Asked by Jamie
A shipwrecked mariner is stranded on a desert island. He seals a plea for rescue in a 1.00 liter bottle, corks it up, and throws it into the sea. If the mass of the bottle, plus the message and the air inside, is 0.454 kg, what percentage of the volume of the bottle is submerged as it bobs away? Take the density of seawater to be 1030 kg/m3. For simplicity, assume the bottle and its contents have a uniform density.
Answers
Answered by
Henry
Convert density of bottle and water to
the same units.
Db = 0.454kg/Liter
Db = 0.454kg/1000cm^3
Db = 454.*10^-6kg / cm^3. = Density of bottle.
Vb = 1 Liter = 1000 cm^3. = Vol. of bottle.
Dw = 1030kg/m^3
Dw = 1030kg/10^6cm^3
Dw = 1030*10^-6kg/cm^3. = Density of the water.
Vs = (Db/Dw)* Vb
Vs=(454*10^-6/1030*10^*1000cm^3=441cm^3
Vol. submerged.
%Vs=(Vs/Vb)*100%=(441/1000)*100%=44.1%.
the same units.
Db = 0.454kg/Liter
Db = 0.454kg/1000cm^3
Db = 454.*10^-6kg / cm^3. = Density of bottle.
Vb = 1 Liter = 1000 cm^3. = Vol. of bottle.
Dw = 1030kg/m^3
Dw = 1030kg/10^6cm^3
Dw = 1030*10^-6kg/cm^3. = Density of the water.
Vs = (Db/Dw)* Vb
Vs=(454*10^-6/1030*10^*1000cm^3=441cm^3
Vol. submerged.
%Vs=(Vs/Vb)*100%=(441/1000)*100%=44.1%.
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