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Question

When a particle doubles its speed from 0.45c0.45 c to 0.90c0.90 c, the relativistic energy will increase by more than a factor of four.

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Answered 2 years ago
Answered 2 years ago
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This statement is true as the connection between the speed and energy is not quadratic but

K=(11v2/c21)m0c2K=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)m_0c^2

So we have that

E1E2=110.921110.4521=10.8>4\frac{E_1}{E_2}=\frac{\frac{1}{\sqrt{1-0.9^2}}-1}{\frac{1}{\sqrt{1-0.45^2}}-1}=10.8\gt4

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