Question

"One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that the air pressure inside of a sealed house is 1.02 atm when a hurricane hits. The hurricane rapidly decreases the external air pressure to 0.910 atm. What net outward force is exerted on a square window of the house that is 2.03 m on each side? (1.00 atm = 1.01 × 105 Pa)"

Answer 1

Answer:

Net outward force exerted on the window = 45.931_KN

Explanation:

In the hurricane, the pressure outside the window = 0.910_atm

Pressure inside the sealed house = 1.02_atm

Size of windows sides = 2.03_m

But pressure = force/area

Therefore, force = pressure × area

Where 1.00 atm = 1.01 × 10^5 Pa

0.910_atm = 1.01 × 10^5 Pa × 0.910 = 0.9220575_Pa

Also 1.02_atm = 1.03 × 10^5 Pa

Area of widow = 2.03^2 = 4.12_m^2

Force on the outside of the window = 0.9220575_Pa × 4.12_m^2 = 379.971_KN

Force on the inside of the window = 1.03 × 10^5 Pa × 4.12_m^2 = 425.9 KN

Net outward force exerted on the window = 425.9_KN - 379.971_KN = 45.931_KN

Net outward force exerted on the window = 45.931_KN