What Archimedean force acts on a 540 g aluminum bar half submerged in water?

Given: m (mass of an aluminum bar) = 540 g (0.54 kg); the bar is immersed in water for 1/2 volume.

Constants: g (acceleration due to gravity) ≈ 10 m / s2; ρw (density of water in which the bar is immersed) = 1000 kg / m3; ρа (density of aluminum) = 2700 kg / m3.

The required Archimedean force acting on the taken aluminum bar is determined by the formula: Fa = ρw * g * Vw = ρw * g * V / 2 = ρw * g * m / 2ρа.

Calculation: Fa = 1000 * 10 * 0.54 / (2 * 2700) = 1 N.

Answer: A buoyant force of 1 N. acts on the taken block.




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