A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question part Part 1 (a

Question

A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question part Part 1 (a) If the ball was in contact with the floor for 0.5 ms (0.5 × 10-3 s), what was the average magnitude of the force exerted on the ball by the floor?

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King 4 years 2021-07-20T21:00:23+00:00 2 Answers 10 views 0

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  1. Sapo1
    0
    2021-07-20T21:01:33+00:00
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    Answer:

    7.5

    Explanation:

  2. dangkhoa
    0
    2021-07-20T21:02:05+00:00
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    Answer:

    The  the average magnitude of the force exerted on the ball by the floor 11.2 \times 10^{3} N

    Explanation:

    Given:

    Mass of ball m = 0.4 kg

    Initial speed v_{i} = -8 \frac{m}{s}

    Rebound speed v_{f} = 6 \frac{m}{s}

    Contact time interval \Delta t = 0.5 \times 10^{-3} sec

    For finding the average magnitude of the force on the ball by the floor is given by,

       F_{avg} = \frac{\Delta P}{\Delta t}

    Here \Delta P = m (v_{f}- v_{i} )

       F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}

       F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }

       F_{avg} = 11.2 \times 10^{3} N

    Therefore, the  the average magnitude of the force exerted on the ball by the floor 11.2 \times 10^{3} N

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