Determine the mass of a piece of aluminum, which, when immersed in water, has a buoyancy force of 1.2 N.

Determine the mass of a piece of aluminum, which, when immersed in water, has a buoyancy force of 1.2 N. The density of water is 1000 kg / m3, aluminum is 2700 kg / m3.

The Archimedes force will act on aluminum in water:
Fа = ρw * g * V, where Fа is the Archimedes force (Fа = 1.2 N), ρw is the density of water (ρ = 1000 kg / m³), g is the acceleration of gravity (g = 10 m / s²), V is the volume of a piece of aluminum (V = m / ρa, where m is the mass of a piece of aluminum, ρa is the density of aluminum (ρa = 2700 kg / m³)).
Let’s express and calculate the mass of a piece of aluminum:
Fa = ρw * g * V = ρw * g * m / ρa.
m = Fa * ρa / (ρw * g) = 1.2 * 2700 / (1000 * 10) = 0.324 kg.
Answer: The mass of a piece of aluminum is 0.324 kg.