Recent content by SmashtheVan

So Polarization of light has always really bugged me. Linear polarization is simple. I understand Elliptical and Circular as well; the Electric Field vector changes as the Ex and Ey(or Ex and Ez, etc) functions grow and recede, periodically. This is easy to show as the propagation of a singular...

Thanks everyone. When I saw the solution it didn't make sense to me based on the previous exercises I got right using Unions/Intersections, so I gave myself that example to work it out mathematically. Must have been just a mixup by the author over the Union and Intersection symbols. Just...

Homework Statement Can't quite figure out the LaTeX for Indexed Sets, so bear with me: From "Book of Proof" Section 1.8 #11 http://www.people.vcu.edu/~rhammack/BookOfProof/index.html Is the UNION of Aa, where a is in I, a subset of the INTERSECTION of Aa always true for any collection of sets...

bring me a non-relativistic photon, and I've got a bridge in brooklyn id like to sell you.

i still think you're doing something a bit off here. the .544 and 2.86 should be summed, not divided. and then divide 13.6 by the new sum for your answer. it should give you n^2=(approx.)4, and n=2 remember that n is a discrete energy level, which are integers, and if your decimal is off by...

it looks like your problem is with the relationship between the energy of the photon and the change in energy levels. E_{photon}= E_{f}- E_{i} where the initial state is the 5th energy level, E of the photon you figured out was 2.86, and the final state is what we're solving for. so, E_{n}=...

That's the correct answer. a simpler way to do this would to consider the Kinetic/Potential energy systems. The train has initial Kinetic energy found by .5mv^2, and when the train stops the buffer has all of that energy transformed into potential energy, which for a spring is .5kx^2. set the...

so does this mean everything is made of cheese? I don't understand :D

i worked it out, checked with the help of Maple, and will help you out here to get what i got. let u= e^{x} then, x= ln\left|u\right| and dx= \frac{du}{u} this should leave you with \frac{\sqrt{u+1}}{u}du and now consult your integration tables

maybe my wording is just not great, because I was trying to convey what you said here.

not quite. ill give you an example. you have two balls traveling towards each other. Ball 1 has a mass 10kg, and Ball 2mass 5kg They both move with velocity 2m/s the momentum of ball 1 is equal to (10)(2)= 20 the momentum of ball 2 is equal to (5)(2)= 10 so you have two different masses...

What i meant was if you look at the vehicles as if they were rigid bodies, they wouldn't APPEAR to conserve momentum like if we look at billiard balls which do appear to conserve like rigid bodies

What you will study in physics is the momentum of rigid bodies. Unfortunately vehicles are not necessarily rigid bodies, they are made specifically to compress upon impact, in order to reduce the acceleration of the occupants in the vehicle(so that they have a better chance to survive). Momentum...

beat me to it! Cerenkov radiation is by far the coolest thing I've read about outside of my undergrad studies.

when you lower the mass gradually, what you are doing is applying an upward force so that the acceleration of the mass is very small, so that at the bottom there is no motion. Therefore, you are applying a force, over a distance, which is Work. Work= change in Kinetic Energy. Normally the...