# A piece of ice with a volume of 5 dm3 floats on the surface of the water Determine the volume of the underwater

A piece of ice with a volume of 5 dm3 floats on the surface of the water Determine the volume of the underwater and surface parts of the water density 1000 kg \ m3 ice density 900 kg \ m3

Given:

V = 5dm³

Pw = 1000kg / m³

Rl = 900kg / m³

Find: V1; V2.

5dm³, tobish the full volume is transferred to the SI system, then it will be 0.005m³

Decision:

Let’s write down Archimedes’ law for two floating bodies:

Rl × g × V = Pv × g × Vvv;

(Vвв – the volume of water displaced by the body),

g to shrink,

And it remains: Rl × V = Pv × Vvv,

Then Vвв = Рл × V / Pв,

This will be the volume of the underwater part.

Substituting the numbers, we have:

Vwv = 900kg / m³ × 0.005m³ / 1000kg / m³ = 0.0045m³ (or 4.5dm³);

Then the above-water part is 5 – 4.5 = 0.5dm³, or 0.0005m³.

Answer: above-water part – 0.0045m³, underwater – 0.0005m³

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