Recent content by swede5670

Homework Statement Determine the specific latent heat of fusion of ice given the amount of time it takes the ice to melt (150 seconds). Energy is supplied to the ice at 530 W. The sample is .25 kg. Homework Equations Q= ml The Attempt at a Solution (Time * 530W) = .25kg * L 150...

Homework Statement It takes 15 seconds to warm ice 15 degrees ( -15 degrees celsius to 0 degrees celsius). The ice has a mass of .25 kg. The energy is supplied at 530 W to the ice. Homework Equations Q = mc delta T The Attempt at a Solution I plugged in all the information for Q...

Would you mind helping me with the follow up question? It is: The resistor R is replaced with an electromagnet and a switch. When the current is switched on, small pieces of iron, initially on the ground below the electromagnet, are attracted to, and stick to, the electromagnet. (A) State...

I was thinking though, normally to find (I) don't you need to have the total resistance in the circuit? Since r and R are in series it should be V = I (R + r) right? 1.4 = .2(6 + r) 1.4/.2 = 6 + r 7 = 6 + r 1 = r is that correct?

alright so using that, 1.2/6= .2 so I is .2. In the equation its then E = .2 + .2r In order to solve for r though I need to know what E is. What does E stand for?

so I found V=IR for the 6 ohm resistor 1.4= I (6) so I equal .2333 so in E = DeltaV + Ir E = .2 + (.2333)(r) Is this correct? What does E stand for again?

Homework Statement The cell supplies 8.1×10^3 J of energy when 5.8×10^3 C of charge moves completely round the circuit. The current in the circuit is constant. A. Deduce that the e.m.f of the cell E is about 1.4 V B. The resistor R has resistance 6.0 Ω. The potential difference between its...

I would need to convert g=9.81 m/s to g=0.00981 km/s to get coulombs correct? If that does work, T= (0.00981 * 1)/cos(11.14) T=.009998389 T*sin(11.14) = F(elec) = qE .009998389 * sin(11.14) =qE [.009998389 * sin(11.14)]/E = q .0019317587 / E = q .0019317587 / 9000 = q 2.146398 e-7 Correct?

Ok so cos-1 (52/53) = 11.14 degrees So can I plug that into the equations I was given up above? T*cos(11.14) = mg T*sin(11.14) = F(elec) = qE

I tried this, but the answer was incorrect. Can you see what I did wrong? | \ | \ | \ 52 | \ 53 | \ | T \ * So the point charge is basically the dobber of a simple pendulum. In this case you have two forces acting on it...

Homework Statement I need some help solving this question, it is the last one I have to do and I'm not sure how to solve it. A point charge (m = 1.0 g) at the end of an insulating string of length 53 cm is observed to be in equilibrium in a uniform horizontal electric field of 9000 N/C...

{(8.99*10^9)(-9*10^-6)}/.015^2 = -359,600,000 {(8.99*10^9)(6*10^-6)}/.015^2 = 239, 733, 333 Adding these two up -359,600,000 + 239, 733, 333 = -119, 866, 666 -119, 866, 666 N/C seems ridiculous, what did I do here?

You say "for the 1q" What is the 1q? Is it the one quadrant? You gave me two equations for the vectors for attractive and repulsive forces, do I need to add these two vectors? I don't know what I need to do. I can plug the numbers in for these equations: q2/d2 i - q2/d2 j q2/2d2*cosθ i +...

so sigma is the sum right?E = k(6x10^-6)/r2 is added with: E = k(-9*10^-6)/r2 once I've added these two Es up is that my final answer?

I'm still a bit confused by what your telling me, I talked to a friend and here is what he said: To calculate the forces you really only need to do it once and then apply the direction as if you were doing it in each quadrant. . K= 9x109. r = .9m for doing the adj and r = √.92+.92 for the ones...