Is there any way to prove that entropy change is independent from the path?
dS=dQ/T
dQ=dE+PdV
d(PV)=PdV+VdP
dS=nCd[ln(T)]+nR[ln(T)]-VdP/T
i go this far, but it is not very different from dS=nCd[ln(T)]+PdV/T
Dear AbdulKadir, that is easy;
l=nx right?
for first one l=(4/3)t
for second one l'=(3/2)t
path difference is (1/6)t, don't get confused, other is not traveling at space it also in a medium
actually, i think it did its time, the very first professional experiment that lead people to believe existence of magnetic force so magnetic field,(the experiment of Ampere's about that force). It is too small to be applied and so small that it is never a problem like eddy currents.
and besides...
actually i advise you to move your question to advanced physics or mathematics forums, there especially at mathematics part maybe experts can help.
i haven't legally studied these stuff much, but i once heard definition of linearity is f(kx)=kf(x)
so f(0)=0 more, it must be like f(x)=ax
both...
actually these are applications of one powerful magnetic field and one magnetic dipole.
I don't know any application of force between two wires, i think it is completely educational and a bit historical.
It stays same, because total energy of all gas molecules are NkT=nN(avagadro)kT=nRT right? this is the start point of PV=nRT. pressure is more or less same for a room if it changes we can't breathe properly. Volume of the room is already constant. so left hand side of the equality is constant so...
1-) Quantum physics is the segment of the physics which can be applied to particles for which v/c-->0 but m is so small that it is comparible with h
so yes.
2-) Actually in quantum physics, there is no distinction of potantial or kinetic energy ratio, you just put potantial energy as a function...
i am sorry cathode-ray but i strongly disagree, this is the last thing to be said someone who is dealing with gauss' law, outer cylinder is perfectly symmetric that it won't have any effective electric field inside and it is clear if you take a cylinderic gaussian area.
there is just one weird...
i don't think i understood your question quite well, if they are glass they will break i think.
and with all due respect, i will answer for dickfore;
when you are at A's reference frame you see B moving to right with a speed of (180/181)c.
when you are at B's reference frame you see A moving to...
depends on what kind of collision it will be, if it is an inelastic collision, they will stop, if it is an elastic collision they will exchange their velocities, fortunately just like in the classical mechanics for this spesific case
it is nothing but rearranging
ACW=CW as ACW-CW=0 or CW-ACW=0
later on you will see that moment is a vector in fact and you will get used to this notation