Question

1. Three identical cubes of side 2 cm and density 4gcm are placed one over other in a container filled with three immiscible liquids A, B and C as shown in the figure. If the densities of A, B and C are 3gcm,2gcm,1gcm respectively
determine the net force exerted by the cubes at the bottom of the container.are placed one over the other in a container.
please help
me please I promise that I mark you as a brain list​

Answer 1

Explanation:

hope you like it..and give thaks if you like

Answer 2

Answer:

given = cubes of side 2cm

             density = 4g/cm

              A= 3gmc

              B= 2gmc

              C = 1gcm

to find =  net force

solution =

let consider the masses of cubes  are m= mA + mB + mC

apply the formula of density

ρ = m /v

total volume of cube is v = 3 x a³  = 3 x a³ = 3x2³= 24cm³

now, 4 = m/ 24

total weight

w= mg

w = 24 x 0.096Kg

w = mg = 0.096 x 9.8= 0.940N

total upthrust force F = vg x ρA +ρB+ ρC

F= 2³ x 9.8 x(3 + 2 + 1)x10⁻³

F = 0.4704N

for finding net force ,

F= total weight - total upthrust

F = 0.940- 0.4704

F= 0.4696N

so,the net force is 0.4696N.