Respuesta :
Answer:
If the lifting force varies jointly with wing surface area and the square of the plane's velocity, then:
F = kAv²
If the lift is 27,800 pounds when the area is 190 sq. ft and the velocity is 230 mph, then the constant of proportionality is:
27,800 = k(190)(230)²
k = (27,800) / (190)(230)² = 139/50255
The lifting force the plane slows down to 200 mph is:
F = (139/50255)(190)(200)²
F = 21,020.79
Rounded to the nearest pound, the lifting force is 21,021 pounds.
Answer:
28,339 lbs
Step-by-step explanation:
If the lifting force exerted on an airplane wing varies jointly as the area of the wing's surface and the square of the plane's velocity then:
[tex]F \propto Av^2 \implies F=kAv^2[/tex]
Where:
- F = lifting force in pounds (lbs)
- A = area in square feet (ft²)
- v = velocity in miles per hour (mph)
- k = some constant
Given:
- F = 10,500 lbs
- A = 180 ft²
- v = 140 mph
Substitute the given values into the found equation and solve for k (variation constant):
[tex]\begin{aligned}F & = kAv^2\\\\10500 & = k \cdot 180 \cdot 140^2\\10500 & = 3528000k\\k & = \dfrac{10500}{3528000}\\\implies k & = \dfrac{1}{336}\end{aligned}[/tex]
Therefore:
[tex]\boxed{ F=\dfrac{1}{336}Av^2}[/tex]
To find the lifting force (F) when the plane speeds up to 230 mph, substitute the given area of the wing (A = 180 ft²) and the new velocity (v = 230 mph) into the equation and solve for F:
[tex]\begin{aligned}F&=\dfrac{1}{336}Av^2\\\implies F& = \dfrac{1}{336} \cdot 180 \cdot 230^2\\& = \dfrac{1}{336} \cdot 180 \cdot 52900\\& = \dfrac{1}{336} \cdot 9522000\\& = \dfrac{9522000}{336} \\& =28339.2857...\\&=28339\;\sf lbs \; (nearest\:pound)\end{aligned}[/tex]
Therefore, the lifting force is 28,339 lbs (nearest pound).