# Tag Info

16

The water gets colder the longer you run it (in the UK at least) because the water mains pipes buried in the ground are colder than the ones in your house, so sadly this isn't evidence for any fundamental physical effect. In principle any fluid flowing in a pipe gets hotter because energy is dissipated in viscous flow. You could in principle calculate the ...

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If they didn't release energy, they wouldn't happen. The alternative, nuclear reactions that require energy, clearly need said amount of energy, which has to come from somewhere, e.g. kinetic energy involved in the collision of two nuclei (even ones that release energy usually have a "barrier" and some amount of initial kinetic energy is needed to overcome ...

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Concerning the factor $\frac{1}{2}$: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy $$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$ Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.$^1$ However we should also take the kinetic energy $\langle ... 8 I think the key conceptual hurdle is that the vacuum state is not nothing. Quantum field theory describes matter as excitations in quantum fields. These quantum fields are very strange things, and I don't know of any easy way to explain to a non-physicist what a quantum field is. The key thing is that the quantum fields fill all of spacetime. So a vacuum is ... 6 Your book may be treating things a little backwards from the way they are usually done. The usual way is to define the momentum four-vector as the combination$(E/c, \vec{p})$, where$E$is already known to be the total energy (the thing that reduces to$mc^2 + \frac{1}{2}mv^2$for$v\ll c$) and then go on to show that it satisfies the properties expected of ... 4 When a cup of coffee is hot, the air molecules directly above it get hot as well. After some time, they reach equilibrium and no heat transfer (or maybe very little transfer) occurs. By blowing, you disturb that equilibrium and replace the hot air molecules directly above the cup with colder air and therefore create once again a steeper temperature gradient. ... 4 The equilibrium vapor pressure of water vapor over ice is well known and easy to google for (http://www.its.caltech.edu/~atomic/snowcrystals/ice/ice.htm is one possible link). It is slightly lower than the equilibrium vapor pressure of water vapor over liquid. Ice does not evaporate - it sublimates under those conditions. The equilibrium vapor pressures ... 4 The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ... 3 To greatly simplify John's answer: the temperature gained by friction and velocity is insignificant compared to the current temperature of the liquid. I expect frictional heating would kick in at higher pressures, but once the water had enough kinetic energy to heat up noticeably on impact with your hands, it would also have enough energy to strip the ... 3 Very interesting question! In chemistry you spend lots of time discussing exothermic and endothermic reactions: when you put your reagents together, sometimes the reaction heats things up, and sometimes the reaction cools things down. Nuclear reactions are very different, in that essentially all spontaneous reactions studied in laboratories are exothermic. ... 3 Can someone explain the atomic process, if it even exists, of how this would work to convert energy in to matter, and what form of energy was initially present, and what is required to cause this change? It is not an atomic process, it is an elementary particle process, atoms are made up of elementary particles in a non trivial way. At the level of ... 3 Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ... 2 Shouldn't the energy gain be greater than this formula describes since the energy from the electric field is applied for so long? The electron gains energy and accelerates until it encounters a collision. This is a statistical process and there's a distribution for the energy loss for many electrons. Then, it can accelerate again from that point on, ... 2 PhotonicBoom is correct in saying that the airflow created by blowing across the top of the coffee will replace the coffee-heated air with cooler air that will absorb more heat from the coffee. It also allows more of the coffee to evaporate (which might seem like a bad thing, but evaporation is simply the hottest molecules becoming gaseous and leaving, so it ... 2 The energy operator is obtained via the so-called correspondence principle. This means that one considers the classical expression for the total energy $$\frac{p^2}{2m}+V(x)$$ and replaces the momentum and position variables (numbers in classical mechanics) by the momentum and position operators.$p^2/2m$is the kinetic energy (it's just another way of ... 