Recent content by steejk

Sorry, Tc is a constant but not T. I can see the partial derivative would hold everything else constant, but I need to find the total derivative dF/dM.

Yes that's the correct equation. Isn't T also a variable or have I just forgotten how to do basic differentiation?

Homework Statement F = 1/2.a(T-Tc)M^2 + 1/4.bM^4 I need to find dF/dM a,Tc,b are positive constants 2. The attempt at a solution I assume this is to do with partial derivatives etc. So I found: ∂F/∂T = 1/2.aM^2 ∂F/∂M = TaM - TcaM + bM^3 And using a chain rule: dF/dM = ∂F/∂T.dT/dM +...

Got it :) Thanks wbandersonjr and gneill

Okay, so from the total KE at bottom v (speed of A before collision) v^2 = u^2 + 4g And from restitution, 0.5v = v2 - v1 Do I subst. v into this then?

Any clues on where I'm going wrong..? The masses cancel in the momentum equation I think. Would the initial velocity then not be u + increase. I did the final velocity wrong, is it just x - y?

A smooth wire forms a circular hoop of radius 1m. It is fixed in a vertical plane. Two beads, A and B, of masses m and 2m respectively, are threaded onto the wire. The coefficient of restitution between the beads is 0.5. Bead B rests at the bottom of the hoop. Bead A is projected from the...

But where is the right angle triangle?

Okay so from my diagram does N = (mu^2)/a + mgcosθ?

Sorry :smile: Hmm. I know its probably something obvious but I'm not getting it. :redface: What would acosθ equal?

Ok so at bottom TE = KE = 0.5mu And at point B TE = 0.5mv² + mgh = 0.5mu² How can I work out h?

Not really sure where to start with this. I know it has something to do with conservation of energy but not really sure how to go about it.

Can it also be said that Pavg = Vrms(across resistor)/R ?

Thanks :)

Hi, Im trying to calculate the Q-factor of a series LCR circuit. I am struggling at how to calculate the average power. I want to find it using this formula: What is the RMS voltage in the formula for - is it across the resistor, or what? Any help is appreciated thanks.