MHB Finding diameter of a sphere, have density and mass

[polskon]

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?

 

[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.

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/

 

[soroban]

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