2 The "Energy operator" in a quantum theory obtained by canonical quantization is the Hamiltonian$H = \frac{p^2}{2m} + V(x)$(with$V(x)$some potential given by the concrete physical situation) of the classical theory promoted to an operator on the space of states. Since the core of the quantization procedure is promoting the classical phase space ... 2 It depends what you mean by diffusion. If you're considering a system at equilibrium, e.g. a sodium chloride solution, and you're trying to extract energy from the diffusional motion of the sodium and chlorine atoms then you can't extract any energy. Your friends idea of a tiny turbine is effectively the same as the Maxwell's demon idea for extracting ... 2 Look at the phase diagram of water: You can see there is a region where ice and water vapor can coexist. Ice turning to vapor without melting is called sublimation. The energy needed for a water molecule to break the bonds holding the ice together is quite large - so the probability that a thermal fluctuation is big enough for a molecule to escape is ... 2 Solid ice does sublimate (go directly from a solid to a gas) over time, but the process is fairly slow. You can see this simply by leaving ice in your freezer for several days. The ice will gradually sublimate away. Wind and dry air both speed up this process, but it is still nowhere near as fast as evaporation under most circumstances. In Antarctica, ... 2 In short, they are hard to separate because even though the molecules are very different, they have properties that attract them to each other. Water is a polar molecule. the oxygen molecule oxidizes the two hydrogen molecules, creating a positive charge on the hydrogen side, and a negative charge on the oxygen side. Meanwhile, salt is composed of Sodium, ... 2$"$If we put a crystal of salt in the water, what will happen? Salt is a solid, a crystal, an organized arrangement of "salt atoms." Strictly speaking, the crystal is not made of atoms, but of what we call ions. An ion is an atom which either has a few extra electrons or has lost a few electrons. In a salt crystal we find chlorine ions (chlorine atoms ... 2 I don't understand why the time component of the 4-vector [$m~\gamma(|\vec v|)~(c, \vec v)$] is being denoted as$E/c$. So the underlying question is two-fold: Why is "energy" considered the time component of some 4-vector at all?, and Why this specific time component expression, among time components of all different 4-vectors imaginable? (Where ... 2 I can think of two reasons why we need a lower bound, one statistical, one inuititive. First, the intuitive: The annihiliation/creation operators represent adding/removing particles (or excitations, or whatever). The vacuum state as lowest energy state represents obviously the empty state from which no further particles can be removed. It is also clear (in ... 2 You can write kinetic energy as $$K=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{(mv)^2}{2m}.$$ You can take the next few steps of rewriting this in terms of momentum$p=mv$. 2 If efficiency is the issue, then definitely parallel TECs (or use a single unit rated for twice the power, same thing). The only reason for stacking TECs is to get a lower temperature. However that comes at great expense to efficiency and overall power consumption. Another point is that paralleling TECs is actually more efficient overall. The reason is ... 2 You'll want a much bigger heatsink!! (and maybe just one TEC) If it's being cooled only by convection then maybe a heat sink area* that is 10 times that of the TEC. (maybe bigger) The classic mistake with a TEC is to make the heat sink too small. With too small a heatsink the hot side of the TEC gets hotter, more thermal leakage through the TEC, it has ... 2 Work. Potential energy exists because of some force that exists, and moving an object relative to that force causes work to be done. And by the work-energy theorem, the work done on an object is equal to the change in kinetic energy of that object. 1 It hinges on the sentences you did not quote, and on whether one is talking of an adiabatic system or not. He assumes two systems adiabatically isolated then their internal energy can be added. Then when brought in contact the adiabatic isolation no longer holds, potential energies and radiations exchanged are not additive in a simple manner ... 1 The velocity of light is 299792458 meters per second. That is the velocity of displacement of a peak to peak as the wave passes. the wave in slow motion It is not clear what you want to do with your experiment but it is not possible to carry it out the way you envisage. There is no technology that could spin anything that fast. Now in your title you ... 1 When a force is applied over a certain distance, that force does mechanical work,$W$. If the force is constant$F$and the object it is exerted on is moved by a distance$\Delta x$, then$W=F\Delta x\$. If the force is not constant but a function of the position, this turns into an integral: $$W = \int_{x_1}^{x_2}F(x)\,\mathrm d x.$$ If you don't know ...

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