A(n) 0.212 kg baseball is thrown with a speed of 18.8 m/s. It is hit straight back at the pitcher with a final speed of 21.2 m/s. What is the magnitude of the impulse delivered to the ball?

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aristeus aristeus · Answer 1 · 2019-10-15T05:10:00+02:00

Answer:

Impulse will be 8.4788 kgm/sec

Explanation:

We have given mass of the ball m = 0.212 kg

Initial velocity u = 18.8 m/sec

We know that momentum P = mass×velocity

Taking initial velocity as negative

So initial momentum = 0.212×-18.8 = -3.9856 kgm/sec

Final velocity is given as v = 21.2 m/sec

So final momentum = 0.212×21.2 = 4.4944 kgm/sec

We know that impulse is given by change in momentum

So impulse = final momentum-initial momentum = 4.4944-(-3.9844 ) = 8.4788 kgm/sec

okpalawalter8 okpalawalter8 · Answer 2 · 2022-02-25T12:25:14+01:00

The magnitude of the impulse is mathematically given as

Impulse = 8.4788 kgm/sec

Question Parameters:

Generally the equation for the momentum is mathematically given as

P = mass×velocity

Initial momentum = 0.212×-18.8

Initial momentum= -3.9856 kgm/sec

Final velocity

vf = 21.2 m/sec

final momentum = 0.212×21.2

final momentum= 4.4944 kgm/sec

Therefore

Impulse = final momentum-initial momentum

Impulse = 4.4944-(-3.9844 )

Impulse = 8.4788 kgm/sec

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