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?

Question

Chemistry

Melvin Young

19 August, 08:58

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?

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Answers (1)

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Veronica Long · Answer 1

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