Question

A boat has a porthole window, of

area 0.00849 m^2, 6.25 m below
the surface. The density of sea
water is 1027 kg/m^3. The air
inside the boat is at 1 atm. What
is the NET force on the window?
(Unit = N)
(1 atm = 101300 Pa)

Answer 1

534.6

Step-by-step explanation:

Answer 2

F = 534.6[N]

Step-by-step explanation:

We must find the pressure exerted by the water at the depth of the boat, by means of the following equation.

`P=Ro*g*h`

where:

Ro = density of sea water = 1027 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = wáter Depth = 6.25 [m]

Now replacing:

`P=1027*9.81*6.25\\P=62967.93[Pa]`

The net force is:

`F = P*A\\F = 62967.93*0.00849\\F = 534.6[N]`