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LiRa [457]
2 years ago
15

Suppose that Salim hits a 100 g fastball moving towards him at a speed of 42 m/s with his bat as shown in the image.the impulse

of the fastball is 8 kg.m/s in the antiparallel direction of the initial velocity. Then the final momentum of the fastball is --------------( Suppose that the direction of the fastball toward the bat is negative and away from the bat is positive)
1.8 kg m/s
0.6 kg m/s
3.8 kg m/s
2.4 kg m/s
Physics
1 answer:
FinnZ [79.3K]2 years ago
3 0

The final momentum of the ball is 3.8 kgm/s.

<h3>Change in momentum of the ball</h3>

The impulse received by the ball is equal to change in momentum of the ball.

J = ΔP

where;

  • J is the impulse
  • ΔP is change in momentum

ΔP = P₂ - P₁

P₂ = ΔP + P₁

<h3>Final momentum of the ball</h3>

The final momentum of the ball is calculated as follows;

P₂ = 8 + (- 0.1 x 42)

P₂ = 8 - 4.2

P₂ = 3.8 kgm/s

Learn more about change in momentum here: brainly.com/question/7538238

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Given

Car 1

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v1 = 20 m/s

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Procedure

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\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

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A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t
navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that v^2=u^2+2as

16^2=24^2+2\times a\times 50

a=-3.2m/sec^2

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So v^2=u^2+2as

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So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration a=-3.2m/sec^2

So time t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec

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