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0

If uranium undergoes fission, its products are too massive to release energy in fusion. Atoms must be smaller than iron to release energy through fusion.


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Perhaps a confusion : fusion and fission are not made with the same atoms. In a Mendeleev table, the most stable atom is iron, which is placed near the middle of the table. Before iron, you must merge atoms ( fusion ) to release energy. After iron, you need to break the atoms ( fission ). Then, chaining fusions and fissions to get pure energy, has not a ...


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It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions. The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 ...


3

My first question is then: Can I have an atom, fission it, then fusion it, then fission it, etc, etc,. And arrive at no mass and pure energy? I know this is wrong, but I don't know why so. No , you cannot. In addition to special relativity where invariant masses of complex objects are the "length" of their energy momentum summed four vector, which ...


-1

I think the following is a different way of saying what has been said above: How can energy be conserved AND uncertain? Remember basic vector space theory: every state can be expressed as a linear combination of other states (which form a basis). So a state which is not a state of definite energy can be expressed as a linear combination of states which ARE ...


0

I am only familiar with QM. Pauli (or maybe Dirac) wrote that there is a symmetry: energy-time is perfectly analogous to momentum-position, and one can think of energy as the momentum a thing has as it travels thru time. Einstein tells us that one man's space is another man's time, so one man's momentum IS another man's energy. How can energy be conserved ...


0

Newton's second law reads: $$F = ma$$ In the case of zero mass, the total force is always zero (although the acceleration may be nonzero, which might be slightly counterintuitive). Since the sum of the forces on the pan is zero, the contact force must be equal in modulus to the (sum of) forces of the springs on the pan. Therefore, the total work done on ...


3

In a curved spacetime, the wave four-vector $k^a$ for a solution to Maxwell's equation (in the appropriate geometric-optics limit) satisfies the geodesic equation: $$ (k^a \nabla_a )k^b = 0. $$ Effectively, this means that "the derivative of $k^a$ in the direction $k^a$ is zero". (Note that the first term is just a directional derivative in the direction of ...


3

Gravitational red/blue-shifting conserves energy. It merely swaps potential energy for kinetic energy. A photon outside a gravity well has zero gravitational potential energy. As it falls into the well, it gains momentum and loses potential energy, similar to anything else falling into a gravity well. At least, that's a simplistic way of looking at what's ...


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 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 ...


0

You are describing a resonant cavity with a discrete set of allowed modes and your thought experiment is a good illustration of the Purcell Effect (see Wiki page here), which is the influence on quantum mechanical photon emission probability amplitudes by an emitter's surroundings. The resonator simply shifts the emission frequency to one where the ...


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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

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 ...


-2

Where U is the intial velocity,g is the gravitational constant,R is the range of the projectile and ∅ is the launch angle between the projectile and the flat ground (imagine the flat ground as the x-axis with the projectile inclined at an angle ∅ on it).


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.


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 ...


8

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 ...


0

There is no conservation of mass. There is conservation of mass/energy. Proton mass does not equal neutron mass.


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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 ...


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 ...


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 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 ...


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 ...


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 ...


0

I just need to check if my understanding of the transfer of gravitational potential energy to kinetic energy is correct. Is anything wrong below? Yes. Gravity is not a force in the Newtonian sense. The potential energy of the body isn't stored in the gravitational field or the Earth, it's in the body itself, in its mass-energy. When you drop the body the ...


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The velocity you have correctly written is independent of the mass of the object, hence any object of same mass follows that velocity law. However, if you count in the drag force then you must add in the drag factor, meaning your velocity of the falling object now would be $$v=(\frac{2mg}{\rho AC})^{1/2}\tanh[t(\frac{g\rho A C}{2m})^{1/2}]$$ where $\rho$ ...


0

Yes, two objects with the same shape but different masses will have the same velocity after being dropped from the same height. Galileo tried this a while ago by dropping two such objects from a tower. Factoring in drag, we know that drag ∝ velocity2. So if the object (falling from the same height) has a greater surface area/size, its drag will be ...


0

I've found an historical detailed description in this book: Controversy and Consensus: Nuclear Beta Decay 1911-1934 by Carsten Jensen , From 1911 to 1934, 23 years, a lot of ping-pong with the experiments and theories went forth and back. I will not try to resume the history and the book deserves a reading. My textbooks, aged, only mention the winning ...


0

The electrostatic force is indeed $$F = \frac{q_1\cdot q_2}{4\pi \epsilon_0 r^2}$$ Now we need to move the two charges closer to each other - so we are doing work against this force. Indeed, there is an initial force needed to accelerate the particle (start the motion); however, assuming that the particle comes to rest again when it gets to $d_2$, then ...


0

We assume $q_2$ does not accelerate, so the net force on it is zero while it is moving. The "extra" forces needed to get it going and to stop it at the end cancel each other out upon integration.


0

The answer depends on how particular you want to be about it. Upon first examination you might say that in the Earth's conservative gravitational field, that the work must be the same, since the start and endpoints are the same and we have ignored any non-conservative forces (e.g. friction, air resistance, etc.). This is more than likely the response that ...


0

Let's call the rightward horizontal direction $+x$, and the upward vertical direction $+y$. Both balls reach point 1 at the same time, going the same speed. They both have the same $x$-component of velocity. At the beginning of the dip in B's path, ball A remains at constant velocity, $v_1\hat{i}$, but ball B gains in $v_x$ until the bottom of the dip. It ...


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 ...


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 ...


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 ...


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 ...


0

Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion $$ F = m g $$ with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$. Therefore, as ...


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 ...


-2

F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it. a better equation to represent the effects of gravity would be: F=(G*m1*m2)/d^2 where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) ...


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

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 ...


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 ...


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 ...


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 ...


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 ...


0

Kinetic energy is the work required to accelerate a mass from rest to a velocity (KE = 1/2 mv^2). Momentum is a measure of the amount of movement a mass has at a velocity (p = mv). Kinetic energy may be considered a process, and momentum may be considered the result of a process. Momentum is conserved, but kinetic energy seems to come and go, as it ...


0

That's easy. Think in a simple example that this happens. Imagine two particles of equal masses moving at $\vec v_1 = \vec v$ and $\vec v_2 = -\vec v$. Their momentum: $\vec p_1 = m\vec v_1$ and $\vec p_2 = m\vec v_2$. The momentum of the system is therefore: $$ \vec p = \vec p_1 + \vec p_2 = m\vec v_1 + m\vec v_2 = m\vec v - m\vec v = \vec 0 $$ The ...


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 ...


0

Conservation of Energy can be derived if one accepts that $F = ma$. I won't include the derivation here unless you ask. This means that to prove Conservation of Energy wrong, one must prove $F = ma$ wrong. This could be attempted in a variety of ways. One such way would be applying a force to an object and noticing the $F = ma$ doesn't give the correct ...



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