Question complete statement:
The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. Without making any swimming movements, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about 1230 kg/m3) in each of these cases?
Answer:
P(in) = 23.17 %
P(ex) = 17%
Explanation:
Given data:
Density after inhaling = 945 kg/m3
Density after exhaling = 1020 kg/m3
Density of water body = 1230 kg/m3
% of human body above the surface in the dead sea = ?
Formula:
The buoyant force is equal to the weight of the body.
P (f) x g x V(f) = P(o) x g x V(o)
Volume of the fluid displaced corresponds to volume of the object beneath the fluid level.
For inhaling:
[tex]\frac{V(DS)}{V(in)}[/tex] = [tex]\frac{P(in)}{P(DS)}[/tex]
V(in) = P(in) / P(DS) x V(DS)
= 945 / 1230 x V(DS)
V(in) = 0.7683 x V(DS)
Calculate what percentage of the human body would be above the surface in the dead sea.
P(in) = (1 - 0.7683) x 100
= 0.2317 x 100
P(in) = 23.17 %
For exhaling:
V(ex) = P(ex) / P(DS) x V(DS)
= 1020 / 1230 x V(DS)
= 0.8292 x V(DS)
Calculate what percentage of the human body would be above the surface in the dead sea
P(ex) = (1 - 0.8292) x 100
= 0.17 x 100
P(ex) = 17%
