6) A sub has a volume of 20 m3 and a density of 1500 kg/m3. It sinks to a depth of 40 meters. The density of sea water is 1025 kg/m3.

a. What is the pressure on the sub at this depth?
b. The sub has a window with a diameter of 30 cm. What is the net force on the window of the sub?
c. What is the buoyancy force on the sub?
d. State the magnitude and the direction of the force necessary to keep the sub at this depth.

a. To find the pressure on the sub at a depth of 40 meters, we can use the formula for pressure in a fluid:

Pressure = Density * Gravity * Depth

The density of sea water is given as 1025 kg/m3, and gravity is approximately 9.8 m/s2. Plugging in the values:

Pressure = 1025 kg/m3 * 9.8 m/s2 * 40 m = 401,200 Pascal

Therefore, the pressure on the sub at a depth of 40 meters is 401,200 Pascal.

b. To find the net force on the window of the sub, we need to calculate the difference in pressure inside and outside the sub.

Since the sub is at a depth of 40 meters, the pressure inside the sub will be the same as the pressure at that depth, which we calculated in part a: 401,200 Pascal.

The pressure outside the sub (sea water) is given by:

Pressure = Density * Gravity * Depth = 1025 kg/m3 * 9.8 m/s2 * 40 m = 401,200 Pascal

Since the pressures are the same, there is no difference in pressure and therefore no net force on the window of the sub.

c. The buoyancy force on an object submerged in a fluid is given by:

Buoyancy Force = Fluid Density * Gravity * Volume of the Displaced Fluid

In this case, the fluid density is the density of sea water (1025 kg/m3), gravity is 9.8 m/s2, and the volume of the displaced fluid is the volume of the sub (20 m3).

Buoyancy Force = 1025 kg/m3 * 9.8 m/s2 * 20 m3 = 200,200 Newtons

Therefore, the buoyancy force on the sub is 200,200 Newtons.

d. The magnitude and direction of the force necessary to keep the sub at this depth is equal to the net force acting on the sub.

Since there is no net force on the sub, there is no additional force needed to keep the sub at this depth.