Question

What is the external net force exerted on a 3.5kg of papaya, which is being pushed across a table and has an acceleration of 2.2m/s² to the left?

Answer 1

Answer:

Given -

Mass (m) = 3.5 kg

Acceleration (a) = 2.2 m/s²

To find -

External net force exerted on papaya.

Solution -

From the second law of motion i.e., F = MA

→ F = 3.5 × 2.2

→ 7.7 kg ms-²

Therefore, the external net force exerted on papaya is 7.7 kg m/s².

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