Recent content by Truman I

Awesome, thanks for the help!

Yeah I definitely had a misconception about current there. So at t=0, the 75 ohm and 25 ohm resistors act like they are in series. 150 Volts will drop across the 75 ohm resistor (Ohm's Law). Therefore, because the 15-mH Inductor is in parallel with the 75 ohm resistor, there will also be a...

At t=0, I believe that the current is instantaneously 0 Amps. If that is correct, then technically at that instant there is no voltage drop across any of the resistors due to Ohm's Law. So I replaced the resistors with wire. Next, I tried replacing all of the capacitors with open circuits to...

It is possible that the problem is not referring to this force. It's hard to know what exactly your physics class is doing right now without being in it myself. Now that I think about it, if you have not learned this equation yet, you may want to solve it the way Orodruin suggested above.

Thanks for your replies. They have been very insightful. We went over a few more things in class today that I can now connect everything together. I can solve the problem now. Thanks again.

You are absolutely correct with kinetic energy. That is exactly how you are supposed to find it. As for the opposing force, it's not as hard as it seems. But first, let me give you some definitions and a new equation. The change in momentum (Δρ) is also called Impulse (some textbooks will use...

So the solution is saying that the wire could break somewhere along where the red here touches the outside of the circle. Perhaps the best way to go about this problem is process of elimination. We understand that C and D are obviously wrong. D because it just doesn't make sense, and C because...

Are you saying like this? Initial: Rod: mass=6kg, velocity=0m/s, ρ=0kgm/s Clay: mass=1kg, velocity=5m/s, ρ=5kgm/s Final: Rod and Clay together: mass=7kg, ρ=5kgm/s, so therefore velocity=.714m/s If that's all I need to do with linear momentum, then I understand that. If there are more steps...

This part makes sense. I can work my way through that. Thanks again.

Thanks for your response, I'm starting to understand this a bit more. Yes, I do mean the center of mass of the whole system. What I don't understand right now is how linear momentum comes into play for this problem. Could you explain that?

Yeah, you guys are right. My bad. I thought he was saying that he did it like this: Σƒy=mac -ƒtension-ƒg=mac If you solve out that equation, you get an answer that is a little greater than 8N. He said that his answer was a bit over 8N, so I figured this was how he did it. Thanks for clearing...

Is the velocity of the center of mass the same before and after the collision? Because then I could just find it with the center of mass formula and use the 5m/s ν0 to solve that part of the problem.

Yeah, there really is not much to go on for this problem. The rod is not fixed. The clay hits at one of the ends of the rod. r is meaningless, it was just me assigning a variable out of desperation because I have no idea what I am doing. To be honest, my attempt at the probably is probably...

I think Rijad understands this. What he is having trouble with is the sign (+/-) for ƒg and ƒTension. Maybe the best way to explain it is that any forces pointing in the direction of the center of the circle contribute to the centripetal acceleration. Therefore, if they contribute towards it...

I know exactly what you're thinking. It's been awhile since I've solved one of these problems, and I did the same thing as you at first. What you have to realize is that when the ball is at the top of the circle, ƒg and ƒTension are both contributing to the centripetal acceleration because both...