With what force does the steel cube push gasoline out if the size of the cube is 2 * 3 * 4 cm?

The force that pushes the steel cube out of the gasoline is the force of Archimedes.

Fa = ρ * g * V, where ρv is the density of gasoline (ρv = 710 kg / m³), g is the acceleration of gravity (we assume g = 10 m / s2), V is the volume of the cube (V = a * b * c, where a is the length of the cube (a = 2 cm = 0.02 m), b is the width of the cube (b = 3 cm = 0.03 m), c is the height of the cube (c = 4 cm = 0.04 m)).

Let’s calculate the buoyancy force:

Fa = ρ * g * a * b * c = 710 * 10 * 0.02 * 0.03 * 0.04 = 0.1704 N.

Answer: The buoyancy force is 0.1704 N.