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15

updated calculations - based on neutrino energy escaping and vapor inhalation risk Your math is close but not quite right. First - the number of tritium atoms. There are 1000/(16+3+3) = 45 moles (as you said) This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s. ...


9

There is an effortless way, if you accept geometrical reasoning. You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write: $$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$ Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot ...


9

In principle, no: a particle may leave a reaction with any kinetic energy you can imagine. However in practice, there's a limit known as the GZK cutoff. A baryon (proton or neutron) with energy above about $10^{19}\,\rm eV$ will see the cosmic microwave background blue-shifted so strongly that the baryon can scatter from the background photons to produce ...


5

For most systems, if you are operating near equilibrium you are at a point where the net force is zero. That means that for small displacements, there will be a small force proportional to the displacement which restores the system to its equilibrium position (Taylor expansion - for small displacements, only first order effects matter). $$F(x + dx) = F(x) + ...


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

When you find the total (squared) value of some vector in an orthogonal basis, such as the Cartesian system $(x,y,z)$ or indeed the spherical system $(r,\theta,\phi)$, what you're doing is simply adding the squared values of each component of the vector. Taking the velocity, let's think about the different components: What is the velocity in the radial ...


4

Yes, the vacuum energy of a spacetime lattice with finite spacing and periodic boundary conditions within a box of finite size is finite. One would not call this "quantizing", though, rather discretizing because we are not carrying out any "quantization procedure" in the sense of going from a classical to a quantum system. In this approach, the finite size ...


3

An important connection to relativity can be made here. Consider the infinitesimal displacement in the Cartesian coordinates: $$ ds^2=dx^2+dy^2+dz^2=dx^a g_{ab}dx^b $$ where $a,b\in\{1,2,3\}$ and $$ dx^a=\left(\begin{array}{c}dx\\dy\\dz\end{array}\right) $$ and $g_{ab}$ the metric, $$ ...


3

There are at least two answers possible to give, but both, in the end, amount to the same thing: There is no "right" way to fix the energy scale of a process, but that doesn't matter, except that your perturbation theory will probably break if you choose the scale badly. The old answer: The renormalization scale is arbitrarily defined to fix some parameters ...


3

One can consider to harvest power at Gibraltar strait (Moon/Sun energy from tides). Any turbine to put in space is not feasible and, if not, then soon it will be locked to the motion of the Moon, and then useless.


3

Electrons occupy shells characterised by the principal quantum number, $n$. The lowest energy shell ($n=1$) is the ground-state. Above that you have the first excitation shell ($n=2$), the second excitation shell ($n=3$), and so on. In the hydrogen atom, the energy states are given by the equation $$ E_n=\frac{-13.6\,\mathrm{eV}}{n^2} $$ So the energy to ...


3

The water droplets that create a rainbow are not emitting the light that you see in a rainbow; if they were, you would see a glowing cloud of consistent color, not a rainbow. The rainbow is formed by sunlight refracting and reflecting through water droplets in the air; the water refracts through the "front" of the drop, reflects off the "back," and refracts ...


3

The person would be accelerated upwards briefly, then continue to float upwards at constant speed until his head hits the ceiling. When standing on a floor, the ground is pushing up on you with a force equal to your weight. Your shoes and your feet aren't perfectly rigid. They are compressed by this force and will act a bit like a spring if this force ...


2

What is your goal - to stop bullets or to generate electricity? The key to stopping bullets is not so much to absorb the energy (although that matters too) but to absorb the momentum. You may know that $$p = F\Delta t$$ In other words, given a certain momentum $mv=p$, you need to apply a force F for a time $\Delta t$ in order to slow it down. The ...


2

Your assessment of the transitions which can occur, and hence the photons which can be emitted, is correct. However, the colliding electron does not go to one of the energy levels in the atom (as Sebastian already correctly pointed out). What happens is that the colliding electron can deposit its energy in the bound electron, 'promoting' it from the ground ...


