A consumer upset with the latest trend of postal rate increases has decided to t

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 destination. A 6.00 × 104 cm3 gas-filled balloon will provide enough lift for a 43.4 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 g = 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.

Explanation / Answer

V =6x10^4 cm^3 = 0.06 m^3

a =2.95 m/s^2

density = mass/volume

mass of displaced air M = air*V = 1.16kg/m³ * 0.0600m³ = 0.0696 kg

buoyant force Fb = weight of displaced air

Fb = air*V*g = 1.16kg/m³ * 0.0600m³ * 9.81m/s² = 0.683 N

mass of gas m = *V = *0.0600m³

weight of gas Fg = mg = *V*g = *0.0600m³*9.81m/s² = *0.5886m/s²

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

Dropping units for ease ( is in kg/m³):

2.95 = (0.683 - *0.5886 - 0.0434*9.81) / (*0.0600 + 0.0434)
2.95(*0.06 + 0.0434) = 0.25725 - *0.5886
*0.177 + 0.12803 = 0.25725 - *0.5886
*0.7656 = 0.12922
= 0.169 kg/m³

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