Aweather balloon is operating at an altitude at which the density of air is 0.90 kg/m^3. at this altitude, the balloon has a volume of 20m^3 and is filled with helium (density of helium is 0.178 kg/m^3). if the balloon skin weighs 88 n, what load can be supported at this level?

54 N

Explanation:

Draw a free body diagram.  There are four forces acting on the balloon.  Buoyant force pushing the balloon up, gravity pulling the helium down, gravity pulling the balloon skin down, and gravity pulling the load down.

Apply Newton's second law:

∑F = ma

B − Wh − Wb − L = ma

When the load is at a maximum, the acceleration is 0:

B − Wh − Wb − L = 0

B − Wh − Wb = L

B − mh g − Wb = L

The mass of the helium is its density times its volume:

B − ρh Vh g − Wb = L

Buoyant force is defined as B = ρVg, where ρ is the density of the displaced fluid (in this case, air), V is the volume of the displaced fluid, and g is acceleration of gravity.  Since the volume of displaced air = the volume of the helium:

ρa V g − ρh V g − Wb = L

(ρa − ρh) V g − Wb = L

Given that ρa = 0.90 kg/m³, ρh = 0.178 kg/m³, V = 20 m³, g = 9.8 m/s², and Wb = 88 N:

(0.9 − 0.178) (20) (9.8) − (88) = L

L = 53.5 N

Rounded to 2 sig-figs, the maximum load that can be supported is 54 N.

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explanation:

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