Question

A solid object that weighs 300 N is placed on a wooden plank and a pressure of 60 N/mºis applied. What is the area of contact between the object and the wooden plank?
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Answer 1

Explanation:

Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium, 

Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium, B=0.5kg

Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium, B=0.5kgThe reading of the spring balance = Weight of water + Buoyant force' reaction pair force downwards

Since the weight of the block must be equal to the buoyant force acting on the block for it to remain in equilibrium, B=0.5kgThe reading of the spring balance = Weight of water + Buoyant force' reaction pair force downwards=1.5kg+0.5kg=2kg

Answer 2

Explanation:

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