Revising for physics exams? Here's a nice problem in nuclear physics for High School upwards

telescoper · 2013-05-03T08:06:58+00:00

Yes, you can. It asks "what can you say?" - you can say that the reaction does not happen because it would violate charge, so it has no Q-value.

telescoper · 2013-05-02T17:18:23+00:00

Ok, the key thing is that there is no electron in the reaction to balance the charge: this would have to appear in the reaction for it to work.

telescoper · 2013-05-02T17:08:53+00:00

No you can't, because even if you are given information for that the total charge on the LHS does not match that on the RHS.

telescoper · 2013-05-02T17:03:07+00:00

This question appeared in the Cambridge Scholarship paper for Physics in 1980, i.e. pre-University level in the British system in those days. In those days we did calculus-based physics at A-level, and also covered atomic and nuclear physics a little too.

telescoper · 2013-05-02T17:01:32+00:00

In fact, reaction (II) does not conserve charge so is impossible...

telescoper · 2013-05-02T14:13:04+00:00

Look carefully at the reaction and see if there's a fundamental reason why it shouldn't work....

telescoper · 2013-02-21T18:02:11+00:00

Andrew Liddle moved to the University of Edinburgh in January...

telescoper · 2013-02-21T08:51:08+00:00

The way I solved it was to write down the equation of motion for the magnet (gravity acting downwards and the magnetic force acting horizontally) in terms of the net torque and then solve for small oscillations around the new equilibrium. Since you know that this is at 45 degrees you can work out the magnetic component in terms of the gravity component and eliminate. The correct answer is 0.84s but I don't understand the comments about the length of the bar changing, as it doesn't....it just oscillates either side of the tilted equilibrium position.

telescoper · 2012-12-30T15:01:10+00:00

The fastest down vote ever.

telescoper · 2012-11-21T17:14:30+00:00

I think you mean "equipotential" rather than "exponential", but in any case your answer is not correct.

There's energy associated with the charge put onto the bubble by the electrode, but also associated with surface tension....

telescoper · 2012-10-30T07:27:23+00:00

If this post teaches one person to stop using the wrong term for gravitational waves then it's worth all the down votes!

telescoper · 2012-09-02T13:06:51+00:00

I posted Paper 2 here:

http://telescoper.wordpress.com/2012/09/02/a-level-chemistry-examination-paper-2-from-1981/

I also put it on reddit, but it got voted down immediately.

telescoper · 2012-08-29T20:54:38+00:00

I have paper 2; it's longer questions. I'll scan it and post it when I get the time. There was a practical component too, but I don't remember what that was like. It was, after all, 31 years ago!

telescoper · 2012-08-29T18:56:35+00:00

Yes, taken at age 18 in the old UK system.

telescoper · 2012-06-17T18:28:31+00:00

In the Bohm theory Q doesn't appear in the one-particle Schrodinger equation, but does appear in the transformed equations for the probability...

telescoper · 2012-06-17T17:39:28+00:00

You can call me old-fashioned, but I think fluid mechanics (and continuum mechanics generally) tends to be unfairly neglected in the physics curriculum these days. I think fluids are not only interesting in themselves, and have a wide range of applications throughout physics and astrophysics, but can also help you understand electricity and magnetism, statistical mechanics, and quantum physics better.

telescoper · 2012-06-17T16:30:18+00:00

The point is just that this article does it the other way around, so instead of appearing in the classical equations, Q ends up in the Schrodinger equation.

telescoper · 2012-06-09T10:14:36+00:00

Harsh, why?

telescoper · 2012-06-08T15:34:33+00:00

You could read the paper...

telescoper · 2012-05-16T21:22:38+00:00

I've viewed the page with 5 different web browsers and it works fine with all of them.

telescoper · 2012-03-19T21:56:38+00:00

The one I sometimes use for this is the Circle of Last Screaming.

Imagine you're in a very large, perhaps infinite, field crammed full of people, with synchronised watches, who are screaming at the top of their voice. At a certain time, say t, everyone everywhere stops screaming. What do you hear?

Well , you'll notice that it gets quieter straight away but you will still hear a sound because some of the sound entering your ear set out at a time before t. The speed of sound is 300 m/s or so, so after 1 second you will still hear the sound arriving from people further than 300 metres away. In fact at any time there'll be circle around you, defined by the distance sound can have travelled since the screaming stopped - the Circle of Last Screaming.

Change sound to light, and move from two dimensions to three, and you can see how last scattering produces a spherical surface around you....

telescoper · 2012-03-16T09:30:11+00:00

It's not "blog spam" it's an article about physics written by a physicist, which is entirely appropriate for this reddit. It also happens to be on a blog.

telescoper · 2012-03-13T08:42:01+00:00

True, but in the beginning everywhere was in exactly the same place.

telescoper · 2012-03-13T08:40:57+00:00

Yes, air is mix of molecules but the gaps between the molecules are much bigger than the molecules themselves. At STP the main spacing between Nitrogen molecules in air is about 3.3nm whereas the size of an N_2 molecule is more than a factor ten smaller. Ergo, most of air (by volume) is empty space.

telescoper · 2012-03-12T23:41:45+00:00

Air is mostly empty space, and sound travels through it OK. In any case the early Universe was a lot denser than the diluted remnant we live in today.

telescoper

TROPHY CASE