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| visits | member for | 2 years, 3 months |
| seen | Jan 26 '12 at 20:37 | |
| stats | profile views | 133 |
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Mar 27 |
awarded | Nice Answer |
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Feb 14 |
awarded | Yearling |
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Feb 14 |
awarded | Yearling |
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Feb 1 |
awarded | Nice Answer |
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Sep 14 |
comment |
Why is the energy density of gasoline so high? If you "only care about" energy density per unit mass, then the densest chemical fuel will be hydrogen (assuming you get your oxygen "for free"). The hydrogen reaction with oxygen isn't especially energetic, but you win because the hydrogen is light. It's not as convenient to store as gasoline, but c'est la vie. If you don't insist on chemical fuel, you can do many orders of magnitude better with nuclear power. This is because nuclear energy scales are MeV, while chemical energy scales are eV. And if you're willing to get crazy, the energy density champion is an antimatter-matter reaction. |
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Sep 13 |
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Why is the energy density of gasoline so high? Re: "we only cared about energy density", are you talking about energy density per volume or per mass? |
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Sep 2 |
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How do we know that C14 decay is exponential and not linear? @Charles E. Grant: I certainly agree: if all you wanted to do was demonstrate that radioactive decay is exponential in nature, other isotopes would be a much better choice. But because the lifetime of C-14 is so important (carbon dating, etc.), folks have gone to considerable effort to nail down the lifetime. And since the original questioner seemed fixated on C14 for some reason, I figured I'd put in a reference to data that showed the decay of C14, and tell the story in terms of that isotope. |
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Sep 2 |
answered | How do we know that C14 decay is exponential and not linear? |
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Aug 31 |
comment |
Do stationary states with higher energy necessarily have higher position-momentum uncertainty? Could you clarify what you mean by "consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left."? I think that's a different situation than what Ron meant by "monotonically increasing". Because if the minimum considered is truly only a local minimum, and the potential decreases continuously to a lower V on the left, you don't have discrete energy levels. |
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Aug 29 |
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Air vs Glass refraction coefficient Yes. Practically impossible. |
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Aug 27 |
answered | Air vs Glass refraction coefficient |
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Aug 25 |
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Question about Rayleigh scattering Just as a followup to Ted Bunn's excellent answer, that tiny difference in the energy of the photon (and the energy of the scattering atom) is the basis of laser cooling. And - to note how laser cooling can be practical with such a small energy change - I'd note that his answer (the "recoil energy") is correct for an atom that is initially stationary. For a moving atom, you can get a much bigger change. |
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Aug 25 |
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What color a transparent object reflect? Until someone invents a perfect anti-reflection coating, EVERY object that refracts light will reflect light (except if you're exactly at the Brewster angle). It's not an all-or-nothing phenomena where it's total internal reflection or nothing. |
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Aug 23 |
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Spin up, spin down and superposition Talk us through your jump from how making the measurement affects the He state. If you make the measurement, what are the possible outcomes and their probabilities, and what are the outcomes for He in each of those cases? If you don't make the measurement, what are the outcomes for He and their probabilities. Also, in your last sentence, do you really mean "subluminally"? Transmission of information at subluminal speeds is A-OK in physics, and happens literally all the time. |
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Aug 17 |
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Physics of Focusing a Laser An excellent free resource for answering a variety of simple gaussian-beam-focussing problems, as well as a lot of other optics stuff, is the Melles Griot Optics guide (now the CVI/Melles-Griot optics guide) which is currently available here: cvimellesgriot.com/Company/CoolTips.aspx Reading this should answer your answerable questions. |
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Aug 16 |
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Why does the weighing balance restore when tilted and released I'm not gonna write up an answer, since there seem to already be a couple of correct ones, but I'll do my best to put it more succinctly. If the center of mass is above the rotation point, horizontal is an unstable equilibrium point. This is how you build a teeter-totter (or seesaw). If the center of mass is below the rotation point, horizontal is a stable equilibrium point. That's how you build a mass balance. For small mass imbalances, the deviation from horizontal is proportional to the imbalance. |
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Aug 16 |
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Practical physics: where does this current come from? Vilx: the fact that you can feel the current flowing through you may contradict my guess. It's true that you can't have voltage without the ability to supply SOME current, but the capacitive coupling here would give rise to a voltage source with very high internal impedance. Hence, if you hook it up to a high-impedance voltmeter you'll see a big voltage. Another test would be to hook up an ammeter between the lines, to see how much current it can source. But please make sure the ammeter has fuse protection on it - otherwise if there is a true short somewhere, you'll blow stuff out. |
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Aug 15 |
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Why do Bell tests give perfect correlations? I'd guess your confusion is coming from the term Sz. There are 3 different Sz's to consider here: The Sz of electron 1 (Sz1), the Sz of electron 2 (Sz2), and the Sz of electron 1 + electron 2 (Sztot). In the system you're talking about, the system is prepared in an eigenstate of Sztot=0, ensuring that a measurement of (Sz1 + Sz2) will always be 0. However, the system is not in an eigenstate of Sz1 (thus its outcome is uncertain) nor in an eigenstate of Sz2 (thus its outcome is uncertain). |
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Aug 11 |
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Practical physics: where does this current come from? Your title says current, but your question says you're not measuring a current, you're measuring a voltage. I would guess that - if the "ground" wire is floating as you say - John McVirgo is correct: it's just some capacitive pickup between that wire and one of the others. The reason you can see a voltage is because the impedance of your voltmeter is bigger than (or comparable to) to the impedance of the capacitive coupling. If you were instead to measure current, you should see extremely little, because the impedance of the capacitive coupling is very large. |
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Aug 11 |
answered | Capacitance of two cocentric spheres, contradicting results |
