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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.


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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 ...


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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.


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'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.


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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 ...


-2

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


2

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. ...


-2

Some kinds of energy are well understood. Heating or cooling a body means the transfer of atomic motion from the body with the higher temperature to the body with the lower temperature. The motion transfer happens through electromagnetic fields. Every body transmit and receive nonstop electromagnetic waves. Whenever you post two bodies near each other the ...


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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 ...


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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 ...


3

A few thoughts to help you on your way. When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work times velocity). If you are changing the velocity of an object, you are changing its kinetic energy: if it's ...


2

As for the helicopter problem, theoretically, arbitrarily low power can be sufficient to float a load, if you push down a lot of air with very low speed. However, you need longer and lighter (and maybe wider) blades for that, so the problem you'll have to solve is that of structural strength. Let me note that recently a muscle-driven helicopter ...


7

Is it possible the: is ke meaning kinetic energy? I ask because the equation for the de Broglie wavelength of a non-relativistic particle is: $$ \lambda = \frac{h}{\sqrt{2m_0 KE}} $$ where $KE$ is the particle's kinetic energy.


4

The energy is conserved but it becomes "lumpy"- more in some places (directions), less in others. Total over all directions is the same.


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1) Yes, work is being done by your muscles. 2) Work is equal to change in Kinetic, Potential, and Heat energy: Kinetic - No change. Potential - Increase in gravitational potential energy accounts for some of the work Heat - Friction in muscles/bones, shoes on the ground, and in the air resistance all produce heat, which is where all the work that didn't ...


3

The work done by a force $\mathbf{F}_1$ when a particle travels a path $\gamma : [a,b]\subset \mathbb{R} \to \mathbb{R}^3$ is defined by $$W(\mathbf{F}_1, \gamma) = \int_\gamma \mathbf{F}_1 = \int_a^b \mathbf{F}_1(\gamma(t)) \cdot \gamma'(t) dt$$ I wrote that way just as a way to make clear that the work depends both on the path and on the force. If ...


0

Since $W = F*d$, you have to be doing work, as the force of friction is being applied over a distance. The reason why your speed is not changing is because the work is mostly being used to increase your gravitational potential energy instead of kinetic energy.


2

The easiest way would be to use an energy monitor device (the most popular one seems to be the Kill-A-Watt but there are others). They simply plug into the wall and then you plug your appliance into it. It displays instantaneous voltage, current, power, power factor, etc. and can keep total energy over time. Another option would be to buy an electric ...


2

You have a rather precise power meter in your home, which is a "gift" of the electrical power company. Turn off every other load that is connected to that power meter and do your measurement. Alternatively, you can invest $20 in an electronic power meter that is available online and in many stores.


7

It's the water itself that forms the lens. Lenses work via refraction. The refractive index of water is about 1.333, which is different from the refractive index of air (about 1.0), so rays of light bend at the junction of the air and the water.


2

The question is what do we need the matter content of the universe for. As I understand it, in the usual case we want to find the conserved quantity associated with a certain conserved current gained by the projection of the energy-momentum tensor into a Killing vector, as for example in the paper by Abott and Deser. The requirement of asymptotical ...


-1

The expansion of the universe is not a force. Forces don't pull things apart at a particular speed; they change the speed by a particular acceleration. The speed itself is just inertia. It's no different in cosmology: there is nothing actively pulling things apart at the speed given by Hubble's law; that speed is just leftover momentum from the big bang, as ...


0

The expansion does lead to a kinetic energy term that can be (at least partially) extracted. For objects that are bound to each other, it would lead to a classic acceleration term, which, of course, is equivalent to a classic pseudo-force. In an expanding universe, any two objects that are bound by a potential, are therefor experiencing an additional (albeit ...


1

The notion of kinetic energy is ill-defined in the spacetimes where you have a time-dependent cosmological expansion. If you somehow attached two galaxies to each other with a spring, however, the expansion of the universe would do "work" against that spring, as there would be a force requied to keep the proper distance of the two galaxies fixed. At ...


1

There is no way one can answer this question without further requirements. In practice, engineering considerations will put wireless charging of vehicles probably somewhere between the 20kHz and 150kHz range (unless the regulator permits a higher frequency). Why 20kHz? Because all wireless charging solutions will involve the generation of some amount of ...


0

For what it's worth (I cannot verify the claims): http://www.j.sinap.ac.cn/nst/EN/article/downloadArticleFile.do?attachType=PDF&id=448 (NUCLEAR SCIENCE AND TECHNIQUES 25, 020201 (2014) - I guess this is a Chinese journal). Abstract: "Considering the mixture after muon-catalyzed fusion ($\mu$CF) reaction as overdense plasma, we analyze muon motion in the ...


2

You may define a conserved total stress-energy tensor (matter + gravitation). The main problem is that a conserved total stress-energy tensor is not covariant, and that a covariant stress-energy tensor is not conserved. Said differently, $\nabla^\mu T_{\mu\nu}=0$, which is a covariant equation, does not represent a conservation law, while $\partial^\mu( ...


0

Eric Angle has it pretty much right. In an inelastic collision some of the kinetic energy is absorbed by the deformation of the material. For example, if two balls of putty collide and stick together, kinetic energy is absorbed by squishing the putty. In a second example, if you shoot a bullet at a log, some of the kinetic energy is absorbed by friction ...


