Hi! I really need help on Part B only.
(a) Calculate the absolute pressure at the bottom of a freshwater lake at a depth of 28.7 m. Assume the density of the water is 1.00 103 kg/m3 and the air above is at a pressure of 101.3 kPa.
(b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 40.9 cm?
So, I try to convert 40.9cm to 0.409*9.8 but, it not right.
40.9cm = 0.409m
diameter/2 = radius
0.409m/2 = 0.2045
area of circle = (pi)r^2
F=PA
F=(what you got for pressure from Part A)((pi)r^2)
To calculate the force exerted by the water on the window of the underwater vehicle, we first need to determine the pressure at the given depth.
(a) Determining the absolute pressure:
At a depth of 28.7 m, the absolute pressure can be calculated using the formula:
P = P0 + ρ * g * h
Where:
P is the absolute pressure at the bottom of the lake
P0 is the atmospheric pressure (given as 101.3 kPa)
ρ is the density of water (given as 1.00 * 10^3 kg/m^3)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the depth (given as 28.7 m)
Substituting the given values into the formula:
P = 101.3 kPa + (1.00 * 10^3 kg/m^3) * (9.8 m/s^2) * (28.7 m)
Calculating the absolute pressure will give you the answer to part (a).
(b) Determining the force exerted on the window:
To calculate the force exerted by the water on the window, we can use the formula:
F = P * A
Where:
F is the force exerted on the window
P is the pressure at the bottom of the lake (calculated in part (a))
A is the area of the circular window
To calculate the area of the circular window, we can use the formula for the area of a circle:
A = π * r^2
Where:
A is the area of the circular window
π is a mathematical constant approximately equal to 3.14159
r is the radius of the circular window
Given that the diameter of the window is 40.9 cm, the radius can be calculated as:
r = (40.9 cm) / 2 = 20.45 cm = 0.2045 m
Substituting the values into the formula, we have:
A = π * (0.2045 m)^2
Once you have the value of A, you can substitute the pressure (P) calculated in part (a) along with the calculated area (A) into the second formula to find the force exerted on the window.
For part B, you want to calculate the force exerted by the water on the window of the underwater vehicle at a depth of 28.7 m.
To do this, let's start by calculating the pressure exerted by the water at that depth. Using the formula for pressure in a fluid, we have:
P = P0 + ρgh
Where:
P is the absolute pressure at the bottom of the lake (what we want to find)
P0 is the atmospheric pressure at the surface (given as 101.3 kPa)
ρ is the density of the water (given as 1.00 × 10^3 kg/m^3)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the depth of the lake (given as 28.7 m)
Substituting the known values into the equation, we have:
P = 101.3 kPa + (1.00 × 10^3 kg/m^3)(9.8 m/s^2)(28.7 m)
Now, let's convert the units so that they are consistent:
P = 101.3 × 10^3 Pa + (1.00 × 10^3 kg/m^3)(9.8 m/s^2)(28.7 m)
P = 101.3 × 10^3 Pa + 282,076 Pa
P = 383,376 Pa
Now that we have the pressure, we can calculate the force using the formula:
F = P × A
Where:
F is the force exerted by the water on the window (what we want to find)
P is the pressure (383,376 Pa)
A is the area of the circular window
To find the area of the window, we use the formula for the area of a circle:
A = πr^2
Where:
A is the area of the window (what we want to find)
π is a constant approximately equal to 3.14159
r is the radius of the window, which is half the diameter (0.409/2 m)
Substituting the known values into the equation, we have:
A = 3.14159 × (0.409/2)^2 m^2
Simplifying further:
A = 3.14159 × (0.2045)^2 m^2
A = 0.1314646245 m^2
Now, let's substitute the values into the formula for force:
F = (383,376 Pa) × (0.1314646245 m^2)
Calculating the force gives us the answer to part B, which is the force exerted by the water on the window of the underwater vehicle at a depth of 28.7 m.
Why would you multiply the diameter by g?
Force equals pressure times area.
You are still mispelling the subject.
A "physic" is a laxative.