Recent content by jr662

thanks

i think i got it but i don't understand why you are adding here to get the force of q4 on q1 here using cos and sin...F 41 = − Q⋅Q4πε 0 ⋅ 2a2 cos 45°iˆ + Q⋅Q/4π ε0 ⋅ 2a2 sin 45°

kind of not really

http://www.webassign.net/hrw/hrw7_21-22.gif Four particles form a square. The charges are q1= q4=Q and q2=q3=q. a) what is Q/q if the net electrostatic force on particles 1 and 4 is zero? B) is there any value of q that makes the net electrostatic force on each of the four particles zero...

Awesome!

Sorry, so i got path a and c: a) delta P = 0...delta V=4.0Vo-Vo and Po = 40 so W=Po(deltaV) = 120 Joules (i don't know how joules is derived, besides the fact that we are looking for W and it is joules usually.. c) delta V= 0...delta p = (Po/4.0) - Po = 30 and W = p(delta v) so W = 30 x 1 = 30...

Ok, so apparently W is always the area under the curve of a p-V graph. So would W for Path (a) be = 160 and (c) 10 and (b) = ?? a) delta p = 0 ... and v increased (4.0 x 1.0 m^3) = 4 x po = 4 x 40 Pa = 160 ? I'll never figure this problem out!

Kuruman, I've read that W = integral pdV is true only if pressure is constant and the question states that the "pressure decreases" and "how much work is done by the gas if its pressure changes with volume." ?? I'm still confused, I guess this is what happens when your textbook doesn't hasnt...

Kuruman,,,sorry for any confusion. Anyhow, could you please confirm if the equation i am suppose to use is => W = nRT ln (Vf/Vi) ? Where n =1 R = 8.31...im not sure how to plug in the different values with the given info..

45. A gas sample expands from Vo to 4.0Vo while its pressure decreases from po to po/4.0. If Vo = 1.0m^3 and po = 40 Pa, how much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C? The p-V diagram can be found at the following addrs on...

thanks a bunch!

[b]1. Calculate the minimum amount of NRG, in joules required to completely melt 130g of silver initially at 15 degrees Celsius. [b]2. ? Q = mC (Tf-Ti) + mHf? saw it online somewhere, not sure what the proper formula for this problem is, and if it is done does everything need to be converted to...