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

im not to sure what it is