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Kitty [74]
4 years ago
5

A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball,

it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?
Physics
1 answer:
Strike441 [17]4 years ago
4 0
Momentum = (mass) x (velocity)
Original momentum before the hit = 
                   (0.16 kg) x (38 m/s) this way <==
               =             6.08 kg-m/s  this way <==
Momentum after the hit = 
                   (0.16) x (44 m/s) that way  ==>
               =           7.04 kg-m/s  that way ==>
Change in momentum = (6.08 + 7.04) =  13.12 kg-m/s  that way ==> .-----------------------------------------------
Change in momentum = impulse.
                                   Impulse = (force) x (time the force lasted)
                          13.12 kg-m/s  = (force) x (0.002 sec)
  (13.12 kg-m/s) / (0.002 sec)  =  Force
             6,560 kg-m/s² = 6,560 Newtons  =  Force    
                            ( about 1,475 pounds  ! ! ! )
 Hoped this helped!! ☺
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2 poles

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Thus, to find the time, we will use the equation;

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t = √(31/9.8)

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<h3>What is work done?</h3>

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