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vichka [17]
3 years ago
13

A(n) 0.49 kg softball is pitched at a speed of 13 m/s. The batter hits it back directly at the pitcher at a speed of 23 m/s. The

bat acts on the ball for 0.014 s. What is the magnitude of the impulse imparted by the bat to the ball?
Physics
1 answer:
garik1379 [7]3 years ago
4 0
<h2>Impulse  = 4.9 kgm/s</h2>

Explanation:

Impulse is given by change of momentum.

Mass of softball = 0.49 kg

Initial velocity of softball = 13 m/s

Initial momentum = 0.49 x 13 = 6.37 kgm/s

Final velocity of softball = 23 m/s

Final momentum = 0.49 x 23 = 11.27 kgm/s

Change of momentum = Final momentum - Initial momentum

Change of momentum = 11.27 - 6.37

Change of momentum = 4.9 kgm/s

Impulse = Change of momentum

Impulse  = 4.9 kgm/s

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Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

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angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

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