The average human has a density of 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling. (a) 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?

20% of the human body float above the surface in the dead sea.

Explanation:

Given data,

The human density after inhaling, d₁ = 945 kg/m³

The human density after exhaling, d₂ = 1020 kg/m³

Therefore the average human density, d = 982.5 kg/m³

The density of the dead sea, D = 1230 kg/m³

The percentage of density of human body to the dead sea is,

                                   P % = (982.5 / 1230) x 100 %

                                           = 80 %

Therefore, the human body has 80% of dead sea density.

Hence, 20% of the human body float above the surface in the dead sea.

this is poiselle's equation from fluid mechanic in physics. mech eng

i believe the answer would be:

d. violin, clarinet, french horn, and tuba