FizzKicks
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The issue is that you have applied a 3-dimensional concept (i.e. Laplace's equation and potential falling with $\frac{1}{r}$) to a 1 dimensional boundary I think. The only solution to Laplace's ...

Yes, air resistance at low velocities is one such example, but I'm sure that there are others. For an object moving at a suitably low velocity, the drag on the object is given by $$\vec{F} = - b \vec{... View answer Accepted answer 3 votes The first boxed equation refers specifically to the ratio of the magnitudes of electric to magnetic fields in a electromagnetic wave propagating through a vacuum. The second equation refers to the ... View answer Accepted answer 3 votes There are lots of cases outside of QM where linear algebra can be of great use. If we have a coupled oscillator, then the equations of motion can be solved using linear algebra with ease. In special ... View answer 2 votes Interesting thought about using gravitational waves to communicate - I find it to be exceedingly unlikely, but I like the creativity. As to whether or not they could be blocked, not really - when ... View answer 2 votes Probably the simplest one is the case of an object which experiences an acceleration which is constant in its own frame i.e. a rocket in deep space, where the change in mass of the rocket due to fuel ... View answer Accepted answer 2 votes This is quite an odd way to introduce the position operator, I have to admit. Both definitions you have used are correct, they're just used in different ways in quantum mechanics. In the the first ... View answer Accepted answer 2 votes Your suggestion is correct in someways but incorrect in others. A better representation of the drag force would be$$\vec{F}_{drag} = \alpha (V_x(t)\space\vec{i}\space +\space V_y(t)\space\vec{j})$$... View answer 2 votes I think you're grossly over-complicating things here. I've listed the convention below for most high-school and undergraduate level treatment of uncertainty.$$\\$$When a measurement is made with an ... View answer Accepted answer 1 votes We can rewrite the differential equation as:$$\frac{1}{\lambda}\frac{dn_{eff}}{d\lambda}-\frac{n_{eff}}{\lambda^2}=\frac{d}{d\lambda}\left(\frac{n_{eff}}{\lambda}\right)=-\frac{n_g}{\lambda^2}$$So$$\...

I think I have solved this by realising that we can define the energy by the invariant $E^{(B)} = \textbf{u}_{obs} \cdot \textbf{p}^{(B)}$, where $\textbf{u}_{obs}$ is the observer who is measuring ...

So for these kind of problems, I'm no programmer by any means, but the easiest way to proceed seems to be to use Newton's second law, $F = ma$, with $F = Force$, $m = mass,$ and $a = acceleration$. ...

I will confess that this I am not certain on this, as this is a quite odd question. but I have formulated what seems like a very reasonable answer. Also the very first thing to mention is that one ...

Okay so this image is a bit daunting, and by no means should you discount Terrell Rotation (it's one of the most interesting parts of relativity in my opinion), but this should help give an idea of ...

Remember the notation; what you have written above it the expectation value of the operator, i.e. $$⟨\psi|Q|\psi⟩ = ⟨Q⟩$$ is clearly a scalar, but we can see from the operation of Q that it is a rank ...

You are forgetting that the weight attached will gain kinetic energy as well, taking away from the energy that the cylinder can have for rotation, and hence lowering the final angular velocity.

If we consider only the radial component in the case when $\dot{\theta} = 0$, then it becomes clear that $\ddot{r}$ is a radial acceleration, i.e. since the angular component of velocity is $0$, the ...

I've put down an answer that should help you but doesn't dive deeply into the mathematics, if you're feeling confident I would suggest trying to solve it using equations of motion and coupled ...

One of the weird things about light is that in vacuum it always moves at the speed of light through space-time, which means that there is nothing we could do to make it go faster. So if you were to ...

I think what you are looking for is an understanding of 4-force, that is the vector which describes force in relativity. Without getting into the details (which are easily accessible by a google ...

I don't see what is necessarily wrong with leaving it in the form with $v_f$ and concluding that it is proportional to $m^\frac{1}{2}$, but nonetheless we can at least come up with some physical ...
We have the general solution $$y(x,t) = \sum_{n=0}^\infty sin(\frac{n\pi x}{a})(b_ncos(\frac{n\pi ct}{a})+c_nsin(\frac{n\pi ct}{a}))$$ For a string constrained to $y(0,t) = y(a,t) = 0$. We can then ...