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...
I would know how to solve this problem if the person had been standing pratically above of the object underwater by using Snell's law and the approximation ##\sin(\theta)\approx\tan(\theta)## fopr ##\theta## small, but in this case I don't see how to find the angles ##\theta_1## and ##\theta_2##...
I get the same result I got in my initial answer: ##i(0)=-\frac{Lh}{R}\frac{dB(0)}{dt}##, ##i(0)LB(0)-mg=0\Rightarrow -\frac{L^2 hB(0)}{R}\frac{dB(0)}{dt}=mg\Rightarrow \frac{dB(0)}{dt}=-\frac{mgR}{L^2 hB(0)}## so ##i(0)=\frac{mg}{LB(0)}##