1)A car misses a turn and sinks into a shallow lake to a depth of 11.4 m. If the area of the car door is 0.56 m2, what is the force exerted on the outside of the door by the water? Note: 1 atm = 101 kPa.

Answer 1

Assume the whole door is at 11.4 m

pressure = rho g h + 1 atm
rho = 10^3 kg/m^3

pressure = 10^3 (9.8)*11.4 + 1 atm

= 112*10^3 + 101*10^3

213*10^3 N/m^2

213*10^3 * .56 = 119*10^3 N

NOTE: You still actually have at least one atm on the inside of the door that you brought down from the surface. That will increase as water leaks into the car and eventually the pressure will equalize inside and out as the car fills with water. You may not wish to wait that long to open the door easily, but right from the start the problem is not quite as bad as implied here.

Answer 2

PS
If you wait until the pressure equalizes inside and out, you will have 1.13*10^5 Pa of pressure above atmospheric in your lungs.
If you then exit the car and swim up to the surface without expelling air from your lungs you will die as your lungs try to expand from the excess of air in them. Divers and submariners know about this and practice expelling air during free ascents using air under pressure (scuba or hard hat). It is called "air embolism".