# Tag Info

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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 ...

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 ...

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 ...

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

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 ...

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

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 ...

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 ...

3

If you setup a perfect cavity where no modes of light are possible, then the light will not be emitted in the first place (0 probability). You run into problems if you consider the emittor as a classical light-source and then combine it with a quantum-mechanical reasoning regarding interference in cases such as this.

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

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

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 ...

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 ...

1

Initial kinetic energy is $K_1=\frac{1}{2} m (v_x^2+v_y^2)$ with potential energy $U_1=0$. At the apogee, the potential energy is $U_2=m g h$ and the kenetic energy is $K_2=\frac{1}{2}m v_x^2$. Equate the two sums to get your answer. $$U_1+P_1 = U_2 + P_2$$ $$0+\frac{1}{2} m (v_x^2 + v_y^2) = m g h + \frac{1}{2} m v_x^2$$  \frac{1}{2} m v_y^2 = m g ...

1

Then just harvest your kinetic energy somehow and keep going indefinitely. But this doesn't result in infinite energy. All "harvesting" of kinetic energy is based on lowering the difference in linear speed or angular velocity between two bodies. Clearly, your starting velocity relative to the planet limits the total kinetic energy you could harvest ...

1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...

1

Short Answer: The contact force (normal force if you like) between the pan and the box is 0 because the pan has negligible mass. Long Answer: The key point in this problem is that the pan has neglible mass. Suppose for a second that the pan had some mass $m_p$. After falling a distance of $0.5 \, m$ the box would have velocity $v_b = \sqrt{2gh}$. In the ...

1

The point source keeps radiating light. Will the light undergo destructive interference completely? Point particle as a source of light is OK, but it would need to move with acceleration to produce EM radiation. Static source of light of zero size seems to be reduction ad absurdum, at least from the standpoint of common theory of light based on EM ...

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

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

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

If we did know the radius of each sphere, would it be possible to skip conservation-ofâ€“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

Energy conservation always applies. Your mistake is in thinking that adding masses will solve the problem instead of clarifying some aspects. In the case of the sigma-zero the decay at rest allows to see that the sum of the constituent masses is larger than the mass of the sigma-zero. For a decay to happen there should be energy left over to go to the ...

1

The three quarks you talk about are usually called the valence quarks of the proton, and their contribution to the mass of the proton is not it. In particle accelerators, when we hit protons with high energy beams, we discover that protons are made of a cluster of smaller constituents (like quarks and gluons, which constantly are created and destroyed in ...

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