# Tag Info

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I won't answer on the energy part of the question, but here a few remarks which in my opinion are worth adding: The laws of physics are not frozen, the evolve (more precisely, they get more and more accurate ; e.g. Newton, replaced with the general relativity, etc.). Maybe one day someone will come up with an even more general theory, a part of which is ...

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Time travel is not impossible because of conservation of energy, time travel is not impossible at all. Entropy does not allow you to go backwards in time, but forwards does not present any problems.

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according to some theories (like big bang) the planets are parts of greater bulks and as gradually they collide and teared apart (after becoming cold) they have gotten a speed due to the explosion and then they have stuck in the gravitational field of bigger planets and stars and rotating in vacuum around them so they don't lose their velocity due to some ...

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Let me answer another component: where the initial energy for their movement came from. Imagine two bodies separated by a large distance. In this case, the gravitational pull is small and the gravitational potential is low. Their relative velocities are just about zero. For all intensive purposes, our energy accounting is zeroed out. KE=0 GE=0 (kinetic and ...

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In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa. Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth ...

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Energy conservation stems from Noether's theorem applied to time (i.e., time-invariance leads to energy conservation, similarly to how spatial-invariance leads to momentum conservation). Since the universe is expanding (and accelerating at that), the state of the universe today is different than it was yesterday and will be tomorrow, hence energy ...

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Here is a simple model as explanation: Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing. When you want to ...

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What we like to call the energy, i.e., the total matter/energy content of space-time, might not be conserved. However, there is a lot of reason to suspect that fundamentally the universe is some big quantum system, and that space-time and particles and fields are emergent from this underlying idea. In that case, we expect there to be a Hamiltonian $H$ and ...

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Provided that there is nothing for the photon to interact with (i.e. we look at it in vacuum), the mean free path will be infinite; that is, it will continue travelling forever in a given direction. There's nothing which will stop the photon's path. Hence, it will go arbitrarily far. Whether you have a single photon or a laser, the answer won't change. The ...

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The H.E.S.S. Gamma ray observatory in the Namibian desert has sufficient collecting area that it is able to detect (via Cerenkov radiation) very rare high energy gamma rays that were simply not available to space observatories with comparatively low collecting areas. H.E.S.S. detects high energy gamma rays from all sorts of astrophysical objects and ...

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Newton's first law states a particle will have constant velocity unless an external force acts upon it. The photon has no mass, but nonetheless the first law still holds true in the case of light. When a ray of light is projected, (say) from the surface of Earth to outside in space. The condition is that, there is no obstruction to it till infinity ...

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The answer depends on many factors, but here are the basic bits of physics that play: The power and wavelength of the laser The reflectivity of the surface (function of wavelength of the laser) The size of the focal spot The thickness of the sheet The thermal conductivity of the sheet The reflectivity of the copper is a particularly important one. If you ...

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Well it all depends on what assumptions you are able and willing to make. For a start you can work out how much of the laser light is reflected. The reflected power will be something like (for normal incidence) $$\frac{P_{r}}{P_i} = \frac{(\eta_m - \eta_{vac})^{2}}{(\eta_{vac} + \eta_m)^2},$$ where $\eta_{vac}=377$ Ohms and $\eta_m$ is the impedance of the ...

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There are so many variables here i think its nearly impossible to give an answer like how thick is the lazer, how far away is the lazer how reflective is the surface of the metal ......... the more variables there are the harder something is to predict if you wanted an answer it would be much more practical to get one by doing the experinment, probably not ...

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The distance that a particle can travel is partly set by its mass. If the particle has a mass less than something like 7 eV, then it could cross the universe without attenuation.

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When do we see particles to be in a superposition of energy states? From the fundamentals of quantum mechanics for "seeing" , i.e. measuring an energy, the result will be one unique energy eigen value "to every observable there corresponds an operator which acting on the state function will return an eigenvalue". So for an individual particle the answer ...

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Whether it be a beam or ray of light, photons will keep traveling until they are absorbed. Photons can't stop because they travel at a constant velocity, the speed of light, i.e., they can't accelerate or decelerate. However, their wavelengths change over time due do the expansion of the universe, i.e, their wavelengths get larger and loose energy as such ...

