Recent content by gomboc

The weak force is the only interaction that can change the flavour of a quark. The strong force can't.

The intrinsic carrier density is usually used in relation to intrinsic semiconductiors; materials that have not been 'doped. It is due PURELY to the thermal excitations of electron-hole pairs in the material. On the other hand the electron concentration and hole concentration are usually...

Makes sense, but I'd take a look at the equation you used to find the length. Do you really want to be using the reduced Planck constant (hbar) there, or simply the Planck constant (2*pi*hbar)? You can look up the de Broglie wavelength of an electron, and you'll see your value is a bit small.

The key here is to look at the quark content of the reactants and products. A positive kaon has an up and an anti-strange quark. Yet none of the pions contain a strange/anti-strange quark. What is the only interaction capable of changing the flavour of quarks?

I think the previous suggestion was looking at the problem wrong. They make a point of wanting you to use the uncertainty principle, so here's what I think they want you to do. Using classical mechanics, you can determine how long each particle is in free-fall before hitting the detector...

Assuming E0 is the zero point energy, you need to determine how many solutions there are to the equation E_0 \left(\frac{8mL^2}{h^2}\right) = n_x^2 + n_y^2 = 2(65) where n_x and n_y are positive integers. For example, one solution would be n_x = 9 and n_y = 7, so obviously n_x = 7 and n_y =...

It certainly looks attenuated - especially above the 790 nm mark. Have you tried pointing the spectrometer at the sun to see if those wavelengths show up? Unless it's broken, the only other thing I can think of is the possibility that your microscope has one of those infrared filters on it to...

So, you're firing various particles at a 14 fm lead nucleus to measure their diffraction. We know that in order to make any measurement in physics, the wavelength of the photon/particle taking the measurement must be similar to, or even better, smaller than the dimension you wish to measure -...

This problem might sound a little harder than it actually is. All you need to notice that an electric field exerts a force on a charged particle given by \vec{F}=q\vec{E} Once you consider that force, the problem can be solved with the usual high school mechanics techniques...

The solution would need to be integrated. Using an average rotational velocity would yield an approximation of the right order of magnitude, but it wouldn't be very accurate - that's because when you think about the disc's area, there is a larger area of the disk moving at a faster speed (the...

You got to \Omega \propto (V-Nv_0)^N , but when you think about it, this doesn't properly represent the situation - this actually represents putting a single particle into the remaining volume of the jar N times. What you have to do is put one particle in, and then progressively add particles...

It looks fine, and yes, that equation for Q-value should work for all nuclear decays. There is one problem though: you seemed to have added the product masses instead of subtracting them in the calculation of the Q-values for the second and third decays.

You just need to use a little ingenuity to solve for A. You know that the mass is projected towards the eq point from x=sqrt(3) with velocity 2. Since you know the spring constant and the mass, you can find the mass's total energy, and thus it's position (A) at maximum extension using...

Thanks, I'm flattered!

Speaking in terms of physics, university is immensely harder than high school for two main reasons. First, the course material is quite a bit more difficult. Many students in high school, myself included, benefited a lot from having a good "physical intuition". It's hard to describe what this...