# Tag Info

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Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ...

4

The heat that makes a filament lamp glow is derived from electrons bashing into the lattice of atoms in the filament and transferring energy to them. The kinetic energy of the electrons becomes vibrational energy of the lattice, and this is exactly what heat is. However the interaction of electrons with the lattice is also what resistance is, and ...

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The whole point of virtual particles is that they do not obey the on-shell physical laws, they are just computational crutches in Feynman diagrams. You should not take the idea of them being exchanged or "filling the vacuum" too seriously, there is no reality to them (hence the name). The energy-time uncertainty relation is one of the most misused results ...

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(Summary: In this post I argue that you need at least an energy of $m v_1^2(1-\cos(\theta))$ in the idealized kinematics situation to deflect an asteroid of mass $m$ and velocity $v_1$ by an angle $\theta$ using rockets, without changing the magnitude of $v_1$.) Energy isn't the be-all and end-all of motion. The problem is that momentum also has to be ...

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Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ...

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In short: you have calculated the maximum displacement of the stone, and your friend has calculated the equilibrium displacement. If you place the stone of mass $m$ on the spring (which I assume to be of negligible mass), the stone ends up oscillating around the equilibrium point. So the sum of the potential energy stored in the spring and the kinetic ...

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We don't know! If we were to observe a situation where energy conservation did not appear to work, that would be a major puzzle. As you say, either we would have to discover some alternative contribution to the energy that we had been neglecting, or we would have to give up on energy conservation. A priori it is not obvious which one of those two resolutions ...

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You might be able to get it to work for quite a while, depending on your skill as an engineer. But there is a critical difference between a well-engineered machine that runs for a while, and a perpetual motion machine that runs forever without input. The latter is impossible.

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To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ...

2

Short answer: The distance the rock falls in your thought experiment is not the same as the spring compression described in the actual question. Long answer: In your thought experiment, the rock is released from rest and it falls some distance before turning around. So the question arises: For an object of mass $m$ released from rest just touching a ...

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To turn thermal energy into useful work completely one would need a thermal bath at the temperature of absolute zero. This is explicitly forbidden by the third law of thermodynamics. The best one can do is given by the efficiency of the (theoretical) Carnot cycle: http://en.wikipedia.org/wiki/Carnot_cycle. Th efficiency of the Carnot cycle only depends on ...

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The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...

1

From the answer... ..my question I have learnt that Newton's 3rd law of motion is a direct consequence of law of conservation of energy. When a body moves in a certain direction and an opposing force acts on it , it exerts a reacting force (by Newton's law) ... Hence, the body loses its kinetic energy. Problem ....gravitational force: ...

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UPDATED: I now think my previous answer was wrong, because the set up would be equivalent to the following question: Is a black body sphere inside a black body shell hotter than the shell? Just change the question to add a carefully crafted lens that focuses all the radiation into the sphere (you could make the shell as large as you want), which of course ...

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The initial potential energy is zero because the ball starts off at essentially ground level, and potential energy is being defined as being zero at ground level. The initial velocity is a vector of magnitude v that points up at an angle $\theta$ from the ground. The components of that initial velocity are $v_x(0)=v \cos\theta$ in the horizontal direction, ...

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$m_2$ will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds leaving the collision as entering it. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision. In order to simplify things and to ...

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If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.

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There is a confusion between the terminology "perpetual" , which means "continuously", devices that almost move forever, and a machine that can produce energy. As the other answers point out energy is conserved and if it looks as if energy is provided from nothing a closer analysis shows the mistake, as in the drinking bird perpetual setup. In the case of ...

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For what it is worth, your formulation of the spherical wave business is flawed. See here for example. These have the form $$u(r, t) = \frac{A}{r} e^{i(\omega t \pm k r)},$$ where the symbols have their usual meaning. So, intensity, which goes as the amplitude squared, will go as $$I = \left(\frac{A}{r}\right)^2$$ falling off as the square of the distance, ...

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A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...

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