The average human has a density of 944 kg/m 3 after inhalingand 1032 kg/m 3 afte

Question

The average human has a density of 944 kg/m3 after inhalingand 1032 kg/m3 after exhaling. (a) Without making any swimming movements, what percentage of thehuman body would be above the surface in the Dead Sea (a lake witha water density of about 1230 kg/m3) in each of thesecases?

After Inhaling After Exhaling % %

(a) Without making any swimming movements, what percentage of thehuman body would be above the surface in the Dead Sea (a lake witha water density of about 1230 kg/m3) in each of thesecases?

After Inhaling After Exhaling % %

After Inhaling After Exhaling % %

Explanation / Answer

The average human has a density of 1=944 kg/m3 after inhaling and 2= 1032kg/m3 after exhaling. (a)The density of water in the Dead Sea is = 1230kg/m3 An object with less density than the fluid will float upwarduntil it reaches the surface of the fluid. At that position, onlypart of the object is submerged.The volume submerged depends on thedensity of the object, as compared to the fluid. The equationis f * Vs = o*Vo or Vs= (o*Vo/f) Therefore,the volume of human body that is submerged afterinhaling is Vs= (o*Vo/f) or Vs= (o*Vo/f) Therefore,the volume of human body that is submerged afterinhaling is Vs= (o*Vo/f) Here,o= 1 andf = or Vs= (1/) *Vo or Vs= (944/1230) * Vo = 0.767 *Vo Therefore,the percent of human body which would be above thesurface in the Dead Sea after inhaling is 76.7 %. Similarly,the volume of human body that is submerged afterexhaling is Vs= (o*Vo/f) Vs= (o*Vo/f) Here,o= 2 andf = or Vs= (2/) *Vo or Vs= (1032/1230) * Vo = 0.839 *Vo Therefore,the percent of human body which would be above thesurface in the Dead Sea after exhaling is 83.9 %. Therefore,the percent of human body which would be above thesurface in the Dead Sea after exhaling is 83.9 %. Therefore,the percent of human body which would be above thesurface in the Dead Sea after exhaling is 83.9 %.

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