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0

I believe your second interpretation is spot on. Whenever anything rolls without slipping, it means that the point of contact of the ball with the frictional surface is instantaneously stationary. Therefore, the frictional force applied to this point does not do any work, and so there is no frictional dissipation. Also, because the ball is still rolling ...


2

From a relativistic point of view, the energy of a particle with mass $m$ and momentum $p$ is given by: $$E^2=m^2c^4+p^2c^2$$ where $c$ is the speed of light. You can clearly see that a massive particle will have some mass energy $E_0=mc^2$, but also some kinetic energy. The photon being massless, the above equation reduces to: $$E_\gamma=pc$$ One ...


-1

Re a recent variant of the question: Physicist using mc² to calculate positive energy and −MmG/R to calculate negative energy to calculate total energy of the universe but I heard that one of them is relatively invariant and other one is not. Is that true? Physicists don't use mc² to calculate positive energy, that's merely a mass-energy equivalence ...


0

As you've indicated in your title, the correct question is "why doesn't this work". The system, as described would continue to produce energy indefinitely without any being added to it (perpetual motion, violation of conservation of energy...) . So you can be sure there's a problem. I believe that part of the trouble here is in the assumption that the ...


0

After asking lots of people, finally the right answer arrived to my inbox from Henry Reich, the creator of MinutePhysics. We can summarize the answer on: You can use lots of energy to deflect an asteroid, but The work done is zero because of the definition of work. Work="change of energy" According to Henry: All physicists mean by “work” is the ...


0

It would, as you predicted, be neither, but it would be closer to the percentage than to the constant. Friction destroys lots of space debris which have a high energy than a dropped ball. The debris loses all of its energy while the ball loses a bit. This is far from a constant by not quite a percent. In the end, it relies on many factors.


1

We have such equation: $$H = \frac{\partial L}{\partial \dot{q}} \dot{q} - L$$ You can show by calculation, that it holds in your special case, too. Now we use chain rule, and Euler-Lagrange equation: $$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}} \ddot{q} =\frac{d}{dt}\left(\frac{\partial L}{\partial ...


5

It's just conservation of energy without being called as such. $$\mbox{Energy}_{\mbox{before}}=\mbox{Energy}_{\mbox{after}}$$ $$\frac{1}{2}m v_1^2+m g h_1=\frac{1}{2}m v_2^2 +m g h_2$$ $$\frac{1}{2}m v_1^2=\frac{1}{2}m v_2^2 +m g (h_2-h_1)$$ $$v_1^2=v_2^2 +2g \Delta y$$ This seems trivial when you have calculus and a concept of energy, but without ...


9

I know that kinetic energy is conserved so $\ \Delta KE=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0$ and that $\ I_{end}=\frac{1}{3}Ml^2$ which in this case is $\ \frac{1}{3}*2kg*1m^2=\frac{2}{3}$ You are looking for $\omega (= y)$ from conservation of $KE = (1*3^2/2)$you know that $\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0\rightarrow 1*3^2 = y^2 ...


1

As with virtually all perpetual motion machines, the reason becomes obvious once you consider the thermodynamic efficiency of the components involved. No turbine is 100% efficient, and also no motor is 100% efficient. This means that out of the initial energy you put in to make the turbine spin, only a certain percentage will be converted to electricity, ...


0

If you hook up a motor to your generator output, the motor would not be able to spin fast enough to produce sufficient current to drive the generator in the first place. This is a typical case of law of conservation of energy. You may also need to check Lenz's law first. It says: "If an induced current flows, its direction is always such that it will oppose ...


1

Your conservation of kinetic energy equation should help you solve the for the stick's initial angular velocity. Think of it this way: the tennis ball has initial momentum since it is moving, right? And the stick is not moving, so it has no momentum. At the end of the collision, the tennis ball stops completely, so it has no momentum, but the stick is ...


1

Use the hint in your second bullet point. The ball has angular momentum about the pivot point before it strikes the stick.


