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"I presumed the source of this energy was not coming from the conversion of other types of energy to dark energy, so it must violate conservation." This is where you go wrong. The positive dark energy is balanced by the negative energy in the gravitational field. As a volume of space expands more dark energy is created in the volume but this is balanced by ...


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By assumption, there is electric current (with density $\mathbf j$) flowing in the wire (think of a circuit or a rod on rails). When placed in the magnetic field, there will be magnetic force $\int \frac{\mathbf j}{c} \times\mathbf B \,dV$ acting on the wire (on the nuclei it is composed of) so some of its parts will begin to move (with velocity $\mathbf ...


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I'm going to assume some things before I answer: This golf ball doesn't feel the effects of air resistance, or is in a place where that does not matter. You know some common equations for kinetic energy and gravitational potential energy. We ignore any effects of weather and assume the only things contributing to the ball's motion is the golfer and gravity ...


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Good question. In my opinion, Banach-Tarski has absolutely no implications for the enterprise of describing physical reality i.e. physics. (It is still a good question though!) Here's why. Banach-Tarski just means that, okay, suppose I build a model of the actions that can be applied to a physical object. If that model has the property that every ...


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First of all, I would like to say that the answer Floris gave is the correct way to do the problem you've set forward, but I thought it worthwhile to note that the result ${5 \over 2}(R-r)$ is the answer if, rather than rolling a marble down the track, you are sliding a cube. If this problem comes from a textbook, there were some unstated assumptions in ...


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Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...


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An energy conservation law only arises when the system studied has a Lagrangian which is invariant under time translations up to total derivatives, due to Noether's theorem. More generally, a quasi-symmetry under spacetime translations gives rise to an entire host of conserved quantities, encoded in the stress-energy tensor. Consider a scalar field; under ...


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If you want, you can pick a theory like e.g. Newton's mechanics and show, that these equation obey the principle of energy conservation. But that's not how energy conservation is typically looked at. Energy conservation is(!) one of the basic principles theories are build on. Energy conservation is more fundamental than any single theory and is not ...


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Well, it depends on what way you want it 'proved'. That is, in what context? If you loo at the first chapter of Landau Vol. 1 it will give you some lovely derivations of the proof. Energy is conserved due to the homogenity of time. That is, because it should not matter if I measure the energy of some mechanical system today, or tomorrow say, or at any ...


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$KE_{\text{initial}}+U_{\text{initial}}=KE_{\text{final}}+U_{\text{final}}$ $KE_{\text{initial}}+U_{\text{initial}}=0+U_{\text{final}}$ $\frac{v_0^2}{2}-\frac{GM}{R}=-\frac{GM}{R+r}$ $\frac{v_0^2}{2}=\frac{GMr}{R(R+r)}$ $v_0=\sqrt{\frac{2GMr}{R(R+r)}}$ This makes sense because if we take the limit: $\lim_{r\to\infty} \sqrt{\frac{2GMr}{R(R+r)}} = ...


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$KE_{\text{initial}}+U_{\text{initial}}=KE_{\text{final}}+U_{\text{final}}$ Set $KE_{\text{final}}=0$


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Let's focus on radiating EM waves first and forget about energy. When you jump into accelerating train and see charge accelerate away from you, this is all in a non-inertial frame. In this frame, electromagnetic theory has to be formulated with modified equations and new appropriate boundary conditions. That being said, nothing forbids static field in ...


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It seems that you are picturing negative energy as something mystical. It is important to note thag kintetic energy is positive, while gravitational potential energy is negative. In case of positron and electron, after creation their kintetic energies are the positive energy, while the potential energy of their system is negative (as one is positive and ...


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Energy is time. In order to provide you with a simplified concept: Energy is what is remaining conserved when time goes by. Any effect has a cause, and any energy state corresponds to a former energy state of the same quantity. Energy conservation means that energy at the end of time will be the same as in the very beginning. And the question: "what was ...


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This is an interview that starts at a more technical overview of the QEG machine: https://www.youtube.com/watch?feature=player_detailpage&v=RJKE5DJRMFQ#t=715 The technical guy James Robitaille is obviously not a sales person, so when he speaks with his pauses, he seems a bit more of a technical geek than a "con artist". I submit he probably isn't a con ...


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Matter possesses energy from Einstein's equation $E = mc^2$. This equation describes how matter is a form of energy as well, and can be converted from one form to the other. This is exactly how the sun is power, in a process called Nuclear Fusion. With this idea in mind, check this hypothesis out, called Zero-Energy Hypothesis which simply says that matter ...


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Energy conservation may be better stated as, The total energy, i.e. energy in the form of mass + all other forms of energy is conserved for an isolated system. This would mean that annihilation is simply an example of interconversion of energy from 'mass-energy' to 'light(electromagnetic) energy'.


