Question

(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of the water is 1.00 103 kg/m3 and the air above is at a pressure of 101.3 kPa. Pa
(b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 34.1 cm

Answer 1

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

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

Where $P$ is the absolute pressure

$P_{o}$ is the atmospheric pressure

$\rho$ is the density

$g$ is the acceleration due to gravity (Take $g = 9.8 m/s^{2}$ )

h is the height

From the question

h = 30.0 m

$\rho$ = 1.00 × 10³ kg/m³ = 1000 kg/m³

$P_{o}$ = 101.3 kPa = 101300 Pa

Using the formula

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

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N