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wariber [46]
3 years ago
11

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds.
a. Determine the force acting upon Logan's body.
b. If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds. Determine what the force on his body would have been for such an abrupt collision.
Physics
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

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<h3>What is the Lorentz magnetic force acting on the proton?</h3>
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                                  F= q (v× B)

                               F= qvBsin\alpha  where, \alpha is the angle between v and B.

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  • Thus, we can find the force acting on the proton as,

                      F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N

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