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

m=0.012

v=700

KE = 1/2mv^2

KE = 1/2(700)(0.012)

KE = 4.2J

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explanation:

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