Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.50 L, 2.6 g/cm3; 0.25 L, 1.0 g/cm3; and 0.40 L, 0.70 g/cm3. What is the force on the bottom of the container due to these liquids

13.524 N

Explanation:

Volume and densities are given as:

ρ1 = 2.6 g/cm³ = > 2600 kg/m³; V1 = 0.50 L = > 0.5 x 10^-3 m³

ρ2 = 1.0 g/cm³ = > 1000 kg/m³; V2 = 0.25 L = > 0.25 x 10^-3 m³

ρ3 = 0.7 g/cm³ = > 700 kg/m³; V3 = 0.4 L = > 0.4 x 10^-3 m³

Next is to calculate force exerted on the bottom of the container due to these liquids:

F = ρ1V1g + ρ2 V2 g + ρ 3 V3g

where,

ρ = density

V = volume

g = 9.8m/s²

F = g (2600 x 0.5 x 10^-3 + 1000 x 0.25 x 10^-3 + 700 x 0.4 x 10^-3)

F = 9.8 (1.38)

F = 13.524 N

Therefore, the force on the bottom of the container due to these liquids is 13.524 N