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One small addition to the other answers: While it is indeed true that the light will never stop if it doesn't hit anything, it will however get red shifted, and thus become less energetic, due to the expansion of the universe. For example, the cosmic microwave background consists of photons which were emitted back when the atoms formed. However, back then ...

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A "ray of light" must be respelled as "photon" because here we are talking physics. Between a single photon and a laser beam, in this case, there is no difference. Every photon will continue his travel until stopped, every single photon is "indistinguible" from others (in the sense that they are no different intrinsecally). The photons of a laser beam are ...

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A ray of light or a laser beam will not stop until it reaches an obstruction. If there is no obstruction, light will NEVER stop. It has no end.

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Note that it is correct that a photon can travel an infinite distance in an infinite time, but it can not reach any desired point in the universe. This is caused by the expansion of the universe, which also leads to the fact that we can not receive information outside of the observable universe.

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Theoretically, the photon (or the beam of photons, there really isn't a difference) can go an infinite distance, traveling all the while at a speed $c$. Since photons contain energy, $E=h\nu$, then energy conservation requires the photon to only be destroyed via interaction (e.g., absorption in an atom). There is nothing that could make the photon simply ...

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*elastic collision occur between atomic particles? inelastic collision occur between ordinary objects? perfectly inelastic collision occur during shooting? super elastic collision occur during explosion?* as John Rehnnie has explained, an elementary particle as the term implies, is not made of other particles, has no lattices ...

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The difference is only in the properties of the material of a body. You can see in this video If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored in the lattices If it is not elastic the body will stay ...

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$$L= 1/2(-f(x)\dot{t}^2 + g(x)\dot{x}^2 + 2l(x)\dot{x}\dot{t})$$ I assumed from the form given that $L$ is a function of $x(\tau), \frac{d x}{d\tau}, t(\tau)$ and $\frac{d t}{d\tau}$ for some parameter $\tau$. Therefore we have the Euler-Lagrange equation for $t$ and one for $x$ \frac{\partial L}{\partial t} = ...

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The only factor is the capacity to return to the original shape when it is deformed by an external force. An elastic object recovers its shape after it has been compressed and deformed, if you crush a plastic bottle and remove the cap it will partially return to his original length, a lead ball is almost completely irreversibly deformed. The Coefficient ...

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I do not see any other possibility than doing the integral along the path. Do your telemetry data include the horizontal position? because if does you can calculate an approximation to the integral. Of course, there is no guarantee that the error will be larger than the effect you want to measure. You should have to do some pre-tests (using situations that ...

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No, the elasticity of the collision depends on a couple of factors. 1) the properties of the materials. The carpet is easily deformable, and the fibers are not elastic, if one removes the ball that fell they return to their original form after some time, or, even don't return. So, the kinetic energy of the falling ball does the work of deforming the fibers. ...

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Intensity has units of watts per area: $$\left[I\right]=\rm\frac{W}{ m^2}$$ where the area is the surface area of the emitting source (in this case, the sun). This tells you the total amount of radiation present (over all wavelengths). The extra factor of 1/nm in your plot gives the spectral irradiance: $$\left[\mathcal E\right]=\rm \frac{W}{m^2\,nm}$$ ...

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$W/m^2$ would be the total energy emitted, regardless of wavelength. When you use $W/m^2/nm$ you are explicitly saying that it corresponds to a specific part of the spectrum (nm is a unit of wavelength). Which is what the graph you posted is showing. The first one is called "irradiance", the one plotted here is called "spectral irradiance". For more details ...

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This follows on from my answer to your previous question: Factors on which Coefficient of restitution depend. The coefficient of restitution of a collision depends on the available degrees of freedom for energy to be lost. If you take your example of the collision of atomic particles, let's say two electrons, then there isn't anywhere for the initial ...

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A purely elastic collision is a theoretical construct, which helps to simplify some situations. In all collisions, some energy is lost as heat, and thus the collision is said to be inelastic. Perfectly elastic collision does not occur between atomic particles Your problem here is that you are assuming atomic collisions can be represented by classical ...

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Most of us will have discovered that a glass rings if you tap it, and the sound of the ring will persist for several seconds after hitting the glass. This shows that elastic energy stored in the glass is not dissipated fast. I would guess that fewer people have tried the experiment with a lead beaker, but I have and I can report that the lead does not ring. ...

