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0

Historically, one motivating experiment for kinetic energy was dropping balls into clay and noting the relation between impact speed $v$ and the impact depth $d$: $$v^2 \propto d$$ I suspect this experiment could be modified to show the role that mass plays and obtain $d\propto mv^2$. A different experiment would be to launch a cart using some type of ...


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There are a lot of things in play here. Certainly ICE(internal combustion engines) are less efficient when accelerating than when running at an optimized RPM. However, the interplay between external drag, gear ratio, and engine speed is rather complicated. Typically there's an optimum speed for maximum efficiency; higher speeds have too much air drag and ...


1

Lets attempt some answers: Both can happen, a quantum transtion can be associated with a photon exchange or a photon exchange can be associated with a quantum transition (this is just 2 ways to state the conservation of energy in these cases) Photons do not have mass but they do have momentum. There are some approaches in physics which associate a virtual ...


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You can apply an infinitesimal force to accelerate your 1kg infinitesimally so that it has a non zero velocity, and allow it to travel 1 metre in an infinite amount of time, by using an infinitesimal amount of energy. Or you can apply a huge force to accelerate your 1kg mass quickly, which consumes a large amount of energy. In other words your question is ...


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I think you're seeing two types of errors here. For low values of $Q$, there is a lot of damping, and thus the values of $\phi$ are low (and decreasing). I'm not sure what the absolute values of $X$ are, but it might be that they become sufficiently small to make machine errors relevant. For higher values of $Q$, higher values of $X$ arise, and ...


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If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.


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We need to create a battery that would instantly store a large amount of electricity at one time. Ex. When a bolt of lighting strikes it gives off a very large amount of power. However a battery needs time to take that energy and change it over to a chemical for storage. Lets say a bolt of lightning is 500 gallons of water and the battery that we ...


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This is not an answer, but is there any contradicción with Noether theorem to say that Kinetic energy gives diferent value according to system of reference?


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We don't know! If we were to observe a situation where energy conservation did not appear to work, that would be a major puzzle. As you say, either we would have to discover some alternative contribution to the energy that we had been neglecting, or we would have to give up on energy conservation. A priori it is not obvious which one of those two resolutions ...


7

Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ...


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The center of mass moves as though all forces act there - so the two cases will result in the same (linear) velocity of the center of mass. However, in the second case you also get rotation. No conservation of energy is violated because it you apply the force for the same (short time), the distance moved in the second case is larger (because the rod starts ...


7

Steam is caused when water vapor condenses. This is caused by the air having too much water vapor for it to hold. When you have a lot of heat under the pan, the air above the pan is quite hot and can hold a lot of water. The water evaporating from the pan disperses into the atmosphere and doesn't condense. When you turn off the heat, the pan and food ...


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As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.


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Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. So, if there is force acting on a unit charge, then electric field does exist. It is the way by which we can prove the existence of electric field (as per definition demands). I don't ...


1

Well, if he doesn't slow down, then he'll hit the Earth at 86,000 mph. So, $$KE = \frac {mv^2}{2}$$ $$KE = \frac {158.3 kg\times (38445 m/s)^2}{2}$$ $$KE = 1.17\times 10^{11} N$$ I have no idea how to calculate the size of the carter it creates. Added: Now for a calculated guess: Assuming that half the energy is used to heat the ground and propel the ...


3

A closed system can not speed itself up, that's the momentum conservation law which is also the key to your problem. As far as I can see you are implicitly supposing the following three equalities to hold $$E_i+A=E_f\\p_i=p_f\\m_i=m_f$$ where subscripts $i$ and $f$ stays for the initial (before acceleration) for the final (after acceleration) states ...


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Short answer: Time. Long answer: The conjugate quanta of energy is time, one can look at energy as the generalized momenta associated with time. With that is mind one can go to QM essentials and proclaim: $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$ As a consequence, one can say: $$ \frac{1}{E}\propto t $$


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One day visiting a science demo years ago, there was a display of the effects of a standard drinking straw put through a block of wood simulating a tree trunk. The display was in regards to the velocity of thrown objects from a tornado. A tornado can increase the velocity of otherwise harmless object to travel through just about anything. It can ...


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All matter is made up from atoms. While it's very hard to accelerate a spaceship to nearly the speed of light we can collide two atoms (well, two nuclei) at speeds approaching the speed of light. This experiment is done at the RHIC and also at the LHC in its heavy ion mode. So what happens when we collide nuclei at nearly the speed of light? Well, the ...


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I believe anything with relativistic velocities and much mass (let alone a spaceship) would quickly be destroyed by "junk" in space (when I say "junk", I include stray hydrogen atoms.) To get an idea of the energy released when something like that hits something (like a planet), in Randall Monroe's latest book What if, he states that a baseball with a ...