2

A state $\psi$ corresponds to an energy $E$ if: $$H\psi = E\psi$$ Clearly, if there is a state $\psi = \sum_i c_i \psi_i$ where $H\psi_i = E_0\psi_i\ \forall i$, then $$H\psi = \sum_i c_i H\psi_i = \sum_i c_i E_0\psi_i = E_0\psi$$ A linear combination of states with the same energy value again has the same energy value. Now consider $\psi = c_1\psi_1 + ...


2

If you have a ground state hydrogen atom, the the first excitation energy is the distance to the lowest unoccupied orbital i.e. it is the lowest energy that can excite an electronic transition. The ground state is with the electron in the $1s$ orbital, and the next lowest energy orbital is the $2s$. So the first excitation energy corresponds to the ...


2

Photons can be emitted when electrons change energy levels. You say that you have worked out where a 12.1 eV difference is. In an ordinary hydrogen atom, the electron will be in the $n_1$ state. Now, what energy state will the electron be in if an ordinary hydrogen atom absorbs 12.1 eV of energy? After absorbing that energy, the electron can lose energy ...


2

There is also, a gas of photons in the right side. It was trapped there when you assembled your box, and since you are assuming a perfectly zero emissivity, these photons must be perfectly reflected from all surfaces. That means they are blue shifted if the wall moves toward the right and red-shifted if the wall moves toward the left. Result: If you ...


2

Yes it is theoretically possible, as discussed in the other answers and indeed we already do a variation of harvesting a planet's orbital kinetic energy in the space navigation manoeuvre called "gravity assist" or "slingshot" to boost a spacecraft's speed without expending propellant. Here one makes one's spacecraft "collide" with a planet (i.e. make a very ...


2

$E=hc/\lambda$ only works for photons, and $E = mc^2$ only works for stationary objects. For moving objects, we use the energy-momentum equation.


2

If you assume adiabatic compression, the equations are $$PV^\gamma = K \text{ (constant)}\tag1$$ Where for air (mostly a diatomic gas), $\gamma=\frac75$. This means that the final volume $V_f$ when the pressure is 50 bar (50x atmospheric) is $$V_f = V_i \left(\frac{P_i}{P_f}\right)^\frac{1}{\gamma} = 61.2 L$$ The temperature of the gas is given by the ...


2

A negative charge attracts a positive charge and repels an also negative charge. If you have many (negative) electrons in one end of a wire and non in the other end, then these electrons will be pushed away from each other towards the end from which they are not pushed away. This principle is the key - the electron-dense end is called the negative end, the ...


1

Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still. But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy? Sort of. ...


1

First off, there are some very misleading answers given above. Introductory quantum courses fail to properly discuss "time." It is a parameter, not an observable. E(operator)=ih(bar) d/dt has no meaning. That operator simply discribes the time evolution of a wavefuntion that is complex. So it does not describe a physical observable. I know this might be hard ...


1

No. Energy conservation always applies. The elastic potential energy will be maximum at a wavetop, since here the rope is stretched the most, $U=½kx^2$. The transverse velocity and thus the kinetic energy is zero at this point $K=½mv^2$ since this part of the rope stops and starts moving back again. $$E_{before}=E_{after} \implies K_1+U_1=K_2+U_2$$ Energy ...


1

Let one end of a very long string is being oscillated transversely so as to generate a sinusoidal wave traveling out along the string. In order to set up a wave on a stretched string, the driving force at the end of the string provides energy. This energy is not retained at the source; it flows along the string at the wave speed. The string transports ...


1

No. As has been said, the raindrop is not emitting the light, it is just acting as an optical device that deflects light emitted by the sun. However, the spectral lines you would expect to see in sunlight refracted by a prism will not, repeat NOT, be seen. The mechanism that produces rainbows is very different than the mechanism that produces a spectrum ...


1

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential: k = -1, $\alpha$ = 1, L = 0.25 k = -1, $\alpha$ = 1, L = 1 What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about ...



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