Answered

Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

A 1.5 kg baseball is pitched at 38 m/s and is hit by the batter. It heads directly back at the pitcher at 27 m/s.

a) Find the impulse applied to the ball.

b) If the bat is in contact with the bat for 0.45 sec, how much force was applied.

Show all your work.

Sagot :

zaniahsims

Answer:

The initial velocity of the ball is +38.0 m/s. The final velocity is -27.0 m/s.  

momentum = mass x velocity

(mass x final velocity) – (mass x initial velocity)

(1.5)(-27) – (1.5)(38) = -97.5

The impulse applied to the ball is -97.5 kg m/s

impulse = force x time interval

force = impulse/time interval

-97.5/0.45 s = - 216.6  

The force applied was 216.6 N.

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.