Absolute pressure at an ocean depth

[jenha14]

a) Calculate the pressure at an ocean depth of 550 m. Assume that the density of sea water is 1022 kg/m3 and that the air above exerts a pressure of 101.3 kPa.
b) At this depth, what force must the frame around a circular submarine porthole having a diameter of 32.0 cm exert to counterbalance the force exerted by the water?

I found the answer for (a) to be 5609880 Pa, but I am stuck on part (b).

 

[mgb_phys]

Assuming the inside of the sub is at atmospheric pressure all you need is the water pressure at that depth.
Since pressure is force / area. You have the pressure,you can get the area of the porthole so hat is the force?

 

[ogb p]

For part (b), we can use the equation for pressure (P = F/A) to find the force exerted by the water on the porthole. The area (A) of the porthole can be calculated using the formula for the area of a circle (A = πr^2), where r is the radius of the porthole (16 cm). This gives us an area of 201.06 cm^2 or 0.020106 m^2.

Plugging in the values for pressure (5609880 Pa) and area (0.020106 m^2) into the equation, we get:

5609880 Pa = F/0.020106 m^2

Solving for F gives us a force of 112.61 kN. This is the amount of force that the frame around the porthole must exert to counterbalance the force of the water at a depth of 550 m.

As a scientist, it is important to note that this calculation is based on the assumptions made in part (a), such as the density of sea water and the pressure of the air above. These values may vary in different ocean environments and can affect the accuracy of the calculation. Additionally, the force calculated is only for the water pressure and does not take into account any other external forces that may be acting on the porthole. It is important to consider all factors and potential sources of error when making calculations and conducting experiments.