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A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.

A. True
B. False

User Ekl
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Final answer:

The average force exerted on the leg is 2.40 × 10³ N toward the leg. The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton's third law) because the change in momentum and the time interval are the same.

Step-by-step explanation:

(a) To find the average force exerted on the leg, we can use the formula:

F = Δp / Δt

Where F is the force, Δp is the change in momentum, and Δt is the time interval. Since momentum is given by p = m * v (where m is the effective mass and v is the velocity), we can find the change in momentum:

Δp = m * Δv

Substituting the given values:

Δp = (1.50 kg) * (-4.00 m/s)

Δp = -6.00 kg·m/s

Now we can calculate the force:

F = (-6.00 kg·m/s) / (2.50 × 10-3 s)

F = -2.40 × 103 N

The average force exerted on the leg is 2.40 × 103 N toward the leg.

(b) If the woman clapped her hands together at the same speed and brought them to rest in the same time, the force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton's third law). This is because the change in momentum and the time interval are the same for both scenarios, so the average force exerted would be the same.

User SheffDoinWork
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