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