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8

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 ...


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 ...


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 ...


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 ...


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 ...


3

Whether energy is or isn't conserved in an expanding universe is a somewhat vexed issue. On the one hand you have an experienced physicist claiming that energy is conserved, and on the other hand you have an experienced physicist claiming that energy is not conserved. The problem is that accounting for energy in general relativity is a complicated business. ...


2

Quoting Sean Carrol's article linked by Симон Тыран, which makes the case for energy not being conserved: Having said all that, it would be irresponsible of me not to mention that plenty of experts in cosmology or GR would not put it in these terms. We all agree on the science; there are just divergent views on what words to attach to the science. In ...


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.


2

Long story short: conservation of energy only holds locally where you can assume a static spacetime. On large scales the expansion of the universe gets relevant, so energy is said not to be conserved universally since the amount of dark energy per volume stays the same while the volume increases, see Sean Carroll's article, from which I quote: The famous ...


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

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 ...


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

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 ...


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 ...


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 ...


1

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


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

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, ...



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