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 find the absolute pressure at a depth of 1000m in the ocean, we need to consider both the atmospheric pressure and the pressure due to the water above.
The pressure due to the water can be calculated using the formula:
P = P0 + ρgh
Where:
P = pressure at depth h
P0 = atmospheric pressure = 101.3 kPa = 101300 Pa
ρ = density of seawater = 1024 kg/m^3
g = acceleration due to gravity = 9.81 m/s^2
h = depth = 1000m
Plugging these values into the formula, we get:
P = 101300 + (1024)(9.81)(1000)
P = 101300 + 10038240
P = 10138340 Pa
Therefore, the absolute pressure at a depth of 1000m in the ocean is 10138340 Pa.
(b) To find the force that the frame around a circular submarine porthole must exert to counterbalance the force exerted by the water, we can use the formula for pressure:
P = F/A
Where:
P = pressure = 10138340 Pa
A = area of the porthole = πr^2 = π(0.15)^2 = 0.0707 m^2
Solving for F, we get:
F = P * A
F = 10138340 * 0.0707
F = 717380.9 N
Therefore, the force that the frame around a circular submarine porthole must exert to counterbalance the force exerted by the water is approximately 717381 N.