Acceleration = (0.2 x g) = 1.96m/sec^2.
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>
<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>
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Answer:
your the class suru and Sahiba and Sahiba are all guddu saifi and Mam aap Aman and Sahiba are not
here we can say that net force on the student vertically upwards will be counter balance by his weight downwards
Let net force F is exerted by each hand
so here we will have
so the force exerted by each hand will be 414.1 N
