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18 May, 13:02

If a m=74.7 kg person were traveling at v=0.990c, where c is the speed of light, what would be the ratio of the person's relativistic kinetic energy to the person's classical kinetic energy? What is the ratio of the person's relativistic momentum to the person's classical momentum?

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  1. 18 May, 14:50
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    ratio of the person's relativistic kinetic energy is 2.486

    the ratio of the person's relativistic momentum is 1.8983

    Explanation:

    Given data

    m=74.7 kg

    v=0.990 c

    c = speed of light

    to find out

    ratio of the person's relativistic kinetic energy and ratio of the person's relativistic momentum

    solution

    we will apply here relativistic constant equation that is

    relativistic constant = 1 / √ (1-v²/c²)

    put here all value

    relativistic constant = 1 / √ (1 - (0.850c) ²/c²)

    relativistic constant = 1.8983

    so

    relativistic kinetic energy = (relativistic constant - 1) mc²

    and we know person kinetic energy = 1/2 mv²

    so ration are = (relativistic constant - 1) mc² / 1/2 mv²

    ratio = 2 (1.8983 - 1) c² / (0.850c) ²

    ratio = 2.486

    so ratio of the person's relativistic kinetic energy is 2.486

    and

    relativistic momentum = relativistic constant mv

    and momentum = mv

    so ration are = relativistic constant mv / mv

    ratio = relativistic constant

    ratio = 1.8983

    so the ratio of the person's relativistic momentum is 1.8983
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