a 5.5 kg box is pushed across the lunch table. The net force applied to the box is 9.7N. what is the acceleration of the box.

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12 votes

Answer:

1.8 m / s^2

Step-by-step explanation:

F = ma ( ignoring friction)

F/ m = a

9.7 / 5.5 = 1.8 m/s^2

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