An astronaut works where the outside pressure is essentially zero, and the gauge on her air tank reads 6.90 x 10^6 Pa. The flat end of the tank is a disk 0.150 m in diameter. Taking pi as 3.14, what force does the air exert on that end?
A1.22 x 10^3 N
B2.44 x 10^5 N
C4.88 x 10^5 N
D1.22 x 10^5 N
Answer & Solution
Correct answer: D. 1.22 x 10^5 N
1. Rearranging the definition of pressure gives force = pressure x area.
2. The end is a disk, so its area is pi times the radius squared.
3. The radius is half the diameter: 0.150 m / 2 = 0.0750 m.
4. Area = (3.14)(0.0750 m)^2 = 1.77 x 10^-2 m^2.
5. Force = (6.90 x 10^6 N/m^2)(1.77 x 10^-2 m^2) = 1.22 x 10^5 N.
6. The value 4.88 x 10^5 N is the trap for using the diameter in place of the radius, which multiplies the area by four.
7. The value 2.44 x 10^5 N doubles the correct answer and 1.22 x 10^3 N drops two powers of ten.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.3 Pressure_
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