Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
Answer:
The answer is: a. 1960 N/m2
Explanation:
The downward force is equal:
[tex]F=mg[/tex]
Where
m = mass = 20000 kg
[tex]F=20000*9.8=196000N[/tex]
Then, the pressure difference is equal to:
[tex]P=\frac{F}{A}[/tex]
Where
A = total area of both wings = 100 m²
[tex]P=\frac{196000}{100} =1960Pa[/tex]