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Not at first, but usually eventually. You should read the Wikipedia article on Annihilation, which is the name of the phenomenon of a particle's colliding with its anitparticle. Particles and antiparticles have opposite-signed, equal magnitude quantum numbers (such as spin and parity to name two). Collisions must conserve these quantum numbers, so the ...

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When a particle collides with its antiparticle, all of the mass in both is converted into electromagnetic energy (usually high energy gamma radiation). Whilst it is certainly possible to then convert this gamma radiation into heat so as to drive a turbine, it may be also possible to convert gamma radiation directly into electricity using a special type of ...

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Sometimes when you're stuck on things, it's helpful to look at the mathematics of what's being asserted. For example, nowhere in Newton's three laws does "energy is conserved" appear. Energy conservation does appear, however, when you have a system that behaves like $m \ddot{x}=-\nabla U$, for some function $U$, where $x$ is a position vector as a function ...

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Derivation that applied work as defined above results, for a particle moving along a straight line, in a change in its kinetic energy (I hope it is not too complex to understand): In the case the resultant force $F$ is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a ...

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To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The charge of an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential. The key to answering the question (part d in the ...

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You could convert h to 4135 e-21 keV/Hz and use that in your calculations, or 4135 keV / (e21 hz), and use that. Your calculation is then 20 keV / 4135 keV * 1e21 Hz or 20 keV / 4.135 *1e18 Hz.

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0.16 aJ is the energy an electron acquires by going through 1 volt. If it goes through 500,000 volts, it gets an energy almost equal to its mass.

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Let me answer your question with a question. How would you use $E$ as it is to calculate $f$? Suppose you plug $E = 20\text{ keV}$ into the formula and solve it for $f$. \begin{align} f &= \frac{E}{h} \\ &= \frac{20\text{ keV}}{4.135\times 10^{-21}\text{ MeV s}} \\ &= \frac{20}{4.135\times 10^{-21}}\times ...

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No. Kinetic energy depends on how much energy you give to an electron. $Volt = Work$ $done$ / $unit$ $test$ $charge$. $1.6 * 10 ^ -19 J$ is the amount of work done to accelerate an electron of charge $1.6 *10^-19$ to $1$ $volt$ potential.

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A lot of the answers get distracted by the "feeling pain" part of the question. So let me start by focusing on that - simplifying a very complex psycho-physiological issue to a simple physical statement (this is a physics forum - we leave the other stuff for other sites): During the impact of two bodies, there will be an exchange of momentum. The force ...

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What is the difference that leads to conservation of kinetic energy in elastic collision ? The difference is only in the properties of the material of a body. If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored ...

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The water inside is a fluid, so it isn't rigidly attached to the walls of the bottle. This means that the bulk of the water will still accelerate at $g$, save for the part of the water close to the bottle walls, which will be dragged along with the bottle. The water isn't really rising up, it's just falling slower than the bottle. In the frame of the ...

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The simple answer is that in an elastic collision (for objects >> in mass than typical molecules) energy moves from kinetic to potential then back to kinetic as long as the "elastic limits" of the materials are not exceeded. In other words, as long as they act like springs. In non-elastic collision the energy goes mostly from kinetic of the colliding masses ...

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When one says that "kinetic energy is conserved in an elastic collision" that means that the total kinetic energy of the system of particles involved in the collision doesn't change. It does not mean that the kinetic energy of each particle is unchanged. For a two particle system, the kinetic energy of each will change, but the sum won't. Also, your ...

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The simplest answer is that this has to do with symmetry. In the immediate vicinity of the droplet, the water is isotropic. Therefore there is no preferred direction. In more rigorous terms, the density-density correlation function for liquid water is rotationally and translationally invariant. Now, if one were to change the geometry of the problem in some ...

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You can also see the exact same phenomenon when you see a bubble. Bubbles always occupy spherical spaces. This occurs due to the fact that a ripple expands in every direction at a constant rate so naturally the resultant shape the ripple encloses is a circle. This might also be related to the fact that for any given area, the smallest perimeter is bounded by ...

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Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.

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