A buoyant force equal to 686N acts on the body of a person immersed in fresh water.

| May 15, 2021 | education

A buoyant force equal to 686N acts on the body of a person immersed in fresh water. What will be the buoyancy of sea water?

Farkh1 = 686 N.

ρ1 = 1000 kg / m3.

ρ2 = 1030 kg / m3.

Farkh2 -?

In fresh water, the buoyancy force of Archimedes Farkh1 will be determined by the formula: Farkh1 = ρ1 * V * g, where ρ1 is the density of fresh water, V is the volume of the human body, g is the acceleration of gravity.

V * g = Farch1 / ρ1.

In seawater, the buoyancy force of Archimedes Farkh2 will be determined by the formula: Farkh2 = ρ2 * V * g, where ρ2 is the density of sea water, V is the volume of the human body, g is the acceleration of gravity.

Farch2 = ρ2 * Farch1 / ρ1.

We take the density of fresh ρ1 and sea water ρ2 from the table of the density of substances: ρ1 = 1000 kg / m3.

ρ2 = 1030 kg / m3.

Farch2 = 1030 kg / m3 * 686 N / 1000 kg / m3 = 706.6 N.

Answer: in seawater, the buoyancy force of Archimedes, which acts on a person, is Farch2 = 706.6 N.



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