Assuming the average composition of air weighs approximately 0.0807 lbs per cubic foot, what is the weight of air in a giant spherical balloon with a diameter of 6 feet?

[tex]V_{sphere} =4 \pi \frac{r^3}{3} = 4 \pi \frac{(\frac{6ft}{2})^3}{3}[/tex]

Volume = 113.097 ft^3

Density=Mass/Volume => Mass=Density*Volume = (113.097 ft^3)(.0807 lbs/ft^3) = 9.12 lbs.

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JeanaShupp JeanaShupp · Answer 2 · 2019-03-28T18:01:07+01:00

Answer: 9.122328 lb

Step-by-step explanation:

Given: The average composition of air weighs (density)= 0.0807 lbs per cubic foot

Diameter of spherical balloon = 6 feet

Then radius of spherical balloon = [tex]=\frac{6}{2}=3\ feet[/tex]

The volume of spherical balloon is given by :-

[tex]\text{Volume}=\frac{4}{3}\pi r^3\\\\\Rightarrow\text{Volume}=\frac{4}{3}(3.14)(3)^3\\\\\Rightarrow\text{Volume}=113.04\ feet^3[/tex]

We know that [tex]\text{Mass}=\text{Density}\times\text{Volume}[/tex]

Therefore, [tex]\text{Mass of air}=0.0807\times113.04=9.122328\ lb[/tex]