Recent content by TiernanW

Why do you figure these things out as soon as you post them? I was honestly at this for ages and just got it. Here is my solution for anyone viewing this in the future: http://i.imgur.com/nnF8t83.jpg

Homework Statement Homework Equations 1/Rt = 1/R1 + 1/R2 + ... The Attempt at a Solution I can do part (i) okay. It is simply 1/12 + 1/12 = 1/6, so R = 6. But I have been trying for ages to figure out part (ii). I tried re-drawing the circuit in different forms but it didn't help me...

This I had also thought about, but I didn't really know how to approach it because the equation looks for a value of h. The water level falls. Is it like the average water level because 5 is half-way between 10 and 0?

Ah I can see where you are coming from. The loss of potential for the water going INTO the turbines would be 10, but for the water leaving it is less than that. We aren't given the total height though. That is odd.

Homework Statement Homework Equations Power = Work Done / Time Work Done = Force * Displacement PE = mass * gravity * height The Attempt at a Solution I got the first part (i) right, so the mass of the sea water is 1.32 x 10^12 kg. Part (ii) is the bit I am stuck on. My thoughts were that...

Shoot. Awkward haha. Thanks! :)

Homework Statement Homework Equations PE = mgh KE = 0.5mv^2 WD = F * s The Attempt at a Solution Its part i. I understand the solution when you look at the whole system. You do not need to consider the tension in the string because they cancel out, but I want to be able to do it just...

I understand it now. -3mgx = -μmg * d as friction acts in the opposite direction. :)

So WD against friction = μmgd. Therefore since all PE is lost to this then 3mgx = μmgd, so d = 3x/μ. That makes a little more sense right now. But what I'm trying to get my head around is that fact that ΔE = WD, so I would have thought work done is actually -3mgx, so -3mgx = μmgd? Forgive...

Surface CD is rough so there is friction.

How is that possible? Doesn't some of the energy go towards movement and sound etc?

The mark scheme has said: 3mgx = μmgd. So if I had taken the directions the other way I would have said -3mgx = -μmgd?

Homework Statement Homework Equations PE at A = 3mgx WD = Fs KE = 1/2mv^2 The Attempt at a Solution The question I am stuck on is part ii. I worked out from part i that the PE at A is 3mgx, so therefore all this must go towards the KE and the sound, and doing work against friction, etc...

Sorry, yes I know. I just meant the guy took that little network out and hand everything flowing clockwise through it, but if I followed that circuit, I would have though the current would have went through the top bulb going to the left and the same with the other bulb. But when he took it...

Homework Statement What way does the current flow through the first light-bulb? Homework EquationsThe Attempt at a Solution This isn't really a homework question but related ot my school work. I'm doing Kirchoff's Laws and normally I would have just thought that the current (if we use...