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The acceleration of gravity is 9.8 m/s 2 . After impact, the block slides 9.15 m before coming to rest. 84 g 10.4 g vbullet 94.4 g 9.15 m µ = 0.251 If the coefficient of friction between block and surface is 0.251 , what was the speed of the bullet immediately before impact?

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anniwuanyanwu1

Answer:speed of bullet = 14.1m/s

Explanation:

Vo=( m+m)/m sqrt(2gdu)

d=9.15m

g=9.8m/s^2

u=0.251

Vo= 84+94.4/84 sqrt(2×9.8×9.15×0.251)

Vo=178.4/84 sqrt(45.01)

Vo=2.1×6.71

Vo= 14.1m/s

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