Recent content by lorenz0

I have solved the problem, but I am not sure if my solution is correct, so I would appreciate if someone would double check it. Thanks (a) Using an inertial reference frame in which the ##x## axis points to the left and the ##y## axis points downwards and computing the torques with respect to...

@kuruman Got it. Thank you very much

That was my initial guess (that the phase shifts cancel out), and in that case it seems to me that ##\Delta=\frac{2n_{soap}d}{lambda_{soap}} lambda_{soap}=\frac{2 n^2_{soap}d}{lambda_0} lambda_{soap}##, so it still is not in agreement with the stated solution

The ray that is reflected at the air-soap interface has an optical path length of ##lambda_{soap}/2##, while the one refracted in the soap has an optical path length of ##2 n_{soap} d+\frac{lambda_{glass}}{2}## due to its traversing the soap film and also reflecting on the soap-glass interface...

You are right.

Corrected. Thanks again.

Thanks!

Done.

Typo. Fixed, thanks.

I think I have completed the exercise but since I have seldom used polar coordinates I would be grateful if someone would check out my work and tell me if I have done everything correctly. Thanks. My solution follows. Since ##\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1## it follows...

Ah, I think I have understood now. Since ##V(t)=V_0\sin(\omega t)## it must be that ##B(t)=B_0\cos(\omega t)## with ##\omega=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{2\delta t}=100\pi\ \frac{rad}{s}## so that ##V(t)=V_0\sin(\omega t)=-N\frac{d\phi(\vec{B})}{dt}=-N\frac{d}{dt}\left( B_0\cos(\omega t)S...

The alternating current will create an induced voltage given by ##V=-N\frac{d\phi(\vec{B})}{dt}=-N\frac{\Delta\phi(\vec{B})}{\Delta t}=N\frac{2 B_0 S}{\Delta t}=\frac{240\cdot 2\cdot 65\cdot 10^{-3}\cdot 10\cdot 10^{-4}}{10\cdot 10^{-3}}=3.12 V## and since this is the effective voltage, the...

ah, of course! I don't know how I didn't see that before, I was thinking I hadn't understood the situation correctly but it was just a typo, many thanks.

Thanks for your interest in my question. I had already included my work, the only thing left to do was to show the formula I had derived with the numbers plugged in, which I have now done. The number under the square root is actually positive, according to my calculator.

Since the forces involved (gravity and electric force) are conservative we can use conservation of energy. The initial energy is ##E_i= k\frac{q_1q_2}{r_0}-G\frac{m^2}{r_0} ## and the final ##E_f=mv^2+k\frac{q_1q_2}{2r}-G\frac{m^2}{2r} ## so from ##E_i=E_f ## we get...