling revelations.
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# ? Oct 10, 2010 00:15 |
Gromit posted:Anyway, consider the topic done unless someone has some startling revelations. I just know there's an airplane-and-treadmill joke in here, somewhere, but it is evading me.
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# ? Oct 10, 2010 00:19 |
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Forgive me if this has already been asked, I'm bad at reading through threads. Is there anything that hasn't been discovered that we KNOW is out there, like another quark or something?
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# ? Oct 10, 2010 04:54 |
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# ? Apr 18, 2024 00:21 |
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Yeet posted:Forgive me if this has already been asked, I'm bad at reading through threads. Is there anything that hasn't been discovered that we KNOW is out there, like another quark or something? That depends on what you mean by "discovered." For example, dark matter is theorized to exist, but we just haven't had someone scoop some of it up in the solar system and bring it back to a lab. And it's one of those things that you can't really "see," but you can observe by how other things are affected by it.
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# ? Oct 10, 2010 05:36 |