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43

This is actually the paradox that led Einstein to the equivalence of mass and energy, and to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up to the Space Station, where they are converted ...


16

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down. This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain ...


13

I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect ...


9

...two roads of the same length. One road is flat, the other road goes up and down some hills. Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one..? The question two roads (between two towns A, B) with a different profile cannot be of same length The flat road (1) is shorter, ...


8

Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ...


4

First, it's not true that energies are generally in the form $\frac{a_1a_2^2}{2}$. Take the gravitational potential energy $U = \frac{GMm}{r}$ as an example. However, it is generally true that kinetic energy takes that quadratic form. Why? Kinetic energy is the energy traded when some agent applies a force on some system that causes it (the system) to ...


4

.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ...


4

Any energy principle is not being violated since the speed of the photon is never less than $c$ and hence the momentum is unchanging (in the classical sense). Why light travels slower than $c$ in a medium is because of the photons being absorbed and reradiated by atoms in the material. In a sense you can make the analogy of light traveling a longer path in ...


4

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


4

If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$ mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...


3

Conservation of energy, as you note, holds for "the system." For instance, if you push on a ball, that ball gains energy, but the energy of the ball is not conserved--only the energy of you and the ball. In this case, the system needs to include more than just "the sloth" because the sloth is not an isolated system--there are external forces at work. Here, ...


3

The link Qmechanic has suggested is a duplicate and does discuss the question you ask. However there is another point that is worth making here. In general relativity we describe the universe as a manifold equipped with a metric, and the metric is the FLRW metric that desciribes expanding spacetime. However the FLRW metric does not include the point(s) at ...


2

You need to use the equations of motion in the expression for $dE/dt$. Lets take a simple example. For the Hamiltonian $$\mathcal{H} = \frac{1}{2}\dot{x}^2 + V(x)$$ (which is equivalent to the Lagrangian $\mathcal{L} = \frac{1}{2}\dot{x}^2 - V(x)$) we have that the time-derivative of the energy is $$\frac{d\mathcal{H}}{dt} = \dot{x}\ddot{x} + ...


2

A nice answer to this question may be found in the 2009 MIT Course on the Physics of Energy (Lecture 3) by R. Jaffe and W. Taylor. They work out the energy balance of car transport as an example of mechanical energy and its conservation. Let me briefly recap their findings in my own words. The two "towns" they consider are Boston and New York, both assumed ...


2

For example, given the initial peak height of the roller coaster, I can predict the velocity at any point, despite the fact that there are various loops and curves It is very simple: if you know the height $h$ of a body you know its potential energy which is $mgh$. This energy is given to the body (transformed into Kinetic energy) at every lower ...


1

This hopefully improves on an approximation made in the other answer, while using the same hypothesis (doubtful in practice) that the scale has no damping or friction no energy is lost when the mass impacts the scale the scale's moving mechanism has negligible mass compared to the mass dropped I use the same notation except for the maximum reading of the ...


1

As the comments to the questions state, this is a question on the research state about the generation of the universe, and the first moment is modeled in the Big Bang model. The real beginning point is not yet known even in this model since gravity has not been consistently quantized within the model, only effective theory is used. Nevertheless existing ...


1

Well your question was not perfect, but acceptable. the idea of energy may sound easy, but deeply it is a very strange Idea. but the answer to your question: no it doesn't mean it "potentially" gonna gain. I see that you read Feynman's lectures, that's very good, but these speeches actually are for people who accepted the idea of energy blindly, and didn't ...


1

Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...


1

For collimated light, E=cu (do you see why?). For light traveling uniformly in every direction, if you have an imaginary plane, cu/4 passes through in one direction and cu/4 in the other. This is the basis for the relationship you mention. (The relevant mathematical fact is that if you have a sphere of radius 1, the average z-coordinate over the z>0 ...


1

The emissive power of a blackbody is $\sigma T^4$ - the power per unit area from its surface. This is derived by firstly establishing that the flux is the integral of the Planck function $B_{\lambda}$ (which is a specific intensity, in units of Watts per square metre per metre per steradian) over the solid angle subtended by radiation outwards into a ...


1

There is no net energy being transferred. The energy each wave has is cancelled by the other wave! You might be confusing this with the "two slit" experiment. In this case you do have both, destructive and constructive interference. This is the result of two simultaneous waves with a given spatial separation causing the waves to "cancel" each other at ...


1

Your question is similar to the common question of putting a wind turbine on an electric car to generate electricity while driving and provide power, and is based on the same conceptual misunderstanding. To understand why it is not possible you should read about why perpetual motion machines are not practically possible. Specifically relating to your ...


1

When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as? Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and ...


1

I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic ...


1

Look at the original, Heinrich Hertz, 1889! If E=B=0 on a plane, S=0 => no energy Transport through that plane. I agree with Annix: "The pictures may be depicting a static wave, but certainly, not a propagating wave."


1

In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means ...


1

From a very fundamental point of view one can see this with respect to Noether's theorem. Every symmetry is related to a conservation law, e.g. the symmetry of space with respect to rotational symmetry (space does not change if we rotate our coordinate system) results in conservation of angular momentum. In this framework energy is related to time symmetry. ...


1

You actually don't need to take any time derivatives here. Since the energy, $E$, is a constant, you only need to know it at one moment in time (say at t=0 as an initial condition), and you know it at all other times. Thus, just solve for $\dot{x}(t)$ in terms of the other quantities ($x,E,...)$ in the second equation that you have to get the equation of ...



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