Answer
(a) $2.7N$
(b) $2.7N$
(c) $1.3m/s$
(d) The water will rise to the top of the container; the heights must be equal.
Work Step by Step
(a) We know that
$F_{base}=\rho gh_1A_{base}$
We plug in the known values to obtain:
$F_{base}=(1000Kg/m^3)(9.81m/s^2)(0.18m)(24\times 10^{-4}m^2)$
$F_{base}=4.24N$
Now the upward force on the ring of the containers is given as
$F_{ring}=P_{ring}A_{ring}$
$\implies F_{ring}=\rho gh_2A_{ring}$
We plug in the known values to obtain:
$F_{ring}=(1000Kg/m^3)(9.81m/s^2)(0.09m)(18\times 10^{-4}m^2)=1.59N$
Now $F=F_{base}-F_{ring}$
$\implies F=4.24N-1.59N=2.7N$
(b) We know that
$W=mg$
$\implies W=\rho Vg$
$\implies W=\rho (h_1A_{base}-h_2A_{ring})g$
We plug in the known values to obtain:
$W=2.7N$
(c) We know that
$v=\sqrt{2gh}$
We plug in the known values to obtain:
$v=\sqrt{2(9.81m/s^2)(0.09m)}$
$v=1.3m/s$
(d) We know that the water rises to the top of the container because the pressure at the top of the water and the top of the spray are equal; hence the heights should be equal.
