A vacationer floats lazily in the ocean with 89% of his body below the surface. The density of the ocean water is 1025 kg/m3. What is the vacationer's average density?

The vacationer's average density is 912 kg/m³

Explanation:

Archimedes principle states that the buoyancy of an object in a liquid is equal to the weight of the liquid displaced, hence since we have

Density of sea water = 1025 kg/m³

Volume of vocationer's body below water = 89 % of total volume

Mass of seawater displaced = mass of vocationer

volume of seawater displaced = 0.89 × volume of vocationer

volume of vocationer = (volume of seawater displaced)/0.89

Density of vocationer = mass/volume = (mass of seawater displaced)×0.89/(volume of seawater displaced)

Therefore deensity of vocationer = 0.89×density of seawater

= 1025×0.89 = 912 kg/m³

dynamic equilibrium is the answer.