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0

You can combat that the same way we avoid getting too cold: apply insulation. The outer surface of the spacecraft may be very cold, but that doesn't mean the internal temperature is that cold. That is what the greenhouse effect does-it insulates the surface from space. Power dissipated inside the body all becomes heat. If you have solar arrays making ...


2

Interesting and complicated question. The things to consider: "Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer ...


1

The force applied to the spring is: $$F = -mg$$ where m is the mass, g is the acceleration due to gravity near the earth's surface (9.8 m/s^2) The equation relating distance and force for a spring is: $$F = -kx$$ where k is the spring constant and x is the distance the spring is stretched from equilibrium. When the mass is attached to the spring and ...


1

You are doing fine. You are correct to say the sun only shines on one side of the spacecraft, which the page you link to misses. Increasing the radiating area by a factor of six will decrease the temperature by a factor $\sqrt[4]6 \approx 1.565$ Dividing their $285$ by $\sqrt [4]6$ and multiplying by $\sqrt [4]{1.36}$ (to correct for your more accurate ...


0

KERS systems (as used in F1 and leMans cars) rely on a flywheel to store kinetic energy. When braking, the transmition is clutched trough a 1:n gearbox to the flywheel. That is, one full revolution of the car's wheels will make the flywheel turn n revolutions. The flywheel system will then store the energy until it is needed. Since a flywheel system rotates ...


7

You are right on track. A great way to think about the Hamiltonian is that it's the thing that "causes translation in time". This just jargon for the fact, which you already observed, that the Hamiltonian tells you how the system moves forward in time, as encapsulated by the Schrodinger equation $$i\hbar d_t |\Psi\rangle = H |\Psi \rangle . $$ My guess ...


5

An experimentalists answer: If you divide the energy of the electromagnetic wave by hv you will have the number of photons that are building up the electromagnetic wave.


1

Part 1. Essentially you're right. You can think of it as subtracting infinity from the energy. A better way to view it is that the convention that the zero of energy should correspond to potential at infinity was always an arbitrary choice. Normally it is a very sensible convention, but if the potential diverges at infinity, as is the case for the harmonic ...


11

You only need to rewrite $\mathbf B$ and $\mathbf E$ in terms of field $A_{\mu}$ (here $\hbar = c = 1$), $$ \tag 1 \hat{\mathbf B} = [\nabla \times \hat{\mathbf A}], \quad \hat{\mathbf E} = -\frac{\partial \hat{\mathbf A}}{\partial t} - \nabla \hat{A}_{0}, $$ which is written as infinite "sum" of photons: $$ \tag 2 A_{\mu} = \sum_{\lambda} \int ...


1

The distinction to be made here is that, for the quantum harmonic oscillator system, there are no unbound states, only bound states thus, there is no benefit to insisting the states have negative energy, no reason to 'subtract infinity' in order to zero the potential at infinity. However, in systems that permit both bound and unbound states, it is ...


1

This is admittedly a late answer. Hopefully it will clear up some of the confusion. If you ignore aerodynamic drag, ignore that the coefficient of rolling friction varies with load, and ignore a number of other factors such as friction between the axle and the bearings that support it, then yes, stopping time / stopping distance is independent of vehicle ...


0

Your question asks "what is the energy..."? So I would consider the energetics of the system rather than forces and torques. If the objects starts from rest at A and end up moving with an angular velocity at point B, then there is a change in gravitational potential energy and a change in rotational kinetic energy. If you then halt the rotation at B then ...


1

Use Friis' formula http://en.wikipedia.org/wiki/Friis_transmission_equation to estimate the received power. Notice that it has the transmit and receive antenna gains and these are nearly ~1 for simple dipole (or monopole) antennas that are omni-directional in a plane perpendicular to the antenna current flow. Given the antennas the received power is ...


2

Section 2.9 (bottom half on page 11) of this document describes why a 5GHz router suffers greater attenuation than a 2.4 GHz router, particularly so in residential and office settings. Attenuation is greater at 5 GHz than at 2.4 GHz for signals that need to pass through walls, doors, and glass. From the above reference: More energy per photon does not ...


0

I was confused about this but it turns out that the context of this equation is important. The energy change from an electron in one energy state to another can be determined by using the formula $\Delta E = h * \nu$ which will give the energy difference of the two states or the energy of the emitted or absorbed photon. However if you are just determining ...


1

For simplicity let's imagine a hydrogen atom, which has one proton and one electron. The energy difference $\Delta E$ you speak of is the difference in energy of the two initial and final states of the electron-proton system. Initially the system is in some state $\psi_i$ with energy $E_i$. Afterward, it's in a different state $\psi_f$ with energy $E_f$. ...


1

There's a sign error. The final potential energy should be less than the initial mechanical energy in the spring by the energy lost due to friction. If you're subtracting the friction energy from the final energy, it's the same as if you're adding the friction energy to the initial energy. I.e., you're saying that you're expecting the system to gain ...


