You have a rubber cord of relaxed length x. It be-
haves according to Hooke's law with a "spring con-
stant" equal to k. You then stretch the cord so it has
a new length equal to 2x. a) Show that a wave will
propagate along the cord with speed...
I failed , sorry vela. The my ******* proffesor put this one test and he knew it was the only one i didnt awnser. Out of 30 homework questions he always puts 6 on the test and he puts the only one i didnt have. Oh well i will still get an A but i wanted 100% cause it makes him mad when someone...
yes i know what your saying:
Kinetic energy energy of moving proton is K1=(gamma1-1)mc2
But this takes me back to the initial quetion i would have to solve for the velocity in the gamma to obtain the kinetic energy just in terms of mass and c2.
Iam stupid sorry, iam not understanding at...
Sorry vela. I thought you mean start by energies and momentum to get velcoties.
Ok i managed to obtain the mass in terms of the K1 using hte euation above and the third relavant equation.
m=(pc)2 - K12/(2K1c2) i dont know if this will help.
When you say the equations for conservation of...
Q=mC C is a constant you can look up for water its different depending on which phase transition you are calculating.
Q=mc(Tf-Ti) for when your raising the temperature c i think its 1cal/gC for water.
Add all the Qs and viola
I got it!! i think..
the proffessors train is larger than it actually it is if they are both measured at rest!!. makes sense.
Now I just have to figure how to make it work ....
Is the students train going to even smaller than it actually it is from the proffessors point of view?
Consider the relativity of simultaneity: A student in a train at rest from his point of view and a professor in a train that is moving in the positive x direction from the student’s perspective. Two lightning of different colors will strike at opposite ends of the trains...
Well you have the proton at rest: E=mc2 K=0 p=0
The moving proton E1=K1+ mc2 p=(gamma1)mu whre u is the speed of the proton
The blob: E2=K2+4mc2 p=(gamma2)mv where v is the velocity of the blob
Equating any of teh energy/momentum equations together just seems like a big mess because...
Well if the proton is going fast enough to overcome the electric repulsive forces it would collide wouldnt it? thats how colliders work dont they?
And i have never heard of 4 vectors sorry =/.
And the velocities i got it from equating the total energies of the "reactancts" and "products"...
A high speed proton collides with a proton at rest, two proton-antiproton pairs were created. That is some of the accelerated protons kinetic energy was enough to be converted into the mass of the new particles. If the proton-antiproton pairs move along together as a single...
Thats what i was trying
a=dv/dt then pass dt to the RHS to integrate and which also has a v=dx/dt which is where iam confused.
And for tiny-tim's response
Iam not seeing where your going with that...
=f(v)dtdv? double integration?