Recent content by jpreed

Lets sort a few things out in this post. When anything gets hot enough it will start to 'glow' and emit light in the visible spectrum. The hotter it gets, the further up the spectrum the peak emittance will be from deep red to orange, yellow, green, blue and then violet. These are the...

You friend is full of ...well you know. If a brown dwarf star was close enough to collide with the sun then we would start to see its effects on the Earth already. We would see huge shifts in the tide and our orbit would start to shift and animals would start acting very strangely. We...

Wow, interesting question.

I use it for all of my analysis, it is incredibly powerful. You should really check into seeing if your college has a deal on the software. A lot of universities can get you a student copy for like $30.

This is correct. Although the air pressure is pushing down on the water at the top of the bucket, it is also pushing on the water coming out the hole in bottom. Sort of like it is trying to push the water back up into the bucket. The effect is that these two pressures from the atmosphere...

Oops, I forgot to normalize by the distance. You are right. It should be I = q v / (2 pi r)

A conservative field is also irrotational. Take the curl of the field and show that it is equal to zero. (It is).

Can't see the attachment yet. While Coulomb interaction dominate the interplay between electrons in graphite, I do believe the force holding the layers together could be referred to as van der Waals.

Your approach is correct, you may want to check your units. e = 1.6*10^-19 coulomb v = 3*10^7 m/s r = 0.5*10^-5 m I = q v m = I pi r^2 I get: I = 4.8*10^-12 m = 3.77*10^-22

To be clear, I realized the ambiguity in the integral as he had it written. If you take the time to read the entire original post he mentions multiplying the conjugate at the bottom. I untangled the ambiguity from what he had written there and then worked out the rest.

There are two problems, one is simple, one is more complicated. First, use "ReplaceAll" instead of "Replace", it uses a deeper algorithm that will properly do what you want it to do. Second, look at the result of "rX1". It is NOT just a number that can be plugged in. It is a list that...

The emf is the electromotive force. This is what pushes the electrons through the circuit and back to ground. Charging a battery sort of provides a bigger emf in the other direction, driving the current backwards. So while the battery is charging the total emf is the other direction but the...

When you charge a battery, you apply a voltage greater than that of battery's in order to reverse the internal chemical reaction. The battery's emf is overcome by the voltage you apply to the leads. The net result is a potential gradient large enough to reverse the chemical flow by pumping...

So you want to find the inverse Fourier transform of \frac{\alpha}{2\pi}\exp(A \omega^2)? It should be: \frac{\alpha}{2\pi}\frac{1}{\sqrt{-2 A}}\exp\left(\frac{t^2}{4A}\right)

From a purely classical standpoint you can ask: -what is energy of a ground state electron in a hydrogen atom? -what is the starting radius of the electron in a hydrogen atom? -we assume, it is moving in a circle, so what is the formula for its kinetic energy? You should be able to...