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48

[5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment] This is actually the paradox that led Einstein to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up ...


19

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...


6

When a radioactive element decays, part of its mass is converted to energy - no obvious need for antimatter anywhere. Instead, the energy is released because the binding energy of the sum of the fragments might be higher than that of the parent nucleus. However, to fully convert matter to energy you do need the antiparticle. Otherwise, you run into ...


6

Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute ...


6

Your question asks why the "current quark masses" [see http://pdg.lbl.gov/2011/download/rpp-2010-booklet.pdf at page 21] of the quarks that make up a proton don't add up to the mass of the proton. The problem is that, for the light quarks, the "current quark masses" are very different from the "constituent quark masses" [see wikipedia]. "Constituent quark ...


5

The mass of a free neutron is 939.566 MeV/c$^2$ (almost 1 GeV/c$^2$, so that's probably where your instructor got the "1" value), and the mass of a free proton is 938.272 MeV/c$^2$. A free neutron will decay into a free proton, free electron ($\beta^-$), and an anti-neutrino, $\bar{\nu}$. The mass of the electron is 0.511 MeV/c$^2$, and of the ...


5

Internal friction in the metal of the bell eventually will bring the ringing vibrations to an end. The bell vibrates when it rings, making its molecules more energetic and creating heat. Bonding between the molecules of the bell resist the vibrations, and eventually the strength of the molecular bonds will create enough friction to bring the vibrations ...


5

To begin, lets go over the basics again. Any ensemble of two body decays in which the parents and children have the same masses in each event has a delta-function energy spectrum, or violates at least one of energy- or momentum-conservation. The fact that the beta decay spectrum is broad and continuous implies that at least one of the pre-conditions is ...


5

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


4

If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$ mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...


4

Anything that "suspends" the bell - whether it be a bolt, a piece of string, or a magnetic field - is applying a force. When the bell vibrates, this vibration will be transmitted. This is because the force of a magnet is a function of position - you can only get magnetic attraction because of a divergence of the field, so if you move, the force changes and ...


4

Energy is never created nor destroyed, and to say "X is converted into energy" is just meaningless. We don't convert things distinct from energy into energy, all we ever do is convert one form of energy into another. The badly posed question from your book probably intends to ask why we cannot convert the mass energy that any chunk of matter contains as per ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


3

You should always apply energy conservation, and it ought to hold in all reference frames, including the frame in which the sigma is at rest. In the sigma's rest-frame, $$ E_{\text{initial}} = E_\Sigma = m_\Sigma $$ and $$ E_{\text{final}} = E_\Lambda + E_\pi \ge m_\Lambda + m_\pi $$ Thus we have that, $$ E_{\text{initial}} < E_{\text{final}} $$ The ...


3

You can distinguish from vertical and horizontal velocity. Both balls have the same horizontal velocity, the difference lies in the vertical component. Up to the second trough there is no difference, but then the second ball accelerates downwards. It can't go straight down, which means that the gravity partially accelerates it horizontally. This difference ...


2

If we don't consider air friction with the ball, then you can see two different effects: an increase in length of the path and an increase in velocity of the ball. Turns out that the second effect more than compensate the first. An exact calculation isn't simple for arbitrary shapes, but we can see a simplification: let's say from (1) to (2) the length is ...


2

To understand this one shall take in quantum-mechanical approximation method namely perturbation theory into account. In perturbation theory, systems can go through intermediate virtual states which often have energies different from that of the initial and final states. This is because of time energy uncertainty principle. Consider an intermediate state ...


2

The equivalence principle tells us that energy and mass are really just two sides of the same coin, and are related by $E = m c^2$. Rearranging, we get that $m = E/c^2$, so instead of asking where all that mass comes from, let's ask where all that energy comes from. In the case of the proton, there are some quarks and gluons that make it up, and those ...


2

I've heard that some physicists think that the net energy of the universe is zero. Me too. They talk about gravitational energy being negative. But see Einstein talking about gravitataionl field energy here. It's positive. For this to happen, I would assume that the negative gravitational energy of a body ought to cancel out its rest energy. That's what ...


2

All you need is 1.) the formula for the velocity of a falling body $$v_C = \sqrt{2g\Delta h} $$ 2.) the formula for the equilibrium in orbital velocity is $$v_C = \sqrt{\frac{GM}{r}} \rightarrow \sqrt{gr} $$ Just set $x = h_C$ and solve the system Inside the cart there are spring scales on which the person is sitting. His limbs are in the air ...


2

For example, given the initial peak height of the roller coaster, I can predict the velocity at any point, despite the fact that there are various loops and curves It is very simple: if you know the height $h$ of a body you know its potential energy which is $mgh$. This energy is given to the body (transformed into Kinetic energy) at every lower ...


1

As the comments to the questions state, this is a question on the research state about the generation of the universe, and the first moment is modeled in the Big Bang model. The real beginning point is not yet known even in this model since gravity has not been consistently quantized within the model, only effective theory is used. Nevertheless existing ...


1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...


1

The bottom of Niagara Falls is in shadow, both because points northward, both for the reflecting fog clouds. Thereafter the rock average temperature is lower there then on top, quickly cooling the fallen water. The experience of ACuriousJim should implies that this effect can be more relevant than the others mentioned above, which just reduce the temperature ...


1

Proposed answer: the stiffness of the spring has influence on the value momentarily displayed; the stiffer, the higher. For unusually soft spring (soft enough that the scale goes down about $5$cm or more when an adult steps on it), the following analysis might allows an estimate of the stiffness of the spring. But with a normal mechanical bathroom scale, the ...


1

1/f spectra have the unique distinction of being "scale invariant" in the sense that the energy in an interval df is proportional to df. The 1/f spectra in fact have the property that the in an interval with width df available energy is proportional to df but not with f. There, namely "scale invariant" attribute for. It is not the energy, but the signal ...


1

Why don't we observe the infinite violations of conservation of energy The reason we don't see e.g. an atom spontaneously turning into a red giant for a fraction of a second is because of the extremely small timescales that such an energy difference would require - not even light could travel a tiny fraction of a proton radius in that time. The heavy ...


1

Explain why the mass of a tree cannot be converted directly into energy. That's a tricky one, because it could turn out that it is possible to turn matter alone into energy. Floris hinted at this with radioactive decay, but there are potentially other methods such as melting hadrons in a quark-gluon plasma (QGP), see for example this report. The interesting ...


1

I am going to make some general comments, expecting that they will lead for further clarification of the question. First - when a boat sails into the wind (as the one in this question does), it is important to realize that (all these are idealized statements... real sailing is a lot more complicated): a) the force of the wind is approximately normal to the ...


1

The rest of the energy is basically emitted as heat energy. Why? You have two capacitors in the circuit, and the connecting wires offer negligible resistance. Hence, when electrons flow from the charged capacitor to the uncharged one, the electrons basically face no resistance, and they collide with high speed with the uncharged capacitor. This collision ...



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