Answer:
The force is [tex]F = 18.33 \ N[/tex]
Explanation:
From the question we are told that
The number of liquids is n = 3
The volume of the first liquid is [tex]V_1 = 0.50 L = 0.0005 \ m^3[/tex]
The density of the first liquid is [tex]\rho_1 = 2.6 \ g/cm^3[/tex]
The volume of the second liquid is [tex]V_2 = 0.25 L = 250\ cm^3[/tex]
The density of the second liquid is [tex]\rho_2 = 1.0 \ g/cm^3[/tex]
The volume of the third liquid is [tex]V_3 = 0.40 L = 400\ cm^3[/tex]
The density of the third liquid is [tex]\rho_3 = 0.80 \ g/cm^3[/tex]
Generally the force at the bottom of the container is mathematically represented as
[tex]F = m_t * g[/tex]
Here [tex]g = 980.665 \ cm/s^2[/tex]
Here [tex]m_t[/tex] is the total mass of all the liquid which is mathematically represented as
[tex]m_t = ( V_1 * \rho_1 )+ ( V_2 * \rho_2)+ ( V_3 * \rho_3)[/tex]
=> [tex]m_t = ( 500 * 2.6)+ ( 250 * 1.0 )+ ( 400 * 0.80 )[/tex]
=> [tex]m_t = 1870 \ g[/tex]
So
[tex]F = 1870 * 980.66[/tex]
=> [tex]F = 1833843.55 \ g \cdot cm /s^2[/tex]
=> [tex]F = 1833843.55 \ g \cdot cm /s^2 = \frac{1833843.55}{1000 * 100} kg \cdot m /s^2[/tex]
=> [tex]F = 18.33 \ N[/tex]
