Question

2

Fig. 21 shows a diver 50m below the surface of the water.
(a) The density of water is 1000 kg/m and the acceleration of free fall is 10 m/s
Calculate the pressure that the water exerts on the diver.

(b) The window in the diver's helmet is 150 mm wide and 70mm from top to bottom.
Calculate the force that the water exerts on this window.

Answer 1

500000N/m²

5250N

Step-by-step explanation:

Given parameters:

Depth(H) = 50m

Density of water = 1000kg/m³

Acceleration of free fall = 10m/s

Unknown:

Pressure the water exerts on the diver = ?

Solution:

Pressure is the force per unit area on a body. In fluids, pressure is the product of density, gravity and height

Pressure in fluids = Density x acceleration due to gravity x height

Input the variables and solve;

Pressure in fluids = 1000 x 10 x 50 = 500000N/m²

B.

width of window = 150mm

height of window = 70mm

Force water exerts on the window = ?

To solve this problem;

Pressure =
`(Force)/(Area)`

Area of the window = width x height = 150 x 10⁻³ x 70 x 10⁻³

= 1.05 x 10 ⁻²m²

Force = pressure x area

Input the variables;

= 500000N/m² x 1.05 x 10 ⁻²m²

= 5250N