A lighter-than-air spherical balloon and its load of passengers and ballast are

Question

A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 4.90 m. Assuming a constant value of 1.29 kg/m3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 106 m in 22.0 s.

An antifreeze solution is made by mixing ethylene glycol ( = 1116 kg/m3) with water. Suppose the specific gravity of such a solution is 1.04. Assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

Explanation / Answer

let (mA) = mass of displaced air = D V

where D is density

V is Volume

so mass mA = 4/3 *pi*(4.9)^3 * 1.29 = 635.72 N

let (mC) = mass of contents after drop

g(mA - mC) = net force = mC*a

g(mA) - mC(g + a) = 0

mC = (mA)/(1 + a/g)

mC*g = weight in Newtons

from our kinematic equations

d = 1/2 at^2 , d = 136, t =18,

a = 2* 106/(22*22)

a = 0.438 m/s^2

so mC g = W = (635.72/((1+ 0.438/9.8)

W = 608.52 N
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