How much oil can a cylindrical tank with a diameter of 22 m and a height of 6 m hold if the oil density is 0.83 g / cm

1. First, let’s translate 0.83 g / cm³ = 83 g / 100 cm³ into kg / m³.
Convert 83 g to kilograms. It is known that 1 kg contains 1000 g, then:
1 kg = 1000 g;
x = 83 g.
By proportion:
x = 83/1000 = 0.083 (kg).
Let’s convert 100 cm³ to cubic meters. It is known that 1 m contains 100 cm, then:
1 m = 100 cm;
x = 1 cm.
By proportion:
x = 1/100 = 0.01 (m).
Since 1 cm is equal to 0.01 m, then 1 cm³ is equal to: (0.01) ³ = 0.000001 (m³). Then 100 cm³ is equal to: 0.000001 * 100 = 0.0001 (m³).
Thus, 83 g / 100 cm³ = 0.083 kg / 0.0001 m³ = 830 kg / m³.
2. Find the volume of the tank. The volume of the cylinder is:
V = πHR ^ 2,
where H is the height of the cylinder, R is the radius of the circle at the base of the cylinder.
The volume of the tank is:
V = 3.14 * 6 * (22/2) ^ 2 = 18.84 * 11 ^ 2 = 18.84 * 121 = 2279.64 (m³).
3. The density of a substance is found by the formula:
ρ = m / V,
where m is the mass of the substance, V is the volume.
Substitute the known values ​​and find the mass of oil:
830 kg / m³ = m / 2279.64 m³;
m = 830 kg / m³ * 2279.64 m³ (in proportion);
m = 1892101.2 kg.
For convenience, we will translate 1892101.2 kg into tons:
1 t = 1000 kg;
x = 1892101.2 kg.
x = 1892101.2 kg * 1 t / 1000 kg = 1892.1012 t.
Answer: 1,892.1012 t