1

(add my comments as an answer) Per @Jim's comment, i would say Einstein's equations are a way of relating energy and geometry (which in turn is interpreted as gravitation). But is important to understand a difference between energy and (underlying) geometry, because when geometry turns into gravitation things can get weird. There seems to be a confusion ...


1

Suppose your initial state is $\lvert 2\rangle$ and that the states $\lvert 0 \rangle$ and $\lvert 1 \rangle$ have lower energies than $\lvert 2 \rangle$. Assuming that there is no so called selection rule that prevents $\lvert 2 \rangle$ from emitting a photon and end up in $\lvert 0 \rangle$ or $\lvert 1 \rangle$, then the final state will be $$ \lvert 2 ...


3

As far as dimensional analysis goes, temperature and energy are separate and independent physical dimensions. However, there is a more or less unique way to translate temperatures into energies and vice-versa, which is by means of Boltzmann's constant $$k_\text{B}=1.380×10^{−23}\:\text{J}/\text K.$$ Any given temperature $T$ has an associated ...


0

In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon be affected by gravity in by Newton's equations? What am I missing? You are forgetting that Newton thought that light is made by particles which at the time were called corpuscles see here, and you surely know that ...


8

The expectation value of energy is something else than the energy in a particular experiment. With your choice of the initial states, the photons emitted (negative difference) or absorbed (positive difference) will have energies either $$ E_1-E_0 \text{ or } E_2-E_0 \text{ or } E_1-E_5 \text{ or } E_2-E_5 $$ If each of the four transitions were equally ...


1

As you stated already, a measurement of the energy of the hydrogen atom must return an energy eigenvalue. Measuring before and after a transition gives us two energies $E_n$ and $E_m$. This is always true, regardless of the fact that the expectation value of the energy before measuring might not be a difference of $E_p$ and $E_q$ for some $p,q$: The actual ...


3

Newton's law does predict the bending of light. However it predicts a value that is a factor of two smaller than actually observed. The Newtonian equation for gravity produces a force: $$ F = \frac{GMm}{r^2} $$ so the acceleration of the smaller mass, $m$, is: $$ a = \frac{F}{m} = \frac{GM}{r^2}\frac{m}{m} $$ If the particle is massless then $m/m = 0/0$ ...


4

While people normally quote Newton's Second law as $\vec F = m \vec a$, it is better written as $$ \vec F = \frac{d\vec p}{dt} $$ Force is a rate of change in momentum. This means that the average force applied when an object undergoes some discrete change in its momentum is $$ F_{\text{avg}} = \frac{\Delta p }{\Delta t} $$ The change in your momentum ...


0

I'd like to answer the first question "Is light affect by gravity? Why?" because I remembered an adorable thought experiment. In the spirit of the equivalence principle consider an observer in a closed box. As we all know, the observer inside that box would be unable to tell (neglecting tidal effects) the difference between a uniformly accelerated box ...


0

1) The bending of light rays is a general relativistic effect, not one due to Newton's law of gravity. 2) It's probably better to think about these things from a field perspective -- a distribution of mass-energy moves along, and it creates a gravitational field. Then, when things enter that field, they interact with it, and this changes their motion. ...


1

A free particle (photon, electron etc) is not restricted to discrete energy levels. Its wavelength and therefore energy can take any value. A plane wave, with any wavelength you choose, can satisfy the Schrodinger equation for a single free particle. A free particle does not have to have a single precise energy. For example, a wave packet (a quantum wave ...


2

There are several things in play here. A photon emitted as a result of a transition of electrons between two well defined orbitals in principle has a well defined state However, the uncertainty principle limits how well that state can be known. You have shown yourself familiar with the energy-time formulation of the uncertainty principle; a transition ...


2

Photons coming from changes in the energy level of an electron in a bound state ( atom,molecule,lattice) come in discrete energy slices. emission spectrum of iron Photons coming from the radiation emitted by accelerating or decelerating charged particles are coming in a continuum spectrum. Bremsstrahlung radiation Spectrum of the X-rays emitted ...


1

Highly recommend you to have a lesson on electrodynamics. It will show you the origin of Einstein's idea of special relativity. And then with the special relativity theory, it can be easily derived that E=mc^2. There's no coincidence in these two formulas. Or, actually you can say that there's always coincidence in physics because, they are just different ...


7

If you take an isolated hydrogen atom then the electron sits in well defined atomic orbitals that are eigenfunctions of the Schrodinger equation. This is a stable system that doesn't change with time. If you now introduce an oscillating electromagnetic field (i.e. light) then this changes the potential term in the Schrodinger equation and the hydrogen ...


1

Transparent media are transparent because the incoming photon does not match any of the available energy levels to transfer its energy to the atom, or molecule, or crystal. A classical analogy is thinking of energy levels as various size sieve holes which allow only certain size of particles to go through. Not any matter of knowing or adjusting, but ...


0

Energy of the electron changes with its interaction with photons, if it reaches more it gets excited, else remains in its position even though interacted with the photon, also it comes back during decay to release the photon. Hence photons don't know the energy level they produce, but it happens by interaction.


0

I would think that the energy spent to make a deadlift is related ( maybe in a non-linear way ) to the work done to move the bar. If you define work as always: $$ W = F\Delta h = mg \Delta h $$ then you can define the energy spent by a machine as $$ E = \frac{L}{\eta} $$ where $\eta$ is the efficiency of the machine. Now, no one can tell exactly what is the ...



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