Recent content by acgold

Is this correct? \lambda L + \rho L = \varepsilon_0 E 2 \pi r L so E= \frac{\lambda + \rho }{\varepsilon_0 2 \pi r} I just noticed that the problem gives the charge per unit LENGTH. Oops. Thanks for the help.

Thanks for the help guys. Unfortunately, I couldn't get this particular problem finished in time to turn it in but I appreciate the help anyway. At least I understand it somewhat now. Thanks again!

Ok, I've been working on this problem for an insane amount of time but I never get 7.2 N/C. Can someone check the expression I came up with and tell me if it's right? This is for the second question... \lambda L + \rho \pi L(b^2-a^2) = \varepsilon_0 E 2 \pi r L so E= \frac{\lambda +...

Suppose that X and Y are independent random variables, where X is normally distributed with mean 45 and standard deviation 0.5 and Y is normally distributed with mean 20 and standard deviation 0.1. (a) Find \ P(40 \leq X \leq 50, \ 20 \leq Y \leq 25). Ans. ~0.5 (b) Find \...

Ok I'm pretty sure I understand the first problem now. Like Krab said for 1a) Q=\int^b_a \frac{k}{r} 2 \pi r L dr = k 2 \pi L \int^b_a dr = k2\pi L (b-a) Since the total charge is asked for you must integrate from a to b. To find the charge for 1b) you have to integrate from a to r...

Hey thanks for the reply; could you give me a quick explanation on how you got that answer? Just understanding the process you went through would really help me a lot. Thanks.

I'm stuck on two problems. I hope someone can help me. Here they are... 1) For 1a I thought Q would be Q=\rho \pi L (b^2-a^2) but since \rho=\frac{k}{r} so Q=\frac {k \pi L (b^2-a^2)}{r}. After being stumped on 1a I'm not sure how to go about 1b. 2) I've derived about 4 equations for this...

Recommendations for Physics book? Hi, I'm taking a college level calculus based physics course and I need some help. My professor teaches his course out of packets that he makes instead of a textbook and the packets are generally horrible. I'm completely lost and the professor isn't available...

Man I need to get some sleep or something. The answer was sitting right in front of me. I already had both equations all setup...all I need to do was solve for r and substitute it into the other equation. Thanks

This one is giving me trouble because I don't know the final radius and I don't know how to find it. 1. A copper sphere of radius 2.000 cm is placed over a hole of radius 1.990 cm in an aluminum plate at 20 deg. C. At what common temperature will the sphere pass through the hole? Ans. 747...

Maybe I'm missing something here but wouldn't I need the area to solve the following problem? 1. Blood plasma has a density of 1026 kg/m^3. What is the minimum height above a vein -- in which the blood pressure is 16 torr -- that a container of blood plasma should be placed in order for the...

Great, answers came out like they should. Thanks a lot.

We were doing simple P=\rho g h problems in class and then all of a sudden I ran into these two when I was doing my homework. We never went over this in class and I have no idea what to do. 1. A whale tank at the zoo is 15 feet deep and filled with sea water. A 6 foot tall by 20 foot wide...

Thanks, you're a much better teacher than my professor. :) I've been working on the second problem for a few minutes now and I'm stuck again. Maybe another typo...I don't know but I'll keep trying it.

Thank you thank you thank you...I've been working on that problem for hours. Yeah I gave you ALL the information I had. My professor is notorious for having typos. I did the first problem the way you showed me and I understand it 100% now. I'm off to work on the last two on my own...thanks for...