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Note that this is an incomplete answer. Imagine an object of mass $m$ at a distance $r$ from the centre of a black hole of mass $M$. The gravitational potential energy is $$U(r)=-G\frac{Mm}{r}$$ This has its highest value when $r=\infty$ and its lowest value when $r$ is at the event horizon of the black hole, i.e. the Schwarzschild radius R = ... 4 Actually, as someone pointed out in the comments, potential/kinetic classification is the only meaningful classification in physics. Potential energy is the energy which comes from interaction, and kinetic energy is the energy which comes from motion. Maybe you stumbled upon terms like chemical energy, thermal energy and so on. But chemical energy is just ... 4 In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ... 3 If I understand your notation right, then by h_L you mean the height of the left weight? In that case your formula doesn't make sense: To account for energy conservation, you would have to take the heights before and afterwards of the right weight. Then you get \begin{align} m_R*g*h_{before} = m_R*g*h_{after} + \frac{1}{2} m_L * v^2 \end{align} You ... 2 If voltage is all that you know, then the answer is No. If you know how much charge Q in Coulombs is added, you only have to divide by the charge e on each electron (in Coulombs). Otherwise, if you have a parallel plate capacitor and you know the capacitance C, you can work it out from Q = CV. Your suggestion that the extra electrons 'push more' ... 1 Yes you are. If a force is conservative, its work does not depend of any path between any points A and B. Since the work integral can depend only on the initial and final points themselves, we defineW_{A\rightarrow B}=\int_A^B\vec F\cdot d\vec r\equiv U(A)-U(B).$$Now define the mechanical energy as E=K+U so that$$dE=dK+dU.$$Suppose there are two ... 1 Force is a vector. Potential energy is a scaler. Forces which have associated potential energy functions as called conservative forces. Conservative forces act in such a direction that, if released from rest, the potential energy function associated with that force will decrease (and the kinetic energy will thus increase) with the velocity increasing, until ... 1 The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ... 1 This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction? 1 If the force is conservative then it is the gradient of the potential:$$ \mathbf F = -\nabla \mathbf V $$If we write this out in component form (in polar coordinates) we get:$$\begin{align} F_r &= \frac{d\mathbf V}{dr} \\ F_\theta &= \frac{d\mathbf V}{d\theta} \\ F_\phi &= \frac{d\mathbf V}{d\phi} \end{align} Since we are told that the ...

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The state of equilibrium is characterised by a minimum in free energy $F=U-TS$ (Helmholtz for simplicity), not a minimum in potential energy. What this means is that while the system is indeed attempting to minimise the potential energy $U$, it is simultaneously trying to maximise the entropy $S$. The balance, i.e. which term dominates, is determined by the ...

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Let $X$ be a sphere of radius $R$ with charge $+Q$ on it. The potential of sphere $X$ is $+\frac{kQ}{R}$. Let $Y$ be another sphere of radius $6R$ with charge $+5Q$ on it. The potential of sphere $Y$ is $+\frac{k5Q}{6R}$. The potential of sphere $X$ is larger than that of sphere $Y$ so if the spheres are connected with a wire electrons will flow from ...

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