Question

What is the kinetic energy in joules of a 0.02 kg bullet traveling 300 m/s?

Answer 1

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Answer...

K = ½mv^2

K = 0.02 x 300 x 300 / 2

K = 0.01 x 300 x 300

K = 3 x 300

K = 900 J

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Answer 2

Answer:

900J

explanation:

K.E of bullet=1/2mv^2

m is mass of bullet=0.02

v is velocity=300m/s

so,K.E=1/2*(0.02)*(300)^2

K.E=900J

so, answer is 900J

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