Recent content by Jamie_Pi

Oh, I think that you're right. That's good, that just clears up the whole thing then.

I'm trying a new formula, but I've come to a point of stagnation: dφ/dt = dI/dt ⋅ 1/σ φ has units vm so dφ/dt must have units vm/s 1/σ is the same thing as Ω, which has units v/A and dI/dt has units A/s vm/s = A/s⋅ v/A the only problem I have now is that unit of length, which I don't know how...

Homework Statement A wire with conductivity σ carries current I. The current is increasing at the rate dI/dt. Show that there is a displacement current in the wire equal to e0/σ⋅dI/dt Homework Equations Id = e0⋅dφ/dt dφ/dt = dE/dt ⋅ A (This is usually true, I'm not sure if it's useful in...

Oh I see, I just wrote it in wrong. If the first two are correct, then the last one should just be: b2/(ax+bt)2 = v2 a2/(ax+bt)2 where v2 = b2/a2

∂2D/∂t2 = -b2/(ax+bt)2 ∂2D/∂x2 = -a2/(ax+bt)2 b2/(ax+bt) = v2 a2/(ax+bt) I think this is correct this time.

Ok, I understand. This is what I have so far: ln(ax+bt) = D(x,t) ∂2D/∂t2 = b22/(ax+bt) ∂2D/∂x2 = a2/(ax+bt) b2/(ax+bt) = v2 a2/(ax+bt) This is true as long as v = b2/a2 Do you think that this is enough for an answer?

Homework Statement Show that the displacement D(x,t) = ln(ax+bt), where a and b are constants, is a solution to the wave function. Homework Equations I'm not sure which one to use: D(x,t) = Asin(kx+ωt+φ) ∂2D/∂t2 = v2⋅∂2D/∂x2 The Attempt at a Solution I'm completely lost on where to start...

Homework Statement One cylinder in the diesel engine of a truck has an initial volume of 640 cm3. Air is admitted to the cylinder at 30°C and a pressure of 1.0 atm. The piston rod then does 500 J of work to rapidly compress the air. What are its final temperature and volume? Homework Equations...

Well, since the pulley is 2kg and it is multiplied by 1/2, it became 1 and wasn't really doing anything. Ahh, but of course! It's being added not multiplied. torque=0.06*(8.1+1)*a 0.242/(0.06*9.1)=a a=0.4432 (looks good so far) time=sqrt(2/0.4432) time= 2.12 Yay! Thanks a bunch! Again.

Ok, that makes sense. I'm not sure what to represent them as, though, because they aren't actually spinning at all. I tried representing them as points rotating around the central axis, and that got me somewhat close: torque=(1/2*2*0.06^2+(4.8+3.3)*0.06^2)*a/0.06 which I simplified into...

Ok, I re did it and I must have done some math wrong before. Torque=0.242 If I understand your suggestions correctly, I think representing the masses as part of the pulley is the best idea, so I'll do: moment of inertia= 1/2(2+4.8+3.3)*0.06^2 So from there, torque=moment of inertia * angular...

Homework Statement The two blocks, m1 = 3.3 kg and m2 = 4.8, in the figure below are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.64 N · m. If the blocks...

Ok, I see. Thanks! You've been a big help.

Ok, I drew this: (sorry, poor quality) And from the top of the hill, it seems that the marble falls as far as h is, which makes sense. So is the potential energy at the top of the hill equal to mgh, but at the top of the loop equal to 2(R-r)? That would make the equation: h=(R-r)(1/2+1/5+2)

Is it mg(h-r)? I'm not really sure what to do other than that.