Recent content by xxkbxx

Sorry, it's initially from point Q, as it leaves the accelerating plates, and it's final point is it entering the deflection plates. I'm not sure how to get vf though, but I have a feeling it's based on a=qE/m

Read what Saketh said: Current in a wire does not "get weaker" as you go along the wire. If it did, charge would accumulate at some points because it would be moving slow at some points and fast at others.

vf is in the final equation because Ki + Ui = Kf + Uf, Kf is 1/2mvf^2, so it's part of the equation. I did mess up on the potentials it should be:v0 = sqrt (mvf^2 + 2(Ub - Ua) ) These would simplify further to: v0 = sqrt (mvf^2 + 2q(V2 - V1) ) But I'm still stumped

Can't really help without knowing what you're looking for

Help! Conservation of Energy The apparatus to be used is a classic one for measuring q/m for electrons. A photo of the cathode-ray tube together with a simplified graphic are shown below. A filament heated with a 6.3-V AC source "boils" off some electrons. These electrons have very little...

Current is the same throughout the wire - so it being in the middle won't have anything to do with it. Remember what current is (by Ohm's law) and remember the formula for Resistance

I believe that the circuit would be the same as the attachment from R1, R2, and R3 - but the last branch would have R5 and R6 in parallel in the "block" as you called it, in series with R4. Would my total resistance than be: Req = R1 + R2 + (1/(R3) + (1/R4) + (1/(1/R5 + R6)) Specifically...

I tried that before, but I was unsure. The eq of R5 and R6 is (1/(R5) + 1/(R6))^-1 Then that would be added to R4, since R4 is in series. Okay, that makes more sense now - I think I have the rest from here!

Homework Statement Okay - the picture is attached, but what I need to find is the current and voltage throughout certain points on the circuit. What I need help on is figuring out how to simplify this circuit in order to find resistance. Homework Equations Parallel Req = (1/R1 +...

It's an online course, that's why I've been utilizing this board lately! I'm not sure whether they're from a book or his own, but he's pretty good with asking questions that the book covers. I just don't always get them on the first couple of tries :).

Ouch, with less than 24 hours, I'd review the chapters basic ideas and make sure you know how to use the formulas. Not much you can do with little time

You spell it out in Hint 3. Ignore all the crap about the car and look at what you have to use. At the beginning, one of the cylinders contains 499 cm^3 of air at atmospheric pressure (1.01 x 10 ^5 Pa) and a temperature of 27.0 C. At the end, the air has been compressed to a volume of...

So, I need to determine the force of the electric field then set it equal to the force of the external object? I know then that W=fd, change in KE is 0 since it begins and finishes at rest, so like you said so W = my DeltaU. The question for the next part is: For the situation of part c...

Okay, I've tried to rethink it. The work will be my force*distance. I know the distance needed, .10m, but I don't know what I should do for the force. It has to be greater than the Force of the electric field. What really confuses me is the follow up question which asks what the work done by...

Okay, the PE at the drop height is mgh while KE is 0. They reverse numbers as it falls, so KE picks up while PE decreases. The entire time Work equals the change in mechanical energy, so W = PE at it's greatest point. I still don't see how that equates to the electric field since the particle...