A passenger-carrying helium balloon consists of an envelope of volume V (it cont

Question

A passenger-carrying helium balloon consists of an envelope of volume V (it contains the gas) and a passenger basket of negligible volume v suspended from the balloon. Let the mass of the helium be M, and the combined mass of everything else (basket, passengers, envelope, etc.) be m= 1200 kg. What must V be in order for the balloon to float in the air at a steady altitude? The densities of helium and air are phe=0.18 kg/m3 and pA=1.3 kg/m3, retrospectively. Hint: think in terms of average density. Begin your final algebraic and numerical solutions this way: V...

Explanation / Answer

Since volume v is negligible. Total volume of the balloon will be V m^3
This is also the volume of air displaced
Buoyant force = volume of air displaced * density of air * g

This buoyant force must balance the weight
Total weight = total mass *g
                         = (M+m)*g
Here m = 1200 Kg
M = mass of helium = volume of helium * density of helium
                                     = V * volume of helium * density of helium

so,
total weight = Buoyant force
(V* density of helium + m)*g = volume of air displaced * density of air * g
(V* density of helium + m) = volume of air displaced * density of air
V*0.18 +1200 = V*1.3
1.12*V = 1200
V=1071.43 m^3
Answer: 1071.43 m^3

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