# Tag Info

## New answers tagged energy

0

The fastest possible way to do a lane change is to fully steer one way on the traction limit and then steer on the opposite way again on the traction limit. The traction limit is $\mu g = \frac{v^2}{r}$ where $g=9.81\;{\rm m/s^2}$ is gravity, $\mu=0.8\ldots0.9$ is the coefficient of friction (half it in the rain), $v$ is the speed in meters per second and ...

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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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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 ...

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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 ...

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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 ...

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The following might help: $H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$ decays exponentially with time along the solution of the damped system. Check by integrating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.

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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 ...

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From an engineering perspective, there are many different fan designs including axial and centrifugal configurations, along with various blade designs including forward curved, backward curved, and radial. There is no formula to calculate the required fan speed, but if a specific fan configuration is known then fan similarity laws can be used to calculate ...

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Using a dimensional analysis indeed can help you find dimensionless numbers as The Dark Side suggested Floris might help with, but beyond that there is no direct analytical method, no closed form solutions that can relate fan speed, shaft torque, flow rate and delta pressure. The issue is that at best the flow behavior is two dimensional, but more likely ...

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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. ...

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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 ...

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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 ...

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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, $$... 1 Energy is required to push something over a distance. In the case of the elevator, when it's not moving a brake can be engaged and the power removed and the elevator will sit just fine. That's because it doesn't take any energy to keep something still. But wait, if all I want to do is keep the helicopter still, then it doesn't require any energy? Sort of. ... 4 When you find the total (squared) value of some vector in an orthogonal basis, such as the Cartesian system (x,y,z) or indeed the spherical system (r,\theta,\phi), what you're doing is simply adding the squared values of each component of the vector. Taking the velocity, let's think about the different components: What is the velocity in the radial ... 9 There is an effortless way, if you accept geometrical reasoning. You know, that T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2. Furthermore, spherical coordinates are orthogonal, therefore you can just write:$$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$Geometrically, one easily finds: v_r = \dot r, v_\theta = r \dot ... 2 A negative charge attracts a positive charge and repels an also negative charge. If you have many (negative) electrons in one end of a wire and non in the other end, then these electrons will be pushed away from each other towards the end from which they are not pushed away. This principle is the key - the electron-dense end is called the negative end, the ... -1 I think it is possible to produce equipment converting kinetic energy from a bullet. We could use a string attached to a generator and a spring. The string then turns the generator and thereby produces electricity but it is to hard to to something that does not get damaged from the bullet. Here we can use momentum. If it is supplied to create thick but empty ... 1 Actually, you are all way over-thinking the issue. There is a trivial material that can slow a bullet down without being damaged at all. Think water. A couple feet of a water shield will do exactly what you need to stop a bullet enough to make it basically harmless with a bit of additional bullet-proofing. Most of the bullet's kinetic energy will stay in the ... 2 What is your goal - to stop bullets or to generate electricity? The key to stopping bullets is not so much to absorb the energy (although that matters too) but to absorb the momentum. You may know that$$p = F\Delta t$$In other words, given a certain momentum mv=p, you need to apply a force F for a time \Delta t in order to slow it down. The ... -4 There is no physical challenge (at first glance). This is a technological issue. Man just didn't developed technology to do this (and they are not trying). 0 Your material has to absorb the kinetic energy before converting it into other forms. A thin, lightweight panel would be unable to absorb the bullet's kinetic energy before it punched a hole in the panel and continued on it's way. Plot the tensile strength of the material against the impact energy of the bullet. Unobtainium, impossibrium, phlebotinum and ... 1 Here's another angle to this: it depends on your latitude. If you are standing somewhere near the equator, you're also experiencing circular motion around the Earth's axis of rotation with an angular frequency of 2*Pi/86400 and a velocity around 463 m/s. In the rotating frame of reference you'll suddenly find a centrifugal force$$|\vec a| = \omega^2 R = ...

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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 ...

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The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at ...

