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.
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