Answer:
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
