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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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A proton has a positive charge so by charge conservation it is not possible to reduce a proton to uncharged radiation particles such as photons (assuming that is what you mean by "pure-energy") Because of gauge invariance charge conservation is likely to hold good in all future physics, but we can't be totally sure of that. It is possible that some charged ...

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The simple answer to the main question is, yes. There are two ways to annihilate matter without using anti-matter. One is called fission, and the other is called fusion. Although only some of the matter is converted into energy in either of these processes, the efficiency of the "annihilation" is not in the main question. If 100% annihilation is required, ...

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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. I m here giving a simple 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 ...

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The mass $m$ in the formula is NOT the rest mass $m_0$ and therefore dependent on the velocity: $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma m_0$$ This means you cannot simply take the normal mass of a nitrogen atom and put it in there, if you assume a speed $v \neq 0$. The $c$ in the formula doesn't mean that the particle travels at light ...

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Your calculation is wrong because $E=mc^2$ doesn't mean that the object has velocity $c$. To make my answer useful i will give a very brief overview of Dynamics at higher velocities which is a consequence of Special theory of relativity. At higher speeds(of order of $c$) Newtonian mechanics is not valid. The linear momentum is defined as: $$\vec p=m_0\gamma ... 1 Usually the Newtonian limit is described as taking v << c but a much better way to express it is saying that the kinetic energy is much less than the rest energy$$ \frac{1}{2}m v^2 << m c^2 $$of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense. This is ... 1 For a particle of fixed mass m moving in a fixed gravitational potential \phi(\vec{r}) the motion is independent of the mass of the particle. The equations are$$ \vec{F}=-m\nabla\phi $$and$$ \vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt} $$It's clear that the m's cancel when combining these equations. So from this point of view it doesn't ... 2 When you say "without altering the actual momentum of it" is that really true?$$ E^2 = p^2c^2 + m^2c^4 $$so for a photon E = pc, since rest mass is zero. Now according to your first "traditional" calculation of m, we would have E = pc = m_1c^2, and therefore p=m_1c, where m_1 is mass according to the first "traditional" calculation. For your ... 13 The definition of an antiparticle is dependent on having the opposite quantum numbers of the particle so that they can annihilate, i.e. the sum of the conserved quantum numbers are zero. Thus the answer by @mpv is adequate. The implication of your question is then: is baryon number conservation a strict law or an emergent law that may be violated at some ... 2 I just started here so I don't have the rep. to comment and I don't have the time for a full answer, but the black hole idea mentioned in the comments above is a fine answer. See, for example, http://arxiv.org/abs/0908.1803v1 and How would a black hole power plant work? 16 I am assuming that by "energy" you mean photons. So you want to transform protons into photons. It is not possible. It would violate several conservation laws - mainly the charge conservation (protons are positively charged), but also baryon number conservation. The antiparticle is necessary to cancel these quantum charges to make the transition possible. 0 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 ... 0 It depends where your energy starts. Isolating two cases should give you the idea. The first is if the energy is already at height h, in the other we'll assume it starts ground level. Case 1: h_0 = h [already at height h] E = mc^2. The gravitational potential energy was stored previously in energy. Energy is not immune to gravity. To create the ... 1 Think logically. Assume that you want to create a mass on the earth, where h=0 (assumption). Therefore:$$E=mc^2$$You as well need to consume some work to take the mass from 0 \to h. So the energy needed is the energy you need to create it plus the one you need to "lift" it. So:$$\sum E = E - W_{W(spent)} = E - (-mgh) = mc^2 + mgh = m(c^2 + gh)$$... 0 This can be solved by adding the non-electromagnetic energy E_{p} of the Poincaré stresses to E_{em}, the electron's total energy E_{tot} now becomes:$$\frac{E_{tot}}{c^{2}}=\frac{E_{em}+E_{p}}{c^{2}}=\frac{E_{em}+\frac{E_{em}}{3}}{c^{2}}=\frac{4}{3}\times\frac{E_{em}}{c^{2}}=\frac{4}{3}m_{es}=m_{em} Thus the missing 4/3 factor is restored when ...

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With regard to mass and energy, nuclear reactions are no different than anything else. Also, mass doesn't turn into energy. Mass is energy. $E = mc^2$ is the shortened version of the equation $E = \sqrt{(mc^2)^2 + (pc)^2}$, where $p$ is momentum. This equation describes any physical system. Take a box and put in whatever you want - compressed springs, balls ...

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