1

When a straw or a hose punches through a tree, that isn't due to the object having a far greater mass due to the tornado, because it doesn't. At the 318 MPH speed of an F-5 tornado, the relativistic mass of an object is only about 1.00000146 times as big as its rest mass, a negligible difference. A more useful explanation is that the ability to punch ...


0

I didn't see anyone mention the practical reason to use an approximation for energy. It is that in most problems you will be computing differences in energy. In that case, for small velocities, you can not only go the approximation ${m v^2\over 2}+m c^2$, but if you are also not converting mass to energy or vice-versa, you can drop the $m c^2$ as well ...


2

A direct answer: Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. The equation $E^2 = (mc^2)^2 + (pc)^2$ does work for all values. As other answers have explained, at low speeds (or equivalently, low momenta), $E = \frac{1}{2}mv^2 + mc^2$ gives approximately the same ...


2

Experimental data is given in http://journals.aps.org/pr/abstract/10.1103/PhysRev.98.889 - unfortunately I only have access to the abstract. It may be worth taking a look. The shape of the cathode does not matter. The material does. Key to solving this problem is knowing the work function of the material - that is the minimum energy that an electron needs ...


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If the body's speed $v$ is much less than $c$, then the equation reduces to That is an imprecise wording, and the cause of your confusion, I believe. The correct description is that the second equation is an approximation of the first, and it will be closer to accurate the lower $v$ is. The approximation is important for understand the connection ...


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Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. Yes, it would be nice if we could find an equation that was "universal". In the case of mass m, energy E, and momentum p, there is such a universal equation: $$E^2 = (mc^2)^2 + (pc)^2$$ Einstein, always wanted to ...


1

When we lift those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture? There are two weights with mass $M$ and $3M$. Initially, both weights are at the same height $h_0$ which we can freely set to zero: $h_0 = 0$. Now, machine B lowers mass $M$ to ...


1

The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is $A=-m\int \sqrt{1-v^2}{\rm d}t$ . Lagrangian is $\cal L$$=-m\sqrt{1-v^2}$, then you can get momentum $p$ from derivative with respect to $v$ (this is the real momentum, not $mv$). Energy $H=E=pv-\cal L$ , just ...


5

If you put $p = \gamma mv $ where $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ and $\beta = \frac{v}{c}$ in Einstein's equation, you get $E^2 = (pc)^2 + (mc^2)^2 = (\gamma mvc)^2+ (mc^2)^2 $ $= (\frac{c^2}{c^2 - v^2})v^2(mc)^2 + (mc^2)^2= (\frac{v^2}{c^2 - v^2})(mc^2)^2 + (mc^2)^2$ $ = (\frac{c^2}{c^2 - v^2})(mc^2)^2 = (\gamma mc^2)^2$ $ \therefore E = ...


2

Special equations of the kind you mention are useful as they elucidate limiting behaviors. Your example is no different from a statement like: the hypotenuse $c$ of a right triangle with legs $a$ and $b$ with $b \ll a$ is given by $c = a + \frac12 b^2/a$. The Pythagorean relation $c^2 = a^2 + b^2$ is valid for any right triangle, but the special case $b ...


3

First, your findings are correct and are found by Einstein. However the fact that a new term appears is only showing that our previous knowledge was incomplete. The new term expresses the Energy related to the rest mass of the body which was not considered before. And it makes sense because before, in equating the Energy of a body only were included those ...


11

First, the non-relativistic equation $$ E= mc^2 + \frac{mv^2}{2} $$ is equivalent to its second power, $$ E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4} $$ If $v/c\ll 1$, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic $$ E^2 = (mc^2)^2+ (pc)^2 $$ which ...


0

OK, I think I've got it, thanks to your comments above as well as this link, which shows how to calculate the temperature of a solar oven. (My situation is very similar to a solar oven, except that the power dumped inside the craft is electrical -- but watts are watts, right?) So, I believe that what I need to do is: Calculate the steady-state ...


0

"If gravitational energy is meaningless in general relativity ..." This is a false premise. Gravitational energy can be understood perfectly well in general relativity. The problem is just that many people who answer here have failed to understand it. "(since it is the geometry)" There is no meaning whatsoever to the statement "gravitational energy is the ...


1

When we think about the state of a hydrogen atom we instinctively think about the solutions to the time independant Schrodinger equation. These are the well known atomic orbitals. However for the time independant Schrodinger equation to apply the hydrogen atom must have existed unchanged for an infinite time and it will then continue to exist unchanged for ...


2

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 ...


3

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 ...


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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 ...



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