Question

My Notes O Ask You (a) Calculate the absolute pressure at 101.3 kPa the bottom of a fresh water lake at a depth of 27.0 m. Assume the density of the water is 1.00 x 10 kg/m2 and the air above is at a pressure of a depth Assume the density of the water is 1.00 x 10 Pa by the water on the window of an Need Help?

Answer 1

Given

Depth of lake (h)= 27 m.

Density of water $(\rho ) =1\times 10^{3} kg/m^{3}$

Atmospheric pressure $(P_{a})=101.3 KP_{a}$

(a) The absolute pressure at the bottom of a fresh water lake(P).

$P=P_{a}+\rho gh$

$=101.3KP_{a}+1\times 10^{3}kg/m^{3}\times 9.81m/s^{2}\times 27m$

$P=3.6617\times 10^{5}P_{a}$

(b) Given

Diameter(D)= 33 cm

Radius(R) is

$R=D/2=\frac{33}{2}cm=16.5 cm=0.165m$

Force exerted by the water,

$F=PA=P(\pi r^{2})$

$F=3.6617\times 10^{5}P_{a}\times 3.14\times(0.165m)^{2}$

$F=3.13\times 10^{4}N$