In: Physics
4. A small plane that travels with constant speed has a mass m =
1563 kg. This includes loading. The combined area of both wings is
16.3 m. Determine the speed of the wind by
above the wing if the wind speed below the wing is 153 m / s. The
air density is 3
1.29 kg / m
answer = velovity at above = V1 =157.689m/s
given vaues
mass= 1563kg
are of both wing =16.3m2
speed of air below the wind = 153 m / s
air density =1.29kg/m3
apply bernoullii theorem
1/2 V12 + g y1 + p1 =1/2 V22 + g y2 + p2
y1 and y2 =horizontal height
p1 andp2 = pressure
v1 and v2 = velocity of air at top and bottom
1/2 V12 + p1 =1/2 V22 + p2
1/2 V12 -1/2 V22 =p2 - p1
1/2 V12 -1/2 V22 =changing pressure
changing pressure =mg/A
1/2 (V12 -V22) = mg/A
1/2 x 1.29x (V12-1532) =1563x9.8/16.3
V12 -1532=(2x1563x9.8/16.3x1.29) = 1456.926
velovity at above = V1 =157.689m/s