How Does a Helium Balloon Achieve Buoyancy at 30 km Altitude?

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A helium balloon with a total mass of 20 kg is designed to lift a 70 kg payload, achieving buoyancy at 30 km altitude where air density is 0.035 kg/m^3. To determine the balloon's volume, the buoyancy force must equal the total weight, which includes both the balloon and the payload. The buoyancy force is calculated based on the weight of the air displaced by the balloon's volume. By subtracting the payload's volume of 0.010 m^3 from the total volume, the necessary calculations can be completed. Understanding these principles is essential for solving buoyancy-related problems in high-altitude scenarios.
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a helium balloon, with a combined mass of the balloon and helium of 20kg, is used to lift a payload of 70kg. At an altitude of 30 km the balloon stops and remains stationary.

a. Explain
b. at 30km, the density of air os 0.035kg/m^3. What is the volume of the balloon assuming that the volume of the payload is 0.010m^3

can someone please help, i need to kinda have a idea for my quiz tomorrow

for the vlume i came up with v=m/density

v = 90/.035

v = 2571?
 
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I think you should use the concept of Buoyancy Force to solve it..
Balance the buoyancy force and the total weight to get the volume of the balloon and payload, then substract the volume of payload..
 
At 30 km, the total force, weight and bouyancy, is 0. You know the weight so you can calculate the bouyancy force- you also know, or should know, that that is equal to the weight of air displaced by the baloon. You are given the volume of the baloon- what would be the weight of that volume of air?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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