Recent content by noob314

Let a represent the area, p the perimeter, d the diagonal, b the breadth, and L the length of a rectangle. One can easily write down from analytical geometry all the various relationships between the above variables, and from these obtain directly a variety of partial differential quantities...

Got it. Thanks for helping me out.

I know that is the potential for one rod, but you stated that I needed to check my limits of integration. I'm also getting conflicting answers from two different sources. In one source, they integrate from -a to a, and have the electric potential as...

Edit: I looked at another source for the same example of the rod and found that my textbook's example is incorrect. The limits of integration should be -a to a. Charge density is Q/12a. And the electric potential of one segment would be V= kQ/12a ln[...] I'm guessing y is a constant, which...

Ah, I see. That makes things a lot more simple. So the integral would be V= -\int^{a}_{0}\frac{k\lambda dx}{\sqrt{x^{2}+y^{2}}} with \lambda being \frac{Q}{6a} and after doing the integral, V = \frac{kQ}{6a}ln\frac{a+\sqrt{a^{2}+y^{2}}}{y} With 6 segments, the total electric potential...

A plastic rod with a total charge Q uniformly distributed along its length is bent into a regular hexagon where each side has a length of 2a, as shown below. Calculate the electric potential at the center of the hexagon (relative to the point at infinity). I wasn't sure how to exactly start...

Can someone explain to me why Cv is used in this example? Determine delta S for the conversion of a monatomic ideal gas from state 1 to state 2. Cv = (3/2)R, Cp = (5/2)R State 1: V1 = 22.0L, P1 = 1.25atm, T1 = 308K, n=0.917mol State 2: V2 = 34.0L, P2 = 0.700atm, T2=325K, n=0.917mol...

Thanks, I understand it now.

Yeah, you're right. I rechecked my calculations and the numbers didn't add up. I was just curious because it was on one of my old exams.

I assume it would be a force that's barely able to overcome the static friction and then disappear after that.

Is it really impossible? I assumed some unknown force allowed it to overcome whatever static friction there may have been which would have allowed mass m to overcome the kinetic friction.

Yes, the problem is copied down correctly.

Homework Statement A large block of mass M = 6kg is on an incline which is at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. The block is attached to a string which runs over a pulley and is connected to a smaller...

Well, the torque from the external force of gravity would act on the center of mass of the disk, which is distance R from the pivot. That would give me to net torque on the object which would allow me to find the angular acceleration, and then the tangential acceleration. But I still don't see...

I'm starting to get frustrated. This was on an exam, but my professor refuses to give out the solutions or even the answers, so I can't even check if I'm close to the answer. This looks like it's suppose to be simple, but I'm just not getting it. For the instant the disk is released...