Aspherical balloon has a radius of 6.35 m and is filled with helium. the density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. the skin and structure of the balloon has a mass of 990 kg . neglect the buoyant force on the cargo volume itself. required determine the largest mass of cargo the balloon can lift.

201 kg.

Explanation:

There are two forces acting (both vertically) on the balloon and the cargo: gravity (downward) and the buoyant force (upward).

If we need that the balloon remain in the air, the buoyant force must be equal to the weight of the balloon and the cargo, as follows:

Fg = Fb

In order to get Fg, we must add the mass of the helium (expressed as a product of the density times the volumen of the balloon, assumed spherical), the mass of the skin and the structure of the balloon, and the mass of the cargo itself, as follows:

Fg = (δHe . 4/3 π r3 + mcargo + mstruct) g (1)

The buoyant force, is equal to the weight of the volumen of the air displaced by the balloon (which is equal to the volumen of the entire balloon as it is completely submerged in air) , which can be written as follows:

Fb = δ air . 4/3 π r3 g (2)

Simplyfing, replacing by the values of δHe, δair, r, and mstruct, and solving for mcargo, we finally get:

mcargo = (4/3 π (6.35m)3 (1.29 kg/m3 – 0.179 kg/m3)) – 990 kg

mcargo =  201 kg.

c) up

hope this

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form