MHB Finding diameter of a sphere, have density and mass

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To find the diameter of a balloon made from a material with a density of 0.310 kg/m² and a mass of 2756 kg, the density refers to the surface area rather than volume. The correct approach involves calculating the surface area using the formula A = Mass/Density, yielding an area of approximately 8890.323 m². From this area, the radius is derived using the formula r = √(A/(4π)), resulting in a radius of about 26.598 m. Consequently, the diameter of the balloon is approximately 53.20 m, highlighting the importance of correctly interpreting density in relation to surface area.
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A balloon is made from material that has a density of 0.310 kg/m2. If the balloon has a mass of 2756 kg and if it is assumed that the balloon is a perfect sphere, what is the diameter of the balloon? Keep the proper number of significant digits.

Mass = Density x Volume

2756kg = 0.310kg/m^2 x Volume

2756kg / 0.310 kg/2 = Volume

Volume = 8890.322581 m^3

8890.322581 m^3 = 4/3(pi)r^3

3(sq)8890.322581 m^3 / (4/3(pi)) = r

r = 12.85117892 m
D = 25.70235784 m

Diameter to 3 significant digits = 25.7 m

Answer is wrong, what is the problem with my answer?
 
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polskon said:
A balloon is made from material that has a density of 0.310 kg/m2. If the balloon has a mass of 2756 kg and if it is assumed that the balloon is a perfect sphere, what is the diameter of the balloon? Keep the proper number of significant digits.

Mass = Density x Volume

2756kg = 0.310kg/m^2 x Volume

2756kg / 0.310 kg/2 = Volume

Volume = 8890.322581 m^3

8890.322581 m^3 = 4/3(pi)r^3

3(sq)8890.322581 m^3 / (4/3(pi)) = r

r = 12.85117892 m
D = 25.70235784 m

Diameter to 3 significant digits = 25.7 m

Answer is wrong, what is the problem with my answer?

1. I assume that the text of the question is correct. Then the balloon is hollow and not solid! The densitiy refers to the envelope of the balloon.

Let A denotes the surface area of the balloon. Then

$ A \cdot 0.310\ \tfrac{kg}{m^2} = 2756\ kg $

That means $ a = 8890.323\ m^2 $

2. The surface of a sphere is calculated by:

$ A = 4 \cdot \pi \cdot r^2~\implies~r=\sqrt{\frac{A}{4 \pi}} $

3. I've got $ r \approx 26.598 \ m $

https://www.physicsforums.com/threa...meter-of-a-sphere-using-a-screw-gauge.991087/
 
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Hello, polskon!

I agree with earboth . . .

A balloon is made from material that has a density of 0.310 kg/m2.
If the balloon has a mass of 2756 kg and if it is assumed that the balloon is a perfect sphere,
what is the diameter of the balloon?
Keep the proper number of significant digits.
Note that the density is given as 0.310 kilograms per square meter.
We are dealing with the surface area of the spherical balloon, not its volume.
. . (And the thickness of the balloon is considered negligible.)

The area of a sphere is: .A = 4πr2

Mass = Density x Area

. . 2756 .= .0.31 x A . . → . . A .= .2756/0.31

Then: .4πr2 .= .2756/0.31 . . → . . r2 .= .2756/1.24π .= .707.4693922

. . . . . r .= .26.59829670 Therefore: .Diameter .≈ .53.20 m
 
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