Please tell me how to solve this AP problem?

Question

While exploring a sunken ocean liner, the principal researcher found the absolute pressure on the robot observation submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025 kg./m^3
a. Calc. the gauge pressure p sub g on the sunken ocean liner.
b. Calc. the depth D of the sunken ocean liner
c. Calc. the magnitute F of the force due to the water on a viewing port of the submarine at this depth is the viewing port has a surface area of 0. 0100 m 6^2
SUppose the ocean liner came to rest at the surface of the ocean before it started to sink. Due to the resistance of the seawater the sinking ocean liner then reached a terminal velocity of 10. 0 m/s after falling for 30 seconds.
Deter. the magnitude a of the average accel. of the ocean liner during this period of time.
e. assume the accel. was constant, calc. the distance below the suface at which the ocean liner reached this terminal velocity
Calc. the time it took to from the surface-bottom.

Answer 1

I can't help with all of these because I don't know what "p sub g" means. If it is a standard pressure gauge reading, it is relative to 1 atmosphere, so the pressure gauge would reas 412 atmospheres.

b) The depth of the sunken submarine is computed from the density of seawater and the fact that 1 atmosphere is about 1.033 kg/cm^2. So if pressure is 412 atmospheres ABOVE surface pressure, then there is 412*1.033 kg of water in a 1 cm^2 column above the submarine. In a column 1 meter high, and 1 cm^2 in area, the weight of the water is .1025 kg (1025/10^4 to convert from cubic meters to a 1-meter column by 1 cm x 1 xm). Therefore, the height of the column must be 412*1.033/.1025.

c) the pressure of the WATER is again 412 atmospheres, or 412*1.033 kg/cm^2 (the extra atmosphere is due to air above the surface), and therefore the pressure is 412*1.033*(area in meters). I can't gell what you mean by .01 m 6^2, so you can fill in the area in meters that is correct to compute this.

d) In the first 30 seconds, the average acceleration must be 10 m/s / 30 s = 1/3 m/s^2
e) If the acceleration is constant, we use simple equations of motion to computed d = v0t+1/2at^2 = 0 + 1/2*1/3*30*30 = 150 meters
After this, use the depth of the sunken vessel, D. The time it took to get to the bottom is the sum of the time it took to get to 150 meters and the time it took to get from 150 meters to D, which is 30 seconds + (D-150)/ 10 m/s