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A golf club strikes a 0.031-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6240 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club

Respuesta :

Answer:

66.5 m/s

Explanation:

m = mass of the golf ball = 0.031 kg

F = magnitude of force applied to the ball = 6240 N

Acceleration experienced by the ball is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{6240}{0.031}[/tex]

a = 201290.32 m/s²

d = distance for which the ball is in contact with the golf club = 0.011 m

v₀ = initial speed of the ball = 0 m/s

v = final speed of the ball = 0 m/s

Using the kinematics equation

v² = v₀² + 2 a d

v² = 0² + 2 (201290.32) (0.011)

v = 66.5 m/s

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