Consider a man whose weight is w=247lb. (a)Find the minimum radius (m) of a heli

Question

Consider a man whose weight is w=247lb. (a)Find the minimum radius (m) of a helium balloon that will lift him off the ground. The density of helium gas is 0.178 kg/m3, the density of air is 1.29kg/m3. Neglect the weight of the deflated balloon itself. Assume that the shape of the balloon is a sphere. Neglect the buoyant force of the air on the man. (b) If we replace the helium with hydrogen gas, how much more weight (lb) can the balloon lift? The density of hydrogen gas is 0.089kg/m3. (Find how much weight in excess of w the same balloon inflated to the same volume can lift.) Caution: the answer is not "w" anymore. The buoyant force is the same. Only the weight of the gas has changed.

Explanation / Answer

Volume of baloon = [4/3]r^3. = V say.

Mass of air displaced V*1.29 kg/m^3

Mass of helium = V*0.178

Mass difference = V* [1.112]


Total mass = 191 lb + V*0.178 = 86.63kg + V*0.178 kg
V* 1.112 = 86.63+ V*0.178
0.934 V = 86.63
V = 92.75 m^3.

[4/3]r^3 = . 92.75

r = 2.81 m .
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Density difference between helium and hydrogen is
0.089 kg /m^3.
The excess mass it can lift is V *0.089
= 92.75*0.089 = 8.25 kg
= 18.2 lb.

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