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This is an experimental physicist's answer: The linked article is careful to state: That means that conservation of energy can appear to be violated, but only for small values of t (time) Italics mine. Conservation of energy is an experimental fact that has been validated in innumerable experiments. This means, as far as the correspondence with ...

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Suppose the ramp wasn't there, then the trajectory of the object would the same as if it fell off a cliff: To get the equation of motion you simply note that the horizontal and vertical coordinates are given by (neglecting air resistance): $$x = ut$$ $$y = \tfrac{1}{2} g t^2$$ So you can get the trajectory by substituting for $t$ to get: $$y = ... 3 There are a couple of issues you might want to consider. Firstly there is the slightly boring one that we physicists measuring the mass of the black hole are outside it, and from this position the photon never reaches the event horizon let alone crosses it. I don't want to go into this here since the subject has been flogged to death in numerous questions ... 3 Please tell me what I did wrong It takes General Relativity (GR) to describe black holes and, in GR, energy conservation is, well, subtle. From John Baez's Relativity FAQ "Is Energy Conserved in General Relativity?": In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved". So, in general, ... 2 Sounds like you're getting at the "coefficient of elasticity," which is a value in [0,1] which represents what percent of the pre-collision kinetic energy is found after the collision. In homogeneous materials, the remainder of the energy is typically lost to deformation or heat (phonons) as you suggest. you could imagine, for the sake of argument, a steel ... 2 The energy is still there in the form of gravitational potential. Think of leaving the earth as a process similar to riding a bicycle up a hill. (When you ride a bike up a hill, you're moving against Earth's gravitational field thereby gaining potential energy, just like what happens to the rock when it moves up away from the surface of the earth.) If the ... 2 Yes, that equation will still give the correct value for the energy of the oscillator system at any point in time, assuming of course that you know dx/dt and x at that time. If there is an external dissipative force on the system (damping) you will find that the value of E decreases with time. But the energy of the oscillator itself is still the sum of ... 2 The two existing answers have both done the correct calculation, but both have forgotten to account for the Sun's gravity. A comet falling from the fringes of the Solar System is accelerated mainly by the Sun's gravity. We can see this from the expression for the potential energy at a distance r:$$ V = -\frac{GMm}{r} $$For the Sun M = 1.9891 \times ... 2 The Earth and the Sun has magnetic fields which shields us from cosmic rays, as a charged cosmic ray particle will loose kinetic energy when its direction is perpendicular to the magnetic field. So what happens to the kinetic energy of the cosmic ray particle? According to the first law of thermodynamics it can't just disappear. It goes to the magnetic ... 2 Your equation is incorrect. The gravitational potential is$$\phi(r)=-GM\frac{3a^2-r^2}{2a^3}$$when you're inside a uniform sphere of radius a with total mass M. This is a quadratic potential in r, which is why it gives rise to harmonic exchange of energy when you oscillate between the planet surface and the core. 1 The force on the pendulum only applies when the pendulum is in the vicinity of the coil. At that moment the harmonic motion of the pendulum is distorted. It 's amplitude is lessened and with it the upward motion. So the kinetic energy of the pendulum is converted into gravitational energy and electric energy. But the gravitational energy is less than without ... 1 As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish. If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would ... 1 but when they bounce off each other again, the energy does not return to being kinetic The potential energy actually does go back to kinetic energy. But often it only partially goes to the translational motion of whole body, while other part of the energy comes into internal motion of atoms. People normally call the latter thermal energy. 1 What you need to do is use the conservation of momentum to get the velocity of the combined system:$$ m_1v_{1,i}+m_2v_{2,i}=\left(m_1+m_2\right)v_f $$This conservation law shows that the final velocity of the two blocks will still be proportional to the initial velocity of the one block (i.e, v_f\propto v_i). Getting this into the fractional change ... 1 Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width \Gamma that is related to its lifetime \tau by$$ \Gamma=\frac{\hbar}{\tau} $$when you measure the mass/energy of such particles in experiments you always get a Lorentzian or Breit-Wigner distribution like this from which you can measure the ... 1 It is not random. If the exact same meteor strikes with the exact same properties then the results will be mostly the same (barring a small amount of chaos). So what is the distribution of energy. Well if you drop a ball on carpet and on a wooden floor it will make different sounds and it will bounce differently. So it is in the details of all possible ... 1 The electrical energy you pay for doesn't just go into the appliances; some amount of it ends up as heat in the electrical path from the meter to the loads. Also, heat from a load such as a heating element can travel along the electrical conductors away from the application (the water tank) - remember: most good electrical conductors are also good thermal ... 1 The gravitational energy of the comet at infinity gets converted into kinetic energy of the comet. Calling m the mass of the comet, M the mass of the Earth, r the radius of the Earth we have:$$G\frac{mM}{r} = \frac{1}{2}m v^2 $$where G is the gravitation constant and v is the speed of the comet when it hits the surface. Thus:$$ v = ...

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There is certainly an interaction there between the optical medium and the photon. Actually, there are two photons in the interaction: the incoming one is absorbed by an electron in the material so that the latter fantastically fleetingly rises to an excited state. A fantastically short time later, another, outgoing, photon is emitted and the electron ...

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Your second therm is still referring to the conservative force acting on the object. For an ideal spring this would mean: $\omega_0^2=\frac{k}{m}$. However in the case of a damped oscillator there will be both conservative and not conservative forces acting on the object with mass $m$. But this does not mean than the potential energy of the conservative ...

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