Recent content by nahya

Thanks, that worked. Weird, though, because I had used my calculator to graph the equation and integrate graphically.

B was approximately 3.326E-4, and emf was 3.23762818E-5 V, which was still incorrect. Am I using the right L for emf = vBL? I'm using 3.13 cm (0.0313 m).

This problem is difficult to describe, so I'll post a picture. http://img71.imageshack.us/my.php?image=pic1ik.gif The figure above shows a rod of length L caused to move at a constant speed v along horizontal conducting rails. The magnetic field B (the magnitude and direction of which are...

i'm recycling the thread. old post was edited to a new question.

doh! i forgot that A is in µC. so... 2 * 9e9 * 4e-6 = 72000, and it's still wrong... the answer's supposed to be in N/C. edit: oh. -72000. hehe. thanks!

ballistic particle and electric field Throughout space there is a uniform electric field in the -y direction of strength E = 450 N/C. There is no gravity. At t = 0, a particle with mass m = 5 g and charge q = -11 µC is at the origin moving with a velocity v0 = 35 m/s at an angle q = 25° above...

well.. the latex thing isn't working atm, i guess. ^^ i solved the problem, however. i just didn't think about the types of the resulting charges.

never mind. sorry :)

doh! thanks. hehe.

A rigid, insulating fiber runs along a portion of the y-axis; the fiber isnot free to move. Gravity acts downward (g = 9.81 m/s2).A charge Qa = -3 µC is fixed to the fiber at the origin. A bead with a hole drilled through its center is slipped over the fiber andis free to move along the fiber...

hmm... h = 1.9 * 2.6 = 4.94 m above 2R, but it says that it's incorrect. bleh TT

A small spherical ball of radius r = 1.5 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.6 m. The ball has mass M = 358 g. How high above the top of the loop must it be released in order that the ball just makes it around the loop? --- i just calculated the...

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.55. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on...

omg... they DID want it in radians..... hahaha.. I'm sorry. i wasn't careful in reading the problem. (in the answer box it says rad/s^2. bleh)

A disk with a radial line painted on it is mounted on an axle perpendicular to it and running through its center. It is initially at rest, with the line at q0 = -90°. The disk then undergoes constant angular acceleration. After accelerating for 3.1 s, the reference line has been moved part way...