Question #346316

An accelerator at a Physics Laboratory can accelerate protons to a Kinetic energy of 10 ^ 12 * eV The rest mass of a proton is m * c ^ 2 = 9.38 * 10 ^ 6 * cV . What is the speed of that proton?


1
Expert's answer
2022-05-30T15:16:50-0400
Ek=mc21v2/c2mc2E_k=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^21012=9.381061v2/c29.3810610^{12} =\frac{9.38 * 10 ^ 6}{ \sqrt{1-v^2/c^2}}-9.38 * 10 ^ 6

1v2/c2=1/((Ek/mc2+1)21-v^2/c^2=1/((E_k/mc^2+1)^2

v=c11/(Ek/mc2+1)2v=c\sqrt{1-1/(E_k/mc^2+1)^2}

v=c11/(1012/9.38106+1)2=cv=c\sqrt{1-1/(10 ^{ 12}/9.38*10^{6}+1)^2}=c

impossible


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS