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12 January, 01:44

What is the kinetic energy of a 0.012 kg bullet traveling at 700 m/s?

Remember the equation for kinetic energy is: Kinetic energy = - mºv2

A) 8.4 Joules

B) 4.2 Joules

C) 2940 Joules

D) 35.3 Joules

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  1. 12 January, 03:20
    0
    m=0.012

    v=700

    KE = 1/2mv^2

    KE = 1/2 (700) (0.012)

    KE = 4.2J
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