Question

A metal piece of mass 160 g lies in equilibrium inside a glass of water (figure 13-E4). The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg/m³, find the normal force exerted by the bottom of the glass on the metal piece.

Answer 1

In the equilibrium position,

Upthrust Force + Normal Force = Weight of Gravity.

∴ Vρg + Normal force = mg

∴ m/density of metal × density of fluid × g + Normal Force = mg.

∴ 1.6/8000 × 1000 (water density) × 10 + F = 0.16 × 10

∴ F = 1.6 - 0.2

F = 1.4 N.

Hence, the force acting is 1.4 N.

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Answer 2

Answer:

Explanation:

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