-2

I believe I've heard Neil deGrasse Tyson say its roughly 17 miles per second to escape Earth's gravitational force in a spacecraft like the Apollo rockets that were used. So I'm assuming we're basically talking about how fast something has to be accelerating to escape Earth's mass/gravity. I believe that is correct - 17 miles per second.


3

Not sure what you mean by "exposed part" of the rope. A diagram might help. But I believe you are overthinking the problem. When the rope finishes falling off the table the center of mass is at a height $L/2$ so the potential energy lost (kinetic energy gained) is $\frac{mgL}{2}$ and therefore $$\frac12 m v^2 = \frac12 mgL$$ And $$v=\sqrt{gL}$$ It ...


0

Assume for simplicity a cavity of length $L$ in which two waves propagate in opposite directions $$y_{+}(x, t) = Ae^{i(kx - \omega t)} \tag{i}$$ and $$y\_(x, t) = Ae^{i(-kx - \omega t)}, \tag{ii}$$ with $k = \pi/L$, i.e. the fundamental mode of vibration. It's obvious that the sum of the intensities of the two waves is $2|A|^2$. Now let's find their ...


5

Virtual particles are not real. Though sounding like a tautology, it is an important one - they are not actual states in the asymptotic Hilbert spaces of a quantum field theory, where particles usually live. They are a name given to internal lines of Feynman diagrams, which, in turn, are mere computational tools in a perturbative approach to QFT. Nothing in ...


1

No, I believe you're right. Escape velocity is the velocity such that if you achieve it, you will escape. If you weigh 500N, and you strap a rocket to your back that provides 501N of force (let's ignore the engineering issues here...) you start accelerating upwards and you keep accelerating upwards until the rocket turns off. You will get to space that way ...


1

The potential energy of the ball is given as $E = mah_1$ where $m$ is the mass of the ball, $h_1$ is the height over the point you set as zero potential and $a$ is the acceleration due to gravity (which is different between Mars and Earth). If you assume there is no energy lost due to heating of the ball or other inelasticities, you have an elastic collision ...


4

I think your confusion is assuming that the energy of two waves add, when in reality there is an interference term. The short, physics-y answer is that it is not that any energy has disappeared, rather the interference has caused some of the energy that 'would have been there' to show up in a different place. The energy got shifted but not destroyed. This ...


0

In a conservative field (no dissipation of energy) whatever way follows the small sphere, i.e. rolling along another sphere, or other, the total energy (kinetic + potential + rotational) is conserved.


0

You say "Now let a particle move under a central force field where the total energy is constant. Now I want to measure the energy of the particle so I have to apply Hamiltonian and I get one of eigenvalues." By applying the Hamiltonian to the state of the particle you can predict its energy, i.e. on the paper, not measure. We measure in the lab, with ...


1

The equation you wrote $$ H|\psi\rangle=E|\psi\rangle $$ is the time-independent Schrödinger equation for an energy eigenstate. I.e., the state you are considering is already an eigenstate of the Hamiltonian with energy $E$. Therefore, as mentioned in the other answer, its time evolution is a simple phase factor, and you will always measure $E$ if you keep ...


2

If the particle is in an eigenstate of the Hamiltonian, you will get the same energy eigenvalue every time. We know that energy is conserved because the Hamiltonian obviously commutes with itself. The only time it is not conserved is if the Hamiltonian depends explicitly on time.


0

Once you measure an eigenvalue, you collapse into a particular eigenstate. Let's call that state $| \alpha \rangle$, such that $\hat{H} | \alpha \rangle = E_\alpha | \alpha \rangle$. Now we use the time-dependent Schrodinger equation: $$ i \hbar \frac{\mathrm{d}}{\mathrm{d} t} | \alpha (t) \rangle = H | \alpha \rangle \\ = E_\alpha | \alpha \rangle \\ ...


1

When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...



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