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The charging current warms the battery as it charges, and because of mass-energy equivalence, an unmeasurably small amount of mass is indeed added to the phone in the process. However, this is not what is measured by your electric utility in order to bill you for energy. The insignificant mass added to your phone when charging is lost partially when the ...

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I've made a small illustration depicting the key idea. If this is in coherence with what you've asked, we could summarize some important points about the case. Total energy of the Earth and Bar Magnet system is given by the equation: $KE + PE = \frac{1}{2}mv^{2} + \frac{GMm}{R}$ While PE is there for both Earth and magnet system (combined), KE is ...

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As of last week, scientists have created a thermoelectric material that increases its efficiency to 15-20%. This is still not good enough to heat or cool your house, but it gives me hope that someday in the future, we have materials so good and cheap that we can build our houses and cars with it. So we may not be able to use the heat inside the house to cool ...

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Finally the bottom line of question is, can we experimentally prove that energy, total energy is actually conserved ? (a Yes answer requires a detailed experiment with complete conservation and no loopholes) Elementary particle physicists have been doing this for more than sixty years. Conservation of energy is one of the main constraints that built ...

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This should go on forever, and current should keep appearing across the load resistance. This is a contradiction. Since there is current through (not across) the load resistance, there is work being done on the load: $p = i^2R$. Let's be clear on this: the coil-load system does no work on the pendulum, the pendulum does work on the coil-load ...

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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 ...

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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 ...

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Actually pendulum wont oscillate forever.Its energy turns into heat in resistor.In other word it's domain would be something like this figure.

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The short answer is that there is a induced force on the magnet. This induced force will make the pendulum loose energy in the same proportions as there is electrical energy being generated. A good experiment to show this effect is by having a small bar magnet and a copper pipe Or solenoid. When you let a small bar magnet drop from a certain height it will ...

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The friction acts only when the contact point slips relative to ground. You can consider speed of lowest point to be sum of $v$ and $\omega r$ with proper directions. Friction acts till there is slipping and condition for no slipping is $v=\omega r$ when v is right and $\omega$ is clockwise acc. to diagram As in first case, the lowest point in always ...

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1st que- as the point in contact of the disk and the plane is not moving, no work is done. and as for the second question; no body is perfectly rigid, therefore due to deformation there is loss in energy. that is why the disk stops.

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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 ...

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Power is, in words, the rate at which work is done and work done equals the amount of energy converted from one form to another. For electric circuits, the power associated with a circuit element is the product of the voltage across and the current through. We can verify this via dimensional analysis: $$v \cdot i = \frac{J}{C}\cdot \frac{C}{s} = ... 0 Starting from the definition of power$$P=\frac{dW}{dt}$$We can solve for the work (after integrating). The thing to know is what power is in terms of current and voltage or resistance$$P=IV=I^2R=V^2/R$$Clearly, P=IV is what we want to use. Last thing is what is V for a inductor? It is V=L\frac{dI}{dt} From here I will let you (and future ... 2 Your equation is incorrect. The gravitational potential is$$\phi(r)=-GM\frac{3a^2-r^2}{2a^3}$$when you're inside a uniform sphere of radius a with total mass M. This is a quadratic potential in r, which is why it gives rise to harmonic exchange of energy when you oscillate between the planet surface and the core. 1 What you need to do is use the conservation of momentum to get the velocity of the combined system:$$ m_1v_{1,i}+m_2v_{2,i}=\left(m_1+m_2\right)v_f $$This conservation law shows that the final velocity of the two blocks will still be proportional to the initial velocity of the one block (i.e, v_f\propto v_i). Getting this into the fractional change ... 0 You are quite right. Energy is not conserved between the reference frames. That is the biggest mystery. It is certainly going to change one's concept of understanding of energy. Here is an example where there is total failure of 'law of conservation of energy' Take an example of spaceship in space. suppose you start the spaceship and accelerate for one ... 1 Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width \Gamma that is related to its lifetime \tau by$$ \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 ... 0 Note that it is not an elastic collision between bob and stick(in fact perfectly inelastic). Hence, some part of initial kinetic energy is stored in form of elastic potential energy which you are not adding and hence the confusion. Also you are not adding the initial potential energy of the rod. Apply conservation of momentum and find velocity of system ... 0 You are not taking into account the initial potential energy of the rod. Using your previous computations, what is the energy of the rod for \theta = 0? 0 Take a simple quantum mechanical potential that describes an atom. The mass of the atom is fixed. Take hydrogen. The electron is in an orbital around the proton which is a probability distribution of its location in time and space: if you measure it, i.e. interact with it, where you may find it. Correspondingly there exists an energy width to the energy ... 0 If you consider things on a microscopic level, the first thing you see are the atoms at the surface of the meteor colliding with the atoms at the surface of whatever it hits, lets call it "the ground". There will be a lot of these initial collisions, each one accelerating the atoms at the surface of the ground material. These atoms in turn accelerate 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 ... 0 Yes, it's true that hard water reduces energy efficiency of boiler. Explanation: The scale buildup has an insulating effect that reduces the element’s ability to heat the surrounding water. The boiler’s thermostat continues to call for heat, so the element heats longer and more often. This causes wasted energy and high electricity bills and ultimately ... 7 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 ... 5 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 = ...

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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.

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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 ...

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Before considering how rest mass of an object turns into energy, it must be considered what relativistic mass is. Relativistic mass is expressed by equation(1) and equation(2). Mass(M) is made of countless relativistic masses which have kinetic energy and small rest masses as it is expressed by equation(2). vi is the velocity of individual mass(mi). When ...

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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 ...

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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, ...

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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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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 ...

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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 $M = 1.9891 \times ... 1 The gravitational energy of the comet at infinity gets converted into kinetic energy of the comet. Calling$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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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 ...

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There would be real problems if as you said " run a really long wire from the array to the ground" if you did that you would have alot of voltage drop. There is a significant voltage drop even for a run of wire of 100 ft. But if you put a pv system on say a mountain top with few sun obstructions and gathered the power there. But you would have to live ...

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I come across this cause I am working on this problem. I see long talk and they don't know the answer. Yes, the energy is lost, when all systems are fully charged and outside the sun is still shining. No meter PWM or MPPT charging system. You just cant put more into batteries when its full. Full means full. Solution is ultracapcitors. Assemble a capacitor ...

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The extra energy comes from the kinetic energy of the moving car. The radiated light is carrying away some momentum and is decreasing the speed (and therefore kinetic energy) of the car.

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