A certain diving bell for use in the ocean has a square window measuring 0.250 m on each side. The inside of the bell is pressurized and maintained at a pressure of 2.00 atm. The mass density of seawater is 1025 kg/m3.
How deep can this bell be lowered before the vector sum of forces exerted by the seawater on the window exceeds 175 kN ?
Average speed
To determine the depth to which the diving bell can be lowered, we need to calculate the pressure exerted by the seawater on the window and then determine the depth at which the force exceeds 175 kN.
Step 1: Calculate the pressure exerted by the seawater on the window.
The pressure exerted by a fluid is given by the formula:
P = ρgh
Where:
P = Pressure (in Pascal, Pa)
ρ = Density of the fluid (in kg/m^3)
g = Acceleration due to gravity (in m/s^2)
h = Depth of the fluid (in meters)
In this case, the pressure is equal to the pressure inside the bell, which is 2.00 atm. We will convert it to Pascal.
1 atm = 101325 Pa
2.00 atm = 2.00 * 101325 Pa = 202650 Pa
Density of seawater, ρ, is given as 1025 kg/m^3.
Acceleration due to gravity, g, is approximately 9.8 m/s^2.
So, we can rearrange the formula to find the depth, h:
h = P / (ρg)
Step 2: Calculate the depth at which the force exceeds 175 kN.
We have the pressure, P, and need to find the depth, h, where the force exceeds 175 kN.
Force = Pressure * Area
The area of the square window is given as 0.250 m on each side.
Area = Length * Width
Area = 0.250 m * 0.250 m
Area = 0.0625 m^2
Now we can calculate the depth:
Depth = Force / (Pressure * Area)
Depth = (175 kN / 1000) / (202650 Pa * 0.0625 m^2)
Depth = (175 * 10^3 N) / (202650 Pa * 0.0625 m^2)
Depth = Approximately 13.6 meters
Therefore, the diving bell can be lowered to a depth of approximately 13.6 meters before the vector sum of forces exerted by the seawater on the window exceeds 175 kN.