Question

a person, having a density of 993 kg/m³, is floating in water. what percentage of the persons body is underwater? ? % ρwater = 1,000 kg/m³

Explanation:

Step1: Recall Archimedes' principle for floating

When an object floats in a fluid, the buoyant force equals the weight of the object. The buoyant force is also equal to the weight of the displaced fluid. So, $F_b = W_{object} = W_{displaced\ fluid}$. Since weight $W =
ho V g$ (where $
ho$ is density, $V$ is volume, and $g$ is acceleration due to gravity), we have $
ho_{object} V_{object} g=
ho_{fluid} V_{displaced} g$. The $g$ cancels out, giving $
ho_{object} V_{object}=
ho_{fluid} V_{displaced}$. The fraction of the object's volume underwater is $\frac{V_{displaced}}{V_{object}}=\frac{
ho_{object}}{
ho_{fluid}}$.

Step2: Calculate the fraction

Given $
ho_{person} = 993\ kg/m^3$ and $
ho_{water}=1000\ kg/m^3$. The fraction is $\frac{993}{1000}=0.993$.

Step3: Convert to percentage

To convert the fraction to a percentage, multiply by 100: $0.993\times100 = 99.3\%$.

Answer:

99.3