An ice floe of uniform thickness floats in water, plunging into it by 0.9 of its volume

An ice floe of uniform thickness floats in water, plunging into it by 0.9 of its volume. the density of water is 10 ^ 3 kg / m ^ 3 to determine the density of ice.

ρw = 10 ^ 3 kg / m3.

Vv = 0.9 * V.

ρl -?

Two forces act on the ice floe, which is immersed in water: the force of gravity Ft directed vertically downward, and the buoyancy force of Archimedes Farkh, directed vertically upward.

Ft = Farch.

The force of gravity Ft is determined by the product of the body mass m by the acceleration of gravity g: Ft = m * g.

The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρv * g * Vv. Where ρw is the density of the liquid, that is, water, g is the acceleration of gravity, Vw is the volume of the part of the body that is immersed in water.

m * g = ρв * g * Vв.

m = ρl * V.

ρl * V * g = ρw * g * Vv.

ρl = ρw * Vw / V = ​​ρw * 0.9 * V / V = ​​0.9 * ρw.

ρl = 0.9 * 10 ^ 3 kg / m3 = 900 kg / m3.

Answer: the density of the ice floe is ρl = 900 kg / m3.