Question

A consumer upset with the latest trend of postal rate increases has decided to try to send letters by balloon even though they may not reach their intended destinations. A 6.00×104 cm3 gas-filled balloon will provide enough lift for a 45.1 g package to be accelerated upward at a rate of 2.95 m/s2 . For these circumstances, calculate the density of the gas the consumer fills the balloon with. The acceleration due to gravity is ?=9.81 m/s2 and the density of air is ?air=1.16 kg/m3 . Neglect the mass of the balloon material and the volume of the package. ?= kg/m3

Answer 1

The mass of displaced air M = ρair*V = 1.16kg/m³ * 0.0600m³ = 0.0696 kg
The buoyant force Fb = ρair*V*g = 1.16kg/m³ * 0.0600m³ * 9.81m/s² = 0.683 N

The mass of gas m = ρ*V = ρ*0.0600m³
The weight of gas Fg = ρ*V*g = ρ*0.0600m³*9.81m/s² = ρ*0.5886m⁴/s²

Now the acceleration a = Fnet / totalmass = (Fb - Fg - m*g) / (ρ*V + m)

2.95 = (0.683 - ρ*0.5886 - 0.0451*9.81) / (ρ*0.0600 + 0.0451)
2.95(ρ*0.06 + 0.0451) = 0.240 - ρ*0.5886
ρ*0.177 +0.112 = 0.290 - ρ*0.5886
ρ*0.7656 = 0.178
ρ = 0.232 kg/m³

Answer 2

The answer for above problem is explained below.