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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 $\psi = c_1\psi_1 + ... 2 Your assessment of the transitions which can occur, and hence the photons which can be emitted, is correct. However, the colliding electron does not go to one of the energy levels in the atom (as Sebastian already correctly pointed out). What happens is that the colliding electron can deposit its energy in the bound electron, 'promoting' it from the ground ... 1 Now I assume that the question is asking the following: When an electron of energy 12.1eV collides with this atom, photons of three different energies could be emitted. Show on the diagram (with arrows) the transitions responsible for these three photons. Because from one single collision the emission of three photons doesn't make much sense to me. The ... 2 Photons can be emitted when electrons change energy levels. You say that you have worked out where a 12.1 eV difference is. In an ordinary hydrogen atom, the electron will be in the$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 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 ... 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 ... 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 ... 1 What you seem to be referring to (a guess from your mention of quantum tunneling) is perhaps near-field magnetic induction. Here's a summary - magnetic coils have a near-field region around them, the primary contriubtion to which involves minimal radiation of power. This region is dominated by the response of the coils (for example) to the fields, and is ... 0 You have to find out what kind of statistical ensemble you are dealing with. As soon as you know that, you can get the corresponding thermodynamical potential from the knowledge of the partition function. When you know the potential, you know everything! EDIT: Since you don't know to which ensemble this partition function$\mathcal{Z}$belongs to you ... 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 ... 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 ... 1 It's not really worthwhile in this type of situation. (It makes sense in other situations however ... like transferring power from the ground to an airplane or satellite.) The two most plausible system types are: (A) Microwaves / radiowaves: Emitted by an antenna, collected by a rectenna (B) Visible / infrared: Emitted by a laser, collected by a ... 1 I think you're referring to the tunneling effect. If you have two states of low energy (here: single nucleus/two nuclei) with a high energy barrier in between (here: highly deformed nucleus), then it's possible to observe transitions from one state to another, even if there is insufficient energy in the system to climb the energy barrier. A non-mathematical ... 1 Most fissile materials have some probability of spontaneous fission. For example in uranium-235, seven out of every billion decays are fissions. These spontaneous fissions are the reason why a critical mass of fissile material may spontaneously develop a fission chain reaction. Alpha particles incident on beryllium-9 will break the Be nucleus into two ... 1 Uranium 235 is naturally radioactive, with a half life of 703.8 million years. So if you take a lump of uranium 235 there will be nuclei decaying and releasing neutrons just due to its normal decay. These neutrons will then cause other nuclei to decay, and off goes your chain reaction. So you don't need anything to start the reaction. All you need to do is ... 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 ... -1 First of all Galilean Transformation laws are not valid in most cases. For the moving reference frames the special theory of relativity is more suitable. In fact it is the most suitable one given the experimental data. In special relativity one defines a so-called energy-momentum four vector, whose magnitude every observer in an inertial reference frame ... 1 I'm going to have to give an answer that's very different to Jimmy360's. Apologies. How does the potential and kinetic energy of a photon relate? They don't. The photon is all kinetic energy. Do they mean the same thing? No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic ... 0 Quote from a webpage a bit over my head :-) greatians.com Photon has linear momentum. Photon travels in vacuum space at the ultimate speed of light. Photon has the quantized energy of hf as given by eq. WD.1.2. E = hf … eq. WD.1.1 where h = Plank’s constant ... 0 Imagine a beam of light, going towards a massive object. It has potential energy in the gravitational field. Of course, the potential energy has to become kinetic energy. This is done by shifting frequency. The energy of a photon is given by$E = hf$so to increase kinetic energy we must increase frequency. If the beam of light was red, it will be a higher ... 1 I understand the problem now from leongz's excellent explanation and patience in the chat discussion. Thank you very much leongz for helping me out! The capacitor is being discharged through constant current. If it starts with charge Q_0, then from the definition of current we know that the charge decreases by It after time t, where I is the constant ... 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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You want to consider the following factors: Maximize initial energy stored in the rubber band. This means you need to be able to twist the band lots of times, and as it unwinds it must continue to produce torque. Minimize internal friction - make the mechanism that converts power from the band to kinetic energy as "direct" as possible. In fact a rubber ...

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If you have a flux of electrons, the transmitted power is proportional to the number of tunneling electrons. The other electrons are reflected from the barrier. Thus, the barrier just redistributes the incident electron flux into two ones. It does not change the individual electron energy.

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Suppose a box is pushed in a vacuum.Due to this push the box at rest starts to move.But in a vacuum there are no other forces which oppose the motion of the box.So it will continue to move.Or to think in energy terms,once we gave the box energy there is no mechanism for the energy to transfer from the box.The energy has been "trapped" in it.

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