During a storm, the wind exerts a 150-N force on a window that measures 1.00 m by 0.50 m. The outside air pressure is 101 kPa. What pressure, in pascals, does the wind exert on the window?
a. 75 Pa
b. 150 Pa
c. 1.5 x 10^3 Pa
d. 3.0 x 10^2 Pa
e. 3.0 x 10^5 kPa
They already have told you the force, so divide it by the window area. The force they have quoted is the NET force (outside - inside), but they didn't tell you that. They should have.
Force/Area = 300 Pa
The normal (static) air pressure (which is much higher) isn't needed
150 Pa
To find the pressure that the wind exerts on the window, we can use the formula:
Pressure (P) = Force (F) / Area (A)
Given:
Force exerted by the wind on the window (F) = 150 N
Area of the window (A) = 1.00 m * 0.50 m = 0.50 m^2
Substituting the values into the formula:
P = 150 N / 0.50 m^2
P = 300 N/m^2
Since 1 Pascal (Pa) is equal to 1 N/m^2, the pressure in pascals is equal to 300 Pa.
Therefore, the answer is:
d. 3.0 x 10^2 Pa
To find the pressure that the wind exerts on the window, you need to divide the force exerted by the wind by the area of the window. Here's how you can calculate it:
1. First, calculate the area of the window by multiplying its length by its width:
Area = length × width = 1.00 m × 0.50 m = 0.50 m²
2. Next, convert the given force of 150 N to pascals (Pa). Remember that 1 pascal (Pa) is equal to 1 Newton per square meter (N/m²):
Force in pascals = Force (N) / Area (m²)
Force in pascals = 150 N / 0.50 m² = 300 N/m² = 300 Pa
Therefore, the wind exerts a pressure of 300 pascals (Pa) on the window.
In the given answer choices:
a. 75 Pa - This is not the correct answer; it is less than the calculated pressure of 300 Pa.
b. 150 Pa - This is not the correct answer; it is less than the calculated pressure of 300 Pa.
c. 1.5 x 10^3 Pa - This is not the correct answer; it is greater than the calculated pressure of 300 Pa.
d. 3.0 x 10^2 Pa - This is the correct answer; it matches the calculated pressure of 300 Pa.
e. 3.0 x 10^5 kPa - This is not the correct answer; it is much greater than the calculated pressure of 300 Pa.