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a 0.149-kilogram baseball, initially moving at 15 meters per second, is brought to rest in 0.040 seconds by a baseball glove on a catchers hand. the magnitude of the average force exerted on the ball by glove is

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carlosego
The net force in this case will be given by:
 F = m * ((delta V) / (delta t))
 where,
 delta V: speed differential.
 delta t: time differential.
 m: mass.
 Substituting the values:
 F = (0.149) * ((15) / (0.040))
 F = 55,875 N
 Answer:
 the magnitude of the average force exerted on the ball by glove is
 F = 55,875 N
fichoh

Using the relationship between force and acceleration, the average force exerted on the ball by the glove is :55.875 N

Given the Parameters :

  • Mass, m = 0.149 kg
  • Velocity, V = 15 m/s
  • Time, t = 0.040 seconds

Recall :

  • Force = Mass × acceleration

  • Acceleration = Δv / t

Force = (mΔv ÷ t)

Inputting the Values into the equation :

Force = [0.149 × (15 - 0)] ÷ 0.040

Force = (0.149 × 15) ÷. 0.040

Force = 2.235 ÷ 0.040

Force = 55.875 N

Therefore, the force exerted is 55.875 N

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