Recent content by gadje

Oh yeah. I didn't notice that 4 in the denominator. Knew it was something stupid! Cheers.

Gauss's Law and Superposition of Fields (edited again, something else wrong) Homework Statement Right. The shape itself has charge Q, so it has charge density \frac{Q}{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (\frac{R}{2})^3} = \frac{6Q}{7\pi R^3} Let's call this \rho. If it were filled in...

\frac{\partial}{\partial a} f(a \mathbf{x}) = pa^{p-1}f(\mathbf{x}) Okay. I'm still clueless. EDIT: Hang on. \frac{\partial}{\partial a} f(a \mathbf{x}) = \frac{\partial}{\partial a} (a \mathbf{x}) \frac{\partial}{\partial \mathbf{x}} f(a\mathbf{x}) = \mathbf{x} \cdot \nabla f (a \mathbf{x})...

Homework Statement We say that a differentiable function f : \mathbb{R}^n \rightarrow \mathbb{R} is homogenous of degree p if, for every \mathbf{x} \in \mathbb{R}^n and every a>0, f(a\mathbf{x}) = a^pf(\mathbf{x}). Show that, if f is homogenous, then \mathbf{x} \cdot \nabla f(\mathbf{x}) = p...

Yeah, I screwed up the units in the OP; had them right in my paper calculation.

Homework Statement A square of side 2cm, in the first quadrant of the x-y plane, with a corner at the origin, is in a magnetic field pointing out of the page of magnitude 4t2y. Calculate the emf around the square at t = 2.5s and give its direction.Homework Equations \epsilon = - \frac{d}{dt}...

I don't get that either. I get -1000N, taking g = 10 N/kg KE= GPE = mgh = 5000J All this is taken from the diver in the 5m of water: F = W/d = 5000/5 = 1000N. But I'm quite drunk at the moment so I might be wrong.

Work done is the change in kinetic energy, yes. Work done is also the average force multiplied by distance across which it is applied...

What is the diver's velocity as he hits the water? What is his kinetic energy? What is the change in kinetic energy when the diver decelerates from that velocity to 0? What is the work done in decelerating the diver? What is the average resistive force on the diver as he decelerates?

Yes, sorry, I misread your post - I've edited it now.

How have you used the formulae you are given? What have you tried to do? Note that to calculate the potential you must integrate from infinity up to the point that you're concerned about - in this case the formula for the electric field changes between infinity and your point.

The one-dimensional wave equation is \frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2y}{\partial t^2} . See how you get on, knowing this.

There's an earlier part of the question which talks about the lake being at 10 Celsius and being of a high specific heat. Dropping the block into the lake introduces mgh of energy into it, changing the entropy of the universe by mgh/T = 10*g*100/(10+273). I think.

I think I've got it, actually. It's not what I have up there, upon thinking about it.

I know this isn't technically the right place to post this, but it's so closely linked to the original point of the thread... I'm doing a problem sheet on entropy now, and the question is "What is the entropy change of the universe when a block of mass 10kg is dropped from a height of 100m into...