0

KERS is also known as regenerative braking. It is a system found on the latest hybrid cars. When a vehicle is stopping, the brakes are applied to the discs to generate friction to slow it down. This generates heat which is energy that is lost or wasted. Regenerative braking creates friction in addition to the brakes, however, some of the energy is ...


2

Well, I suppose that you are well aware that neglecting air friction the work done on the bar will be $MgL$. This much energy is spent by the person on lifting the bar. This question might be helpful. It says the efficiency for a human body on average for cycling is 20%. So it is a safe bet to say that the man used $5MgL$ energy, spent $MgL$ on lifting ...


1

In this case, the eigenstates of the Hamiltonian are not useful to solve the problem, and one has to work with the Schrödinger equation directly: $$ i\hbar \, \partial_t \psi(x,t)=\frac{-\hbar^2}{2m}\partial_x^2\psi(x,t)+P\,t\,\psi(x,t) $$ Using a Fourier transform in the variable $x$ you can show that the general solution is $$ \psi(x,t) = ...


0

For a particle in a time dependant external potential any Hermitian the Hamiltonian will still have a complete set of eigenstates, and the corresponding eigenvalues will still be the allowed energies of the system. These energies and eigenstates will however generally be time dependant and this means that they are of far less use for solving the TDSE. For ...


5

This is unphysical. There is no force or momentum that would make something unstoppable. First of all, any moving object can be made to diverge from a straight line path simply by applying a small force perpendicular to its motion. Secondly, no matter how large a momentum an object has, it can be completely halted by a similar object with the same momentum ...


2

In flat space the surface area of a sphere is $4\pi r^2$. In positively curved space the surface area of a sphere is less than $4\pi r^2$ and in negatively curved space the surface area of a sphere is greater than $4\pi r^2$. By $r$ I mean that if you start at the centre of the sphere with your trusty (infinitesimal) ruler and measure the distance to the ...


1

This addresses the broader question of potential energy minima, not the specific situation of a conductor/magnet/current. You often see statements that lower potential energy states are preferred in some way. Personally I think there are a few issues with such statements, especially for systems in which only conservative forces act. Nonetheless, I think ...


0

I was looking for an answer to the same question as that asked here but I wans't fully satisfied with the given answers, so I kept looking and a good answer that I stumbled upon is the one given here: http://www.gaussianwaves.com/2013/12/power-and-energy-of-a-signal/. Hope it will serve to others well as it served to me. Cheers


0

I spent some time pondering this question many years ago, when I had run up a pretty steep mountain and found that my Garmin gave me a very disappointing "calories burnt" result ("you didn't go very fast, so you didn't do a lot of work"...). My reasoning went like this: if I climb a mountain, I will be doing work against gravity. Muscles are about 25% ...


1

The problem is that the main reason for the consumption of power is the inefficiency of human muscles. If I run a marathon on level ground then I shouldn't have used any energy at all because my potential energy hasn't changed (there will be some loss of energy to air resistance, but at the speed I run that's negligable). However experience suggests that ...


3

Remember how work is defined. The key word is displacement. I think that when you wrote $Fd$, you considered $d$ to define a single point, and not displacement. Now back to your problem. The work done by a net force is $$W=Fd=mad=ma{(}\frac{v_{f}^{2}-v_{i}^{2}}{2a})=\frac{mv_{f}^{2}}{2}-\frac{mv_{i}^{2}}{2}=\Delta E_{c}$$ The single diference is that I ...


4

In principle, sure. That's what microphones are, as ACuriousMind points out. But if you want to power anything substantial, an important issue to overcome is the relatively small amount of energy contained in sound waves. According to this website, the front rows of a rock concert have a sound intensity of $10^{-1}~\text{W m}^{-2}$. So even if you had a ...


8

Most of the electromagnetic energy is in x-rays which means it is deposited in the bulk material of the bomb and in the surrounding air over a few meters (or tens of meters at most). All that stuff heats up, needs to expand but piles up against other stuff also trying to expand. You get a massively energetic shock-wave of very hot material which is in ...


0

firstly you should not make an analogy with a wheel since electrons are not orbiting the nuclei with equal angular velocity. it is more like satellites orbiting the earth because in both cases the force is proportional to $r^{-2}$. farther the satellite, it has less velocity. to compare the energies of the shells you should consider electric potential energy ...


-1

An electron cloud, is a zone of high probability. If you have an electron orbiting a proton, the cloud is an area where it is extremely likely to find an electron. You can't stick your finger in and pull out an electron. The cloud is just an area where the electron is likely to be found. More on that here. The electron(- charge) wants to be as close as ...


