Recent content by L_ucifer

No I mean I thought that since time is slowed for the muons when looking at them from a ground frame, the muons would actually take longer to reach the Earth's surface. This is why in my solution the time for a muon to reach the Earth's surface is greater in the relative solution than in the...

I put gamma there because I thought it'd take more time for the muons to reach the ground, instead of thinking of the muon's lifetime becoming dilated.

Edited the question and adding the same changes here: Here's the overall equation for the correct solution: No = 300,000 t0 = 2 * 10^-6 s t = distance / speed = 10,000 / 0.98c N = No * e ^ -t/(t0 * gamma) Here's the overall equation for my solution (which is wrong): No = 300,000 t0 = 2 *...

TL;DR Summary: I got this question on a quiz for a Coursera course on special relativity, and I'm confused about the answer. I've detailed my thinking below any help on where I went wrong would be greatly appreciated. Question. Muons are unstable fundamental particles. In its own rest frame...

They haven't mentioned that in the question

Summary: How do I solve the derivative of (((x^{x})^{x})^{x})... How would I solve this derivative?

That makes sense, thanks.

Here is the question: We know the equation \epsilon = \frac{d\phi }{dt} = BAcos(\theta ). This means that the only way we can create an induced voltage is if we change the magnetic field, change the area of the loop in the magnetic field, or change the angle between the normal vector to the...

Well, this has gone horrendously for me and I do apologise for these silly errors. You live you learn, I guess. Ignoring all my logistical errors, is there any solution to my question?

Yes, I am not sure why R and a have been switched, but the radius of the sphere is a.

I apologise, I meant E = (ρ0 * a3) / (6 * ε0 * r2) for r ≥ a

I understand part (a) of this question, and my answer for that part is: *For r < a* E = (ρ0 * r4) / (6 * ε0 * a3) * For r ≥ a* E = (ρ0 * a3) / (6 * ε0 * r2) Now, for part (b), I understand one solution is, for r < a, find the work done to bring a point charge q from infinity to a and then from...