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Photons aren't pure energy - they are a particle like all other particles. Admittedly photons are massless but then so are gluons, and indeed above the electroweak phase transition temperature so are all particles. So pair production from photons and annihilation into photons is just a scattering process like any other particle interaction. However if is ...


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Conservation of energy refers to systems looked from the same reference frame, it does not make sense to require that energy of the same system to be the same in different reference frames. As a consequence of time translational symmetry, energy conservation is usually true unless we drive the system externally which may break this symmetry. Similarly, ...


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Something must be proven or understood. So would beware from such answers like @Anand did. As example Nuclear reactor could be accepted like Perpetuum Mobile device, would it be invented some time ago. But definitely chances, that someone just invents something from future, without proper knowledge's , equipment, theory etc, are low (as my opinion I ...


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See Anand's answer: I'm not sure whether this one is simply misguided or instead subtle fraud (as calls for money to fund research are involved). Actually the claims made in the article are true in one sense, which gives the idea the whiff of sophisticated fraud. In the linked article, it is claimed that the device is powered by a 1kW source and then ...


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According to the 1st law of thermodynamics, energy cannot be created or destroyed, it can only change form and is thus conserved. If it did work you would be going against 200 year of scientific consensus. Good luck!


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There follows my try to decompose the solution into a minimal amount of calculation and apart from that only geometrical considerations. The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center. We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure). The conservation ...


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Most, if not all, scientific analysis of real situations involves approximations. If some of the kinetic energy gained from falling was converted to kinetic energy of downstream flow (like a more sliding board shaped waterfall) it could affect the calculation. The environment could affect the temperature of the pool of water at the bottom of the falls, ...


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In fact, thermal energy is heat. Think about heat as atoms vibrating: The more vigorously they vibrate (i.e. the more kinetic energy they have), the hotter it gets. The relation between thermal and kinetic energy is summarized by the simple equation $k_B T=m\langle v_x^2\rangle$, where $k_B$ is Boltzmann's constant and $\langle v_x^2\rangle $ the expectation ...


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A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy. The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal ...


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First, in the most common model of dark energy, $\Lambda$CDM, dark energy is a constant energy density, which means that the "energy" from dark energy does increase as the universe expands. Second, the law of conservation of Energy is only valid in a static universe. Because our universe is expanding, it is no longer the same at every moment of time and so ...


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The potential energy $ P $ of the car doesn't change (the car stays on the ground the whole time), and because it moves uniformly in a circle, its speed $\left | \mathbf{V} \right |$ doesn't change. But the kinetic energy $ K $ of the car is $\frac {1}{2} m\left | \mathbf{V} \right |^2 $, so it doesn't change aswell. Hence, the total energy $ E= K + P $ ...


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When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


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$\def\om{\omega}\def\vr{{\vec r}}\def\l{\left}\def\r{\right}\def\ve{{\vec e}}\def\vom{{\vec\omega}}\def\ds{\,'}$ Let the car move in the (x,y)-plane, let $m$ be the car's mass and let $J$ be the moment of inertia for the rotation about the axis through the center of mass aligned parallel to the z-direction. If you have a straight line and a circle with ...


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No, it does not gain energy. The confusion arises because there's a force that does no work. If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $ \vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which ...


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Its funny you should ask this as I recently ran several simulations on matlab regarding the same thing except with atoms. Effectively, I had a diatomic molecule (H-H for example) and an atom (F lets say). The atom and diatomic both had some momentum relative to each other and the collision was setup to be perfectly collinear. Now, what I noticed is that ...


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I will give you a starting point: You could solve this problem by equating the change in potential energy with the change in kinetic energy, $\Delta P.E. = \Delta K.E.$ You could solve the problem with calculus starting from $$ F=ma=mdv/dt=mvdv/dr=F_{gravity}(r)\Rightarrow\\ \int mv dv = \int F_{gravity}(r) dr $$ Even better: solve it both ways and check ...


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We can't create noting from something and we can't create something from nothing. Nothing or nonexistence doesn't exist, it is it's definition. The only thing that exist is energy in different forms. So energy exists and cannot become nonexistent, as there is no such thing as nonexistence. Energy can neither come into existence because it has nowhere to come ...


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Say the density of the string is $\mu$ and the tension is $T$. It's clear that the kinetic energy of an infinitesimal piece of string is $$dT = \frac{1}{2} (\mu \, dx) u_t(x)^2$$ The length of the infinitesimal piece of string from $(x, u(x))$ to $(x + dx, u(x + dx))$ is \begin{align} d\ell &= \sqrt{dx^2 + (u(x+dx) - u(x))^2} \\ &= \sqrt{dx^2 + ...


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yes you can say that they energies are the same. And with energy we usually mean that the potential plus kinetic energy always remains constant. so kinetic energy can change and so does potential but the sum of those 2 is always constant unless you add friction.



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