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lutik1710 [3]
3 years ago
10

If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th

e ratio of the person's relativistic kinetic energy to the person's classical kinetic energy?
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

\frac{K.E_r}{K.E}=2.875

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

K.E_r=(\gamma-1)mc^2 ...........(1)

where,

\gamma = relativistic factor, given as; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Now, the classical kinetic energy is given as:

K.E = \frac{1}{2}mv^2    ..........(2)

Dividing the equation (1) by (2) we get

\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}

or

\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}

substituting the values in the equation we get,

\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}

or

\frac{K.E_r}{K.E}=2.875

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