Question

A large, open at the top, upright cylindrical tank contains fresh water with a density of 1.00 ✕ 103 kg/m3. If the air pressure is 101.3 kPa, determine the absolute pressure (in Pa) in the fluid at a depth of 29.0 m. Pa (b) Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.90 cm. N

Answer 1

Answer:

(a) 385500 Pa

(b) 460.5 N

Explanation:

101.3 kPa = 101300 Pa

3.9 cm = 0.039 m

(a)The pressure at the depth of d = 29 m is

$P = P_0 + \rho gh$

where $P_0 = 101300 Pa$ is the air pressure at surface,  ρ = 1000 kg/m3 is the water density, g = 9.8 m/s2 is the gravitational acceleration, h = 29 m is the height difference.

$P = 101300 + 1000*9.8*29 = 385500 Pa$

(b) The area under pressure of the instrument probe is:

$A = \pi (d/2)^2 = \pi * (0.039 / 2)^2 = 0.0012 m^2$

So the force exerted on the pressurized area would be

$F = AP = 0.0012 * 385500 = 460.5 N$