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A 5.5 kg box is pushed across the lunch table. The acceleration of the box is 4.2 m/s2. What was the net force applied to the box? A) 1.3 N B) 1.31 N C) 9.7 N D) 23.1 N

Respuesta :

The sum of all forces=m(mass) times a(acceleration). F=5.5(kg)*4.2(m/s^2)= 23.1N or D.

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