Question

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

Answer 1

Answer:

please add all the given marks

excreted of 3.5 kg papaya pushed = 2.2

then you have to add the the following

3.5+2.2

Answer 2

G=Force 3.5kg

    acceleration 2.2 m/s2

R=mass

E=3.5kg.m/s2  

    2.2m/s2

S=3.5kg.m/s2

    2.2kg.m/s2

A= 7.7N to the left