Let Vp be the required plastic volume.
sgp = plastic specific gravity
sgm = man's specific gravity.
Vm = man's volume = 75000g/1.04 g/cc
= 0.0721 m^3
M = man's mass = 75 kg
sgw = water specific gravity = 1.00
(Total weight) - (buoyancy) = 0
(Vm*sgm + Vp*sgp) - [(Vp*sgw)+0.8*Vm*sqw] = 0
Vp[sgw - sgp] = Vm[sgm - 0.8*sgw]
Vp[0.42] = Vm[1.04 - 0.8]
Vp = 0.57 Vm = 0.0412 m^3
A foam plastic ( sg = 0.58 ) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 % (by volume) of a 75 kg man above water in a lake? Average density of a man is equal to 1.04 g/cc. Assume the foam plastic’s top surface to be level with the water surface
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