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New answers tagged potential-energy

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I believe the term you're looking for is "Magnetostatic Energy". Magnetostatics is the field that studies static (constant) magnetic fields, much like electrostatics. For a uniform material the magnetostatic stored potential energy is: $$E_{\mathrm{ms}} = \frac{1}{2}\mu_0 \int_V \mathbf{M} \cdot \mathbf{H}_{\mathrm{ms}} d^3 r$$ You can find a full ...

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I think the correct answer is it store "lower entropy state", not the energy

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Both definitions are fine as long as you're careful with signs. Here's a derivation of your gravitational potential energy using the idea of the negative work done "by the field." (Note the initial negative sign.) $$U(r)=-W_\text{by field}=-\int\vec{F}_\text{field}\cdot d\vec{s}=-\int_{r=\infty}^{r=x}\underbrace{\frac{-GMm\,\hat{r}}{r^2}}_\text{Toward ... 2 There are some very interesting subtleties here. Let's analyze the situation very carefully. Let's choose our system to consist of the block, spring, and Earth. By choosing the Earth and block to be in our system, we will have a change in gravitational potential energy. In the beginning, the (massless) spring hangs vertically with a block of mass m ... 0 You don't do any negative work. All that gravitational potential says is that an object at a higher altitude (say, a ball in your hand) is at a relatively higher potential than the same object when it's at the Earth's surface. Mind you, I meant "relatively" higher potential, because it's still negative - gravitational potential is negative everywhere, ... 1 The difference in potential energy is due to different definitions of what x=0 means. Since from the perspective of the spring this would be when the spring is not compressed nor stretched (rest length). However in the case of a mass-spring system in a gravity field (assumed to be a constant acceleration, g) this position is often chosen to be the ... 0 The energy stored in the spring is the one that is give \frac{1}{2}{kx^2}. As you mention, by conservation of energy there also a reduction of potential energy, but that reduction is not energy that it's stored by the spring but the complete change of energy of the whole system. Take into consideration, that the problem just states what's happening with ... 3 Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with n degrees of freedom and Lagrangian of the form:$$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q}) where $T$ is quadratic in ...

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Since you're talking about thermal physics, it's probably worth talking about this purely in a thermal context. What does it mean for something to have a high amount of energy? It means that we can do a lot of work with it. So let's say we have some sort of machine that uses something to do work as efficiently as possible. Since we're talking about solids ...

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Combustion is a chemical reaction where the molecules rearrange themselves and is VERY VERY different from change of phase. The energy stored in the solid phase is not necessarily more than that stored in the liquid phase. For example, it takes more energy to convert water into stream ($2260 KJ/kg$) than ice into water ($334 KJ/Kg$).

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The Landau Pole is not a problem for QED because at scales much smaller than it (the Planck scale, which is smaller than the Landau pole by 260 orders of magnitude) the (negative) gravitational self-energy of the particle will more than cancel out its electromagnetic self-energy. So string theory is not necessary in this case, just gravity.

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Interesting compactification, although I admittedly have no idea whether there are any mathematical pitfalls that one must avoid when doing such a geometric manipulation. However, assuming the apparatus is reasonably well-separated from nearby objects and that the apparatus isn't floating at a significantly different voltage than the surroundings, the choice ...

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There's a blog post that may be of some interest to you here: http://motls.blogspot.com/2013/11/the-expansion-is-accelerating-due-to.html Basically, the universe is constantly expanding at an accelerated rate due to "negative pressure". This is better understood with the Second Friedmann Equation: $\frac{\ddot a}{a} =-\frac{4\pi G}{3} (\rho + 3p)$ The ...

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