Question #267039

A 3.5 kg papaya is pushed across a table. If the acceleration of the papaya is 2.2 m/s² to the left, what is the net external force exerted on the papaya?


Expert's answer

The net force:


F=ma=7.7 m/s2F=ma=7.7\text{ m/s}^2

to the left.


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