# Determine the mass of a piece of aluminum on which, when immersed in water, a buoyant force of 1.2 N acts

**Determine the mass of a piece of aluminum on which, when immersed in water, a buoyant force of 1.2 N acts. The density of water is 1000 kg / m3. Aluminum – 2700 kg / m3**

1) Let’s calculate the volume of a piece of aluminum.

Fа = ρ * g * V, where Fа is the active force of Archimedes (Fа = 1.2 N), ρ is the density of the liquid in which the piece of aluminum is located (water, ρ = 1000 kg / m3), g is the acceleration of gravity ( let us take g = 10 m / s2), V is the volume of a piece of aluminum.

V = Fa / (ρ * g) = 1.2 / (1000 * 10) = 0.00012 m3.

2) Let’s calculate the mass of a piece of aluminum.

m = V * ρ, where ρ is the known density of aluminum (ρ = 2700 kg / m3).

m = 0.00012 * 2700 = 0.324 kg = 324 g.

Answer: A piece of aluminum has a mass of 324 grams.