Determine the maximum volume in gallons [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of feet [ft] so the total pressure at the bottom of the container will not exceed pound-force per square inch [psi]. Assume the height of the tank is sufficient to store the amount of olive oil required, and the surface pressure is 1 atmosphere [atm]. The specific gravity of olive oil is 0.86.

This question is incomplete, the complete question is;

Determine the maximum volume in gallons​ [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3feet​ [ft] so the total pressure at the bottom of the container will not exceed 26 ​pound-force per square inch​ [psi]. Assume the height of the tank is sufficient to store the amount of olive oil​ required, and the surface pressure is 1 atmosphere​ [atm]. The specific gravity of olive oil is 0.86.

A maximum of 1602.6 gallons of olive oil can be stored in the closed cylindrical silo

Explanation:

Given the data in the question;

Using SI units

Diameter = 3 ft = 3 × 0.3048 = 0.9144 m

pressure at bottom Pb = 26 psi = 26 × 6.89476 × 1000 = 179263.76 pascal

surface pressure Ps = 1 atm = 1 × 101325 = 101325 pascal

specific gravity = 0.86

we know that; density of water = 1000 kg/m³

so, density of oil P = density of water × specific gravity

= 1000 × 0.86 = 860 kg/m³

pressure at depth can be related to pressure at surface;

Pb = Ps + Pgh

acc. due gravity g = 9.81 m/s²

we substitute

179263.76 = 101325 + ( 860 × 9.81 × h)

8436.6h = 179263.76 - 101325

8436.6h = 77938.76

h = 77938.76 / 8436.6

h = 9.238 m

Now;

Volume of the cylinder = π/4 × D² × h

we substitute

Volume of the cylinder = π/4 × ( 0.9144 )² × 9.238  

Volume of the cylinder = 6.0665 m³

we know that; 1 m³ = 264.172 gal

so; Volume of the cylinder = 6.0665 × 264.172 gal

Volume of the cylinder = 1602.6 gallons

Therefore, A maximum of 1602.6 gallons of olive oil can be stored in the closed cylindrical silo

your answer is, orbit around the nucleus

answer; how far out of place which the object finished relative to the starting location;