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lidiya [134]
3 years ago
12

A(n) 0.5 kg soccer ball approaches a player horizontally with a speed of 10.6 m/s. The player illegally strikes the ball with he

r hand and causes it to move in the opposite direction with a speed of 19.9 m/s. What is the magnitude of the impulse delivered to the ball by the player? Answer in units of kg · m/s.
Physics
1 answer:
Llana [10]3 years ago
6 0

Answer: - 25.2 kgm/s

Explanation: The mass of the ball is 0.5kg, and the initial velocity = 10.6m/s.

The final velocity is in opposite direction of the initial hence final velocity (v) = - 19.9 m/s

Impulse = change in momentum = final momentum - initial momentum.

Final momentum = mass × final velocity

Final momentum = - 19.9 × 0.5

Final momentum = - 9.95 kgm/s

Initial momentum = mass × initial velocity

Initial momentum = 0.5 × 10.6 = 5.3kgm/s

Change in momentum = final momentum - initial momentum = - 19.9 - 5.3

Change in momentum = - 25.2 kgm/s

The negative sign implies that the change in momentum is the opposite direction relative to the first.

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Read 2 more answers
A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500
mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

                                   Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

                                   Dm = Tm*Vm

                                   Dm = 20*60*3

                                   Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

                                   Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

                                   Dw = Tw*Vw

                                   Dw = 15*60*4

                                   Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

                                   s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

            dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

                                   2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

                                   s*ds/dt = (dm + dw)*(Vm + Vw)

                                   ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

                 ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

                 ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]  

                ds/dt = 6.98 ft/s

                 

           

7 0
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