The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

To draw a free-body diagram (FBD), we must identify all the forces acting on the human body floating in seawater. These forces are the buoyancy force (\(F_b\)) acting upward and the gravitational force, or weight of the person (\(W\)), acting downward.

Since the person is floating, the buoyancy force must equal their weight: \(F_b = W\) The buoyancy force is given by \(F_b = \rho_{water} \cdot V_{submerged} \cdot g\) and the weight of the person is given by \(W = \rho_{body} \cdot V_{total} \cdot g\), where \(\rho_{water}\) is the density of seawater, \(\rho_{body}\) is the density of the human body, \(V_{submerged}\) is the submerged volume, \(V_{total}\) is the total volume of the person, and \(g\) is the acceleration due to gravity. To find the percentage submerged, set these forces equal: \(\rho_{water}\cdot V_{submerged}\cdot g = \rho_{body}\cdot V_{total}\cdot g\) Since we are looking for the percentage submerged, we can define a ratio \(R = \frac{V_{submerged}}{V_{total}}\). By substituting this into the previous equation and solving for \(R\), we get: \(R = \frac{\rho_{body}}{\rho_{water}} = \frac{985 \mathrm{~kg} / \mathrm{m}^{3}}{1020 \mathrm{~kg} / \mathrm{m}^{3}} \approx 0.9657\) The percentage submerged is therefore approximately 96.57%. #b) Inhalation and Exhalation Percentage#

Repeating the same process as before, we can calculate the new percentage submerged after the person inhales air, with the new density of the human body being \(945 \mathrm{~kg} / \mathrm{m}^{3}\). \(R_{inhalation} = \frac{\rho_{body(inhalation)}}{\rho_{water}} = \frac{945 \mathrm{~kg} / \mathrm{m}^{3}}{1020 \mathrm{~kg} / \mathrm{m}^{3}} \approx 0.9265\) The percentage submerged after inhalation is approximately 92.65%.

The difference in the percentage of the body submerged before and after inhalation will give us the percentage of volume moving up and down. \(\Delta R = R_{inhalation} - R = 0.9265 - 0.9657 = -0.0392\) Therefore, approximately 3.92% of the volume moves up and down as the person inhales and exhales. #c) Density of Dead Sea Seawater#

The problem states that two-thirds of the volume of the body is submerged in the Dead Sea, so the current ratio of submerged volume is: \(R_{DeadSea} = \frac{2}{3}\)

Using the same relationships as before, we can set up the equation: \(R_{DeadSea} = \frac{\rho_{body}}{\rho_{DeadSea}} \rightarrow \rho_{DeadSea} = \frac{\rho_{body}}{R_{DeadSea}} = \frac{985 \mathrm{~kg} / \mathrm{m}^{3}}{2/3} \approx 1477.5 \mathrm{~kg} / \mathrm{m}^{3}\) The density of the seawater in the Dead Sea is approximately \(\boxed{1477.5 ~\mathrm{kg} / \mathrm{m}^{3}}\).