Question

a constant force acts on an object of mass 8kg for a duration of 4s .it increases from 5m/s to 10/ms .find the magnitude of applied force​

Answer 1

Answer:

answer in the attachment unnie/sis points waste mat karo

Answer 2

Given,

mass=5kg

t 1=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t2=5s

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So,

Let the Force be F

Let the acceleration be a

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So,

a= (v−u)/t = (7−3)/2 =2m/s ²

So the magnitude of the applied force is 10N

And the final velocity after 5s is v

______________________

So,

v=u+at

v=3+2×5

v=13m/s

The final velocity after 5s is 13m/s .