o 5/2/2016 1 1 :00 PM O 58.3/1005/1/2016 10:30 PM Gradebook Print Calculator Per
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o 5/2/2016 1 1 :00 PM O 58.3/1005/1/2016 10:30 PM Gradebook Print Calculator Periadic Table Question 18 of 18 Map sapling learning 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 66400 cm gas-filled balloon will provide enough lift for a 45.1 g package to be accelerated upward at a rate of 2.65 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 pair 1.16 kg/m3. Neglect the mass of the balloon material and the volume of the package Number kg/m3 Previous 8 Give Up & View Solution D Check Answer Next Exit | Hint Begin by finding the mass of the gas which the balloon is filled with. This can be accomplished by summing the forces acting in the balloon in the vertical direction. Keep in mind that the buoyant force, Fe, is equal to Remember tha the density in this equation is the density of the gas which is acting up on the balloon not the gas the balloon is filledExplanation / Answer
Given data
rho = 1.16kg/m^3
V= 66400cm^3
a = 2.65 m/s^2
F_b = pgV
F_b=ma_y+mg
pgV=ma_y+mg
1.16*9.81*(66400*10^-6)=m(2.65)+m(9.81)
m=0.06064249 (this is the mass of both the air in the balloon and the package)
0.06064249-0.0450=0.015642kg (mass of air in the balloon)
density=mass/volume
0.015642kg/0.0664m^3=0.2355kg/m^3
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