Recent content by masterjoda

They here use approximation with two monopoles but in my case I have a sphere and I think I sound me approximation with a one dipole but I don't know how to write flux then?

I think that in my country you don't need license but i'll check, this with pointed antenna can be little tricky, maybe this radio will be better for beginning, thanks.

Thanks for posts, but our project is to make live video so we can directly watch what is happening up there. Because we planed to make our balloon fly at certain altitude, and it is going to stay up there for some time so we wanted live. It is going to some kind of engine that use to "fly" to...

Yeah but they don't really talk about that radio transmitting thing. They talk more about helium and filling the balloon that I knew already.

My friend and I had an idea to put camera on the weather balloon and we wanted to have like a live picture from camera. But because we don't know so much about electronics we don't know how to do it. Fe figured out that we can do it via radio waves, maybe there is some more efficient way. That...

I think of the batteries too, but what when they discharge, I want somehow to charge them. But I don't know if solar panel would be enough.

Hi, I have a weather balloon and I would filled it with He and attach some weight and I would release He as balloon climbs and it will eventually stops at some altitude. So now I wanted compressor to acts like an engine because air is thin at 20-30km and I would put solar panels to provide...

Homework Statement I have a magnetic ball with magnetization M, mass m and radius R that starts to fall from rest through conducting tube of radius a little big larger than R, thickness \Delta and conductivity \sigma. The ball is not rotating, it said that magnetization is oriented vertically...

I have used them and I got that last equation for I.

No just R=X_L.

Homework Statement Find the current in them in this circuit, if we know R=X_L, X_C and u=5sin(314t) The Attempt at a Solution First , 5=U_0, 314=\omega and voltage we can write as u=U_0cos(\omega t + \frac{\pi}{2}) and u=U_0 e^{i\frac{\pi}{2}}=iU_0. U is the voltage at the source U_1 in...

Yes they do, potential at x have the same value as the x coordinate, V(x)=|x|, unit of that x is the same as the unit of E.

So far I have this: -\frac{\hbar^2}{2m}\frac{\mathrm{d^2}\psi }{\mathrm{d} x^2} + |x|\psi=E\psi c=\frac{2m}{\hbar^2} and k=\frac{\sqrt{2mE}}{k} solution is \psi(x)=C_1\text{Ai}(\frac{cx-k^2}{c^{\frac{2}{3}}})+C_2\text{Bi}(\frac{cx-k^2}{c^{\frac{2}{3}}}) because x is always > 0, Bi goes to ∞...

I've just realized that the right potential is V(x)=\left | x \right | \text{for } x\in \left \{ -\infty, \infty \right \}

I need to find ψ and allowed energies inside the well.