Recent content by dt19

Hi, Could anyone explain to me why it is that cosmic radiation supplies slightly more positive than negative muons? I know that negative muons disappear a little faster than the positive ones as they can decay or be captured, but I don't know if this would have any impact on what I'm asking...

Well I'm no expert but I've heard people speak highly of Riley, Hobson and Bence. It's recommended for the Cambridge NatScis so it can't be all that bad :)

Ah OK I understand that now. Any ideas on the second part?

so that would mean impulse = 8.4/0.6 = 14, if I understadnd you correctly, but that's still the wrong answer. Also, what do the arrows above F, r etc mean?

Oh I see, the change in height is 0.6m, silly me ! But I still have the wrong answer. I get \omega = 7 which gives impulse = 8.4 Ns

This is the integration in LaTeX, if anyone else can't see it: v = \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} s = \int v dt s = \int \frac{e^{\frac{t - 1205.525}{-100}}-142000}{30} dt Do you know how to integrate exponential functions? EDIT: I don't know what is up...

By energy changes: At the top, E_p = mgh = 5 \times 9.8 \times 0.3 = 14.7 J And rotational kinetic energy = 0 Just before it stops, all the potential energy is converted to rotational kinetic, so: 14.7 = \frac{1}{2} \times \frac{6}{5} \times {\omega}^2 so \omega = \sqrt{24.5}

I think I can solve it, providing I have understood what you wrote in the right way. This is what I think your differential equation is: \frac{dx}{dt} - \frac{x}{1+t} = 1 \times 10^{-5} This is a differential equation of the form \frac{dx}{dt} + P x = Q where P and Q are functions...

Homework Statement A uniform square board ABCD had mass 5kg and sides of length 60cm. The board is hinged at A so that it can rotate freely in a vertical plane. A light string of length 1 metre is attached to B and to a point E, 80cm above A. The bards rests in equilibrium with AB horizontal...

do you mean a voltmeter? what sort of relationship do you mean - current through it, or what?

sound ok to me...

i'm doing further maths at the moment - last year i did maths A level and this year I'm doing FM A level. So I have twice as much maths as everyone else. At the start of last year, the workload took a bit of getting used to - remember, you're doing 2 years' worth of work in half the time, and...

can anyone explain to me, where does the equation R=\frac{0.61\lambda}{sin\theta} come from when determining the approximate radius of the nucleus by electron scattering?

thanks for all the advice :smile: i reckon i'll probably go there!

my second choice is liverpool at the moment, but i haven't seen durham yet so it might be there. when i first started at my current college i felt sooo stupid - it's very selective about the students it takes from outside the catchment area, and it didn't help that i was doing further maths! i...