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Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of āˆ’8.4Nā‹…s to the baseball? Enter your answer numerically in meters per second using two significant figures.

Respuesta :

Answer:

V2 = -25.93 m/s

Explanation:

First of all, we know that;

Momentum = mass x change in velocity

Thus;

Momentum = m(v2 - v1)

Also, we know that;

Impulse = force x time

And also that;

Momentum = Impulse

From the question, m = 0.145kg and V1 = 32m/s while impulse = -8.4 N.s

Thus;

0.145(v2 - 32) = -8.4

Now,

0.145v2 - 4.64 = -8.4

0.145v2 = -8.4 + 4.64

0.145v2 = -3.76

v2 = -3.76/0.145

= -25.931 m/s and to two significant figures gives -25.93 m/s

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