Recent content by laser

volume is 10 m^3

I think it is correct

Source: Shankar Yale OCW physics I have three questions here: 1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole). 2. When doing the "typical velocity" derivation, I noticed that they used...

Is this what you mean?

I'm not sure what you are asking me.

I am a bit confused on the marking scheme as attached above. P1V1 is a constant by Boyle's Law. If the volume increases by a factor of 3, then the pressure decreases by a factor of 3. This means that the pressure at the top is 1/3 the pressure at the bottom, right? The pressure at the top is...

Yes

I am uncertain if this represents situation (a) or situation (b). i.e. is 2 = E-iR or is 2 = E? Or is the question ambiguous? Thanks!

Thanks guys, for some silly reason I got mixed up with di/dt and i! The general pattern of the decrease in the induced emf makes sense now, and the fact that the decrease is exponential is from the DE.

I might have misphrased my original question... "So what happens then? IDK! My guess is that the two emfs cancel out, (which seems reasonable), and that means current is 0. Okay, fine, the book agrees with me. But what happens next? I know what is supposed to happen - the induced emf goes from...

Yes, the maths behind deriving this makes sense to me. But not the conceptual understanding. Because if di/dt = 0, then surely induced emf = 0 too! But no, the induced emf exponentially shrinks to 0...

But if di/dt = 0, then the induced emf = 0 too? As induced emf = L*di/dt. Do you see my point?

Ya but my question was that why doesn't the induced emf go to zero. Because di/dt is zero.

Let's say I have an LR circuit. I understand the mathematical derivation of this but would like to understand it conceptually. Okay, so a battery is introduced (let's say 12 V). The inductor hates the change, so it induces 12 V in the opposite direction, opposing the battery. So what happens...

So, you could just adjust the frequency of your hand movement... I don't understand what you are trying to say.