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A softball pitcher throws a 0.17 kg ball toward a batter who is 20 m away. The
ball is traveling at 22 m/s when it leaves the pitcher's hand. What is the ball's
kinetic energy at this point?
OA. 4 J
OB. 82 J
C. 41 J

Respuesta :

Answer:

C

Explanation:

Kinetic Energy formula :

K.E. = 1/2mv²

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Given :

⇒ mass = 0.17 kg (needed)

⇒ distance = 20 m (not needed)

⇒ velocity = 22 m/s (needed)

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Solving :

⇒ K.E. = 1/2 × 0.17 × (22)²

⇒ K.E. = 11 × 0.17 × 22

⇒ K.E. = 1.87 × 22

⇒ K.E. = 41.14 ≈ 41 J
c fss trust good luck !!