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?
γ=11−(v/c)2=11−(0.4c/c)2=11−0.16=1.091\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(0.4c/c)^2}}=\frac{1}{\sqrt{1-0.16}}=1.091γ=1−(v/c)21=1−(0.4c/c)21=1−0.161=1.091
Ek=(γ−1)mc2=(1.091−1)×1.67×10−27kg×(2.998×108ms)2=1.37×10−11JE_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}JEk=(γ−1)mc2=(1.091−1)×1.67×10−27kg×(2.998×108sm)2=1.37×10−11J
Answer: 1.37×10−11J1.37\times10^{-11}J1.37×10−11J
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