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 batt

Question

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 −8.4N⋅s to the baseball? Enter your answer numerically in meters per second using two significant figures.

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Nho 4 years 2021-08-18T02:34:42+00:00 1 Answers 274 views 0

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  1. yenoanh
    0
    2021-08-18T02:36:24+00:00
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    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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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )