A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a veloc

Question

A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?

in progress 0
Adela 4 years 2021-08-26T11:40:52+00:00 1 Answers 28 views 0

Answers ( )

  1. Sapo1
    0
    2021-08-26T11:42:14+00:00
    Reply

    Answer:

    About 1.795 \times 10^{-3} seconds

    Explanation:

    \Delta p=F \Delta t, where delta p represents the change in momentum, F represents the average force, and t represents the change in time.

    The change of velocity is:

    37-(-27.1)=64.1m/s

    Meanwhile, the mass stays the same, meaning that the change in momentum is:

    64.1\cdot 0.14kg=8.974

    Plugging this into the equation for impulse, you get:

    8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s

    Hope this helps!

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )