Recent content by rosedog09

I know i'm wrong because it is graded by a computer program and it told me I was wrong. Flipping the negative back gave the correct answer. I evidentially failed to properly copy the question, but it asks for q.

You are correct, that was a silly oversight. However, the more I think about it, the more I get confused. If the two spheres have different charges (q and -q), how can we use an integral that has only one q value. Additionally, I still don't see why my initial method did not work. The problem...

Thinking about it now, I realize that....although we havent learned capacitors yet. It seems to me that the capacitor formula you provided is the exact same as mine except a negative is flipped somewhere. It even says "The voltage between the spheres can be found by integrating the electric...

I have attached a copy of my work. I know that the voltage for a sphere (assuming a reference distance of infinity) is V = kq/r I also know that I am trying to find the charge given potential difference or voltage between the inner sphere (charge +q) and the other shell(charge -q), so I...

All good. I can relate atm. All my AP exams are coming up and that doesn't stop my teachers from assigning lots of homework, so lots of late nights.

Teacher says Im wrong on both these questions. I have consulted with other teachers and they say im correct. What do you guys think? 3. As mass increases, so does terminal velocity. 5. Fnet = FDrag - Fg ma = FDrag - mg FDrag = 79(8) + 79(10) FDrag = 632 + 790 = 1422 N Teacher Comments 5...

This makes sense. I didnt think about the timing of it.

Ok I think I am having trouble because I keep bouncing between force, momentum energy, etc The fan "adds" (yes I know its a conversion) energy to the system, so Energy final =/= Energy Initial within the system. Since the objects of momentum come from outside the system (air) momentum is not...

I dont know. It bothers me, but I dont know.

Ok. I think I've figured it out. The sail is reflecting the air (in this case with an unrealistic level of efficiency) such that momentum goes from p to -p, a total change of -2p. Since the fan initially transferred p to the air we are left with -p, so it is essentially the same as having the...

Ok so what your suggesting is that the momentum change (impulse) on the sail is greater than the fan because the particle is reversing directions. I can understand this, but I still find it hard to understand how there is enough momentum from the fan for it to transfer more momentum to the sail...

Ok I can see that this works, but I want to understand why. I am not understanding how the air could possibly enact a greater change in momentum on the cart after bouncing back than it did when it hit the blade of the fan (pushing the cart backwards) initially.

I understand that the sail will enact a forward force on the cart due to newtons third law from bouncing the air back, but I dont understand why this force would not be counteracted by the force which air exerts on the fan (Newtons third law).

Thank you for the helpful information. Judging by your profile and signature Im guessing your a ham radio operator. I got my technician license about a year ago, but Ive been too busy to get my radio properly setup and use it. Anyways thats really cool :) See you around.