Question #294068

What is the kinetic energy of a


proton (proper mass is m0 = 1.67 x 10^-27 kg) travelling at 40 % of the speed


of light?

1
Expert's answer
2022-02-04T18:00:30-0500

γ=11(v/c)2=11(0.4c/c)2=110.16=1.091\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(0.4c/c)^2}}=\frac{1}{\sqrt{1-0.16}}=1.091


Ek=(γ1)mc2=(1.0911)×1.67×1027kg×(2.998×108ms)2=1.37×1011JE_k=(\gamma-1)mc^2=(1.091-1)\times1.67\times10^{-27}kg\times(2.998\times10^8\frac{m}{s})^2=1.37\times10^{-11}J


Answer: 1.37×1011J1.37\times10^{-11}J


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