0

Imagine you have a hot tub, and you heat it up to a nice toasty temperature. Then the power goes out. The metabolism of the hot tub environment is now zero. It won't get any warmer. But if you get in the tub, you'll still warm up. The thermal mass of the water won't be cooled much by you entering. You're taking advantage of the heat that was produced ...


3

The collapse of the wavefunction is not a real physical process. It's a feature of a particular interpretation of quantum mechanics, the Copenhagen interpretation (CI). Other interpretations, such as the many-worlds interpretation (MWI), don't have such a collapse. The different interpretations make the same predictions about all observables, and therefore ...


0

Kinetic energy $K$ depends upon speed $v$ as $K=\frac12mv^2$. The energy of the electron is the sum of its kinetic energy $K$ (which is positive, $K>0$) and its potential energy $U$ (which is negative, $U<0$): $$E=K+U$$ Inner, core electrons have a larger (positive) kinetic energy and also a larger (negative) potential energy than outer, valence ...


9

As you say, the power produced per cubic metre of the Sun's core is surprisingly low. This is because proton-proton fusion is a very slow process, as has been discussed hereabouts before. The core is so hot because conduction of heat through the core is slow. The average speed with which a photon escapes the core is the astonishingly low value of about ...


0

If you were to "magically" place a planet in the sun's core, I'm fairly sure that is would not be there long. The ambient temperature of the sun's core is somewhere around 15.7 million K, as you said. You should think about why its so how there before you think about melting planets. The density of the core is something like 150 times denser than water. When ...


0

The entire system will be balanced and not move, as the momentum is conserved. There however will be an small shake back and fourth. First when it fires so the momentum move the rocket and the next one where the gas will hit the other side and conserve momentum and move it back to original location of starting point.


0

Are you wondering why your two equations $W=\frac{1}{2}kx^2$ and $W=Fx$ don't match? That's because the latter equation, $W=Fx$, is only true for a constant force. The more general expression is $W=\int \vec F \cdot d \vec x$, similar to what you wrote in differential form $dW=F\,dx$. The expression with $\frac{1}{2}$ in it is correct. The expression with ...


1

You can only measure a part of EU compliance. By using your utility meter and measuring the difference between the power consumed over say 10 minutes with the appliance on and off (with everything else in the house as off or steady as possible), you can measure consumed power. However, that's just one part of EU compliance. The other is power factor. ...


2

Your requirement that the measurement be made with equipment available in a kitchen is a severe constraint as I can't think of any way of measuring the electrical power supplied. If it's impossible to measure the electrical power in then the only other approach is to measure the thermal power out - i.e. measure the heat produced by the appliance. Given that ...


0

In general relativity, the energy content of a region is given in terms of a stress-energy tensor. The elements of this tensor are not given by general relativity itself and can differ depending on what matter and fields are present. To try to draw general conclusions about what is allowed and forbidden in general relativity, physicists have tried to place ...


0

Negative energy occurs when the energy level for a given space is below that which is considered zero energy. A zero energy space is not realy zero but is always full of some virtual particles popping in and out of existence.


1

'The mirror is given a momentum twice that of the incoming photon. As a mirror is typically quit heavy, lets say one gram. Its kinetic energy due to momentum it received will be extremely small. However, the photon will actually change its energy by the same amount, thus its wavelength changes, but not much.


3

Since the photon reflects, its momentum changes: $p_{ph}'=-p_{ph}$. But total momentum of the system is conserved: $p_m+p_{ph}=p_m'+p_{ph}'$. Thus, the mirror will change its momentum. But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass $m_2$, the energy of ...


-3

the waves will obliterate each other but they will still exist, they just won't be moving they would just change form (energy cannot be destroyed it can only change form) so when the waves meet they will cancel each other so sound will change to potential and kinetic will change to sound or whatever


3

If the bell is still vibrating when you let air inside it, then the answer is yes. If the bell was damped just before the door is opened, then the answer is no. Sound is transmitted through compression / decompression waves (pressure waves) in a medium (e.g. air, water, wall). This necessitates contact of the vibrating source of sound with such a medium. ...


-3

If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called "photon" or "wave packet". Linking the appropriate formulas from QM and E&M waves, you get the diameter of the wave packet (about λ/2) but not the length. The radius and the direction of propagation do not change as long as the wave packet is not disturbed. ...


0

In General Relativity, energy momentum flows from one region of spacetime to another. But there isn't necessarily a natural "total energy of the universe." It might help to contrast General Relativity with other theories. In Newtonian mechanics, a particle might gain kinetic energy while a corresponding gravitational potential energy decreases, thus you ...


1

In Newtonian mechanics, a particle might gain kinetic energy while a corresponding gravitational potential energy decreases, thus you get that kind of conservation of energy. The total energy is the same before and after any event. However, the amount of energy depends on who's looking. In Special Relativity a transfer of energy has to happen at an event ...



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