Answer:
2.65 MPa
Explanation:
To find the normal stress (σ) in the wall of the basketball we need to use the following equation:
[tex] \sigma = \frac{p*r}{2t} [/tex]
Where:
p: is the gage pressure = 108 kPa
r: is the inner radius of the ball
t: is the thickness = 3 mm
Hence, we need to find r, as follows:
[tex] r_{inner} = r_{outer} - t [/tex]
[tex] r_{inner} = \frac{d}{2} - t [/tex]
Where:
d: is the outer diameter = 300 mm
[tex] r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm [/tex]
Now, we can find the normal stress (σ) in the wall of the basketball:
[tex]\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa[/tex]
Therefore, the normal stress is 2.65 MPa.
I hope it helps you!
