What is the total mass of the Earth’s atmosphere? (The radius of the Earth is
6.37x106m, and atmospheric pressure at the surface is 1.013x105N/m2.)
(a) Calculate the absolute pressure at an ocean depth of 1000m. Assume
the density of seawater is 1024kg/m3 and that the air above exerts a
pressure of 101.3kPa. (b) At this depth, what force must the frame
around a circular submarine porthole having a diameter of 30.0cm
exert to counterbalance the force exerted by the water?
(a) To calculate the absolute pressure at an ocean depth of 1000m, we first need to determine the pressure due to the water and the pressure due to the air above the surface.
The pressure due to the water can be calculated using the equation P = ρgh, where P is the pressure, ρ is the density of seawater, g is the acceleration due to gravity, and h is the depth. Plugging in the values, we get:
P_water = (1024 kg/m^3)(9.81 m/s^2)(1000 m) = 10,050,240 Pa
The total absolute pressure is the sum of the pressure due to the water and the pressure due to the air:
P_total = P_water + P_air
P_total = 10,050,240 Pa + 101,300 Pa
P_total = 10,151,540 Pa
So, the absolute pressure at an ocean depth of 1000m is 10,151,540 Pa.
(b) The force exerted by the water on the submarine porthole can be calculated using the formula F = PA, where F is the force, P is the pressure, and A is the area of the porthole.
First, we need to convert the diameter of the porthole to meters:
D = 30.0 cm = 0.30 m
The area of the porthole is:
A = (π/4)(0.30)^2 = 0.0707 m^2
Now, we can calculate the force exerted by the water on the porthole:
F = P_total * A
F = 10,151,540 Pa * 0.0707 m^2
F = 717,944.987 N
Therefore, the frame around the circular submarine porthole must exert a force of approximately 717,945 N to counterbalance the force exerted by the water at a depth of 1000m.