Recent content by huh

umm... I'll try reading this again tomorrow... I understand part of what you are saying, but it's not "sticking" in my head right now. Thanks though...

It's in series? It looks like it is in the same position as the first two are... OK so how do I get it for each individual charge (and capacitance)?

I see what you mean in calculating it as a series. Whoops :) So would you just add the values for capacitance together? (3.15e-6)+(5.5e-6)=8.65e-6 8.65e-6+6.85e-6= 1.55e-5

(1/3.15e-6)+(1/5.5e-6)=4.99278e5 (1/4.99278)= 2.00289e-6 2.00289e-6+ (1/6.85e-6)= 1.45985e5 (1/1.45985e5)= 6.85e-6

Would the equivalent capacitance for the whole thing be 6.85E-6 I calculate C1 and C2 together in parallel and then took that answer and did it again with C3.

Homework Statement In the circuit shown in the figure, the applied potential is V_ab = 18.0V . For C_1= 3.15(microFarads) , C_2 = 5.50microF , and C_3 = 6.85microF , Find the charge q_1 on the capacitor C_1 (and the same for q2 on c2 and q3 on c3) Picture is attached... Homework...

Wow. That's cool. Thanks :)

From what I've seen, standing waves (which are really cool looking) have nodes (no displacement) at either end. What would happen if antinodes were at the ends instead of nodes? What could you do to make this happen? hmmm...

Okay let's try again: (G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below use cos(0) and sin(0) x-component: 1.23 x 10^-3 y-component: 0 use cos(90) and sin(90) x-comp: 0 y-comp: 1.23 x 10^-3 (G*600*600)/[(.14^2)+(.14^2)]= 6.13 x 10^-4 for the mass diagonal from...

law of gravitation- is this better? Could someone check this please? Four identical masses of mass 600 kg each are placed at the corners of a square whose side lengths are 14.0cm (.14m). What is the magnitude of the net gravitational force on one of the masses, due to the other three...

Wouldn't there be an extra m on the bottom, and how could I get rid of theta and be left with phi, or does theta count for much (is it negligible)?

Thanks. okay, so... m2v+ mv sin(phi) divided by 2m cos(theta) is v(final)? how can I get v(final) with only v and Phi? I can't have m or theta in the equation.

Please help me understand.

almost there-inelastic collision at an angle Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle phi south of east. After the collision, the two-car system travels at...

A box of textbooks of mass 24.4 rests on a loading ramp that makes an angle a with the horizontal. The coefficient of kinetic friction is 0.230 and the coefficient of static friction is 0.360 . I figured out that the angle is 19.8 and acceleration is 1.2m/s^2. I would like to know how to...