# Tag Info

6

Update: According to this paper, "On the Interpretation of the Redshift in a Static Gravitational Field", the answer I give below is a common but misleading interpretation. The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in ...

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 ...

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 ...

2

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, ...

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

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

When you pluck the string, you impart energy into it that's slowly radiated as sound. There are ways to radiate the energy faster, in which case the string loses energy faster. You're increasing the power and decreasing the time, so energy stays constant.

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

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 ...

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

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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 ...

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