Question

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, 0.80 g/cm³. What is the force on the bottom of the container due to these liquids? One liter = 1 L = 1000 cm³. (Ignore the contribution due to the atmosphere.)

Answer 1

Answer:

The force is  $F = 18.33 \ N$

Explanation:

From the question we are told that

    The number of liquids is  n =  3

      The volume of the first liquid is $V_1 = 0.50 L = 0.0005 \ m^3$

      The density of the first liquid is  $\rho_1 = 2.6 \ g/cm^3$

      The volume of the second  liquid is $V_2 = 0.25 L = 250\ cm^3$

      The density of the second liquid is  $\rho_2 = 1.0 \ g/cm^3$

      The volume of the third  liquid is $V_3 = 0.40 L = 400\ cm^3$

      The density of the  third  liquid is  $\rho_3 = 0.80 \ g/cm^3$

Generally the force at the bottom of the container is mathematically represented  as

    $F = m_t * g$

Here $g = 980.665 \ cm/s^2$

Here  $m_t$  is the total mass of all the liquid which is mathematically represented as

             $m_t = ( V_1 * \rho_1 )+ ( V_2 * \rho_2)+ ( V_3 * \rho_3)$

=>         $m_t = ( 500 * 2.6)+ ( 250 * 1.0 )+ ( 400 * 0.80 )$

=>         $m_t = 1870 \ g$

So

       $F = 1870 * 980.66$

=>   $F = 1833843.55 \ g \cdot cm /s^2$

=>    $F = 1833843.55 \ g \cdot cm /s^2 = \frac{1833843.55}{1000 * 100} kg \cdot m /s^2$

=>    $F = 18.33 \ N$