Final answer:
The buoyant force exerted on an empty storage tank by air at STP conditions is calculated using Archimedes' Principle, which implies that it is equal to the weight of air displaced by the volume of the tank. After converting the volume to cubic meters and using the density of air at STP, the buoyant force is found to be 1,336.26 N.
Step-by-step explanation:
The question posed involves calculating the buoyant force exerted on an empty storage tank by the air. According to Archimedes' Principle, the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
At standard temperature and pressure (STP), which is 0°C and 1 atm, the density of air is approximately 1.2 kg/m³. The volume of the tank is 4,010 ft³, which would need to be converted into cubic meters to use SI units (1 ft³ is approximately 0.0283 m³).
Firstly, the volume of the tank in cubic meters is calculated:
- Volume (m³) = Volume (ft³) × 0.0283
- Volume (m³) = 4,010 ft³ × 0.0283 m³/ft³
- Volume (m³) = 113.483 m³
Next, the weight of the air displaced is found using the density of air and gravity (g = 9.81 m/s²):
- Weight of displaced air (N) = Density of air (kg/m³) × Volume (m³) × g
- Weight of displaced air (N) = 1.2 kg/m³ × 113.483 m³ × 9.81 m/s²
- Weight of displaced air (N) = 1,336.26 N
Therefore, the buoyant force exerted on the tank by the air is 1,336.26 N at STP conditions. This force is equal to the weight of the air that would occupy the space of the tank if the tank were not present.