How Does Water Pressure Affect a Submarine at Various Depths?

[biomajor009]

Homework Statement

Your homemade submarine has a circular hole with a diameter of 14 cm in its side. You decide to plug the hole by taping over it with watertight duct tape. This works pretty well while you are exploring the ocean at shallow depths, but you find that as soon as you reach a depth of 9.7 meters the duct tape is dislodged.
(a) What is the pressure of the ocean water at this depth?
(b) What net force was needed to dislodge the duct tape? You may assume that the interior of the sub was at atmospheric pressure.
(c) When the duct tape is dislodged, ocean water begins to shoot into the sub at a speed of 7.1 m/s. You grab something to put over the hole, but all you have handy is a funnel. The large end of the funnel is big enough to cover the hole, while the small end has an opening with a diameter of 2.2 mm. If you place the funnel over the hole, how fast will water shoot out of the small end of the funnel? Would you want to plug up the small end by putting your hand over it?

Homework Equations

The Attempt at a Solution

I'm having trouble getting past the first part of the question. I tried using the equation
P2 = P1 + pgh, but I don't know what P1 should be. Am I even using the right equation?

 

[mjsd]

P1 is the pressure at the surface

 

[ogb p]

I would like to clarify that the homework statement provided does not specify the units for the diameter of the hole. For the purpose of this response, I will assume that the units are in centimeters (cm).

(a) To determine the pressure at a depth of 9.7 meters, we can use the equation P = ρgh, where P is the pressure, ρ is the density of the fluid (in this case, seawater), g is the acceleration due to gravity (9.8 m/s^2), and h is the depth. First, we need to convert the depth from meters to centimeters: 9.7 m = 970 cm. We also need to know the density of seawater, which is approximately 1025 kg/m^3. Converting this to g/cm^3, we get 1.025 g/cm^3. Plugging in these values, we get P = (1.025 g/cm^3)(9.8 m/s^2)(970 cm) = 9.89 x 10^4 g/cm^2. This is equivalent to 9.89 x 10^8 Pa (Pascal).

(b) To calculate the net force needed to dislodge the duct tape, we can use the equation F = PA, where F is the net force, P is the pressure, and A is the area of the hole. The area of the hole is given by A = πr^2, where r is the radius of the hole. Since the diameter of the hole is 14 cm, the radius is 7 cm. Plugging in the values, we get F = (9.89 x 10^8 Pa)(π)(7 cm)^2 = 1.5 x 10^10 g cm/s^2. This is equivalent to 1.5 x 10^5 Newtons (N). Therefore, a net force of 1.5 x 10^5 N would be needed to dislodge the duct tape.

(c) To calculate the speed at which water would shoot out of the small end of the funnel, we can use the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height of the water column. In this case, h is the same as the depth, which is 9.7 meters