Recent content by Adam564

The conclusion of my attempt I am listing below is that there do exist entropies for both but I am not sure. $$dU=TdS-pdV$$ $$dS=\frac{dU}{T}+\frac{p}{T}dV$$ Therefore, gas A: $$S=\frac{{\Delta}U}{T}+\alpha_A(\frac{-N}{{\Delta}V})$$ Gas B...

I just got the answers, everything is correct, except for some of c) and whole d). t2' is in fact -2.25*10-10 s. this is because t2'=[t2-((vx2)/c2)]/√1-((v/c)2) therefore centipede is not intact because the centipede passes before the cleavers strike the table which is in fact what I wrote so I...

i think the solution for c) is like this (x1',x2')=(0,0.1125) (x1,x2)=(0,0.09) (t1',t2')=(0,0.0675) (t1,t2)=(0,0) x1'=γ(x1-vt1) x2'=γ(x2+vt2) t1'=γ(t1-(v/c2)x1) t2'=γ(t2-(v/c2)x2)

b) L=L0√(1-((v/c)^2)) L0=0.09m v=0.6c L=0.072 m c) S: cleavers are 9 cm apart speed of centipede is 0.6c centipede measures 8 cm S': centipede is 10 cm long cleavers are 0.072 m apart d) the centipede will be cut

Updated solution to part a): L=L0√(1-((v/c)^2)) L=0.08m L0=0.1m v=0.6c

Ok, thanks. I suppose once I am done all the parts I'll make a new post altogether.

@kuruman As far as I know, γ=1/√(1-(v/c)^2) Are you suggesting that γ can be determined some other way? What I am getting from your comment is it could be 2 cm, since that is the change in length of the centipede between the reference frames.

Hello, I am currently a Physics student and I often get stuck on questions wishing I was able to counsel someone quickly so I can learn more about Physics and my studies. This forum is exactly what I am looking for.

Homework Statement A high-speed centipede in S’ is 10.0 cm long measured at rest in S’. A butcher in S holds two cleavers (A and B) 9.00 cm apart (measured in S). The centipede runs at such high speed v across a chopping block that the butcher measures the length of the centipede to be 8.00 cm...