An airplane in a horizontal circle at a speed of 130.5556 m/s. If its wings are tilted 40 degress to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface?

I have this equation that i used, its all that i have..but it has gravity in it..so i don't know if i did it right

tan angle = v^2 / g x R

r is radius
g is gravity
v is velocity
angle would be the 40

thats exactly how you have to do it!!!!

To find the radius of the circle in which the plane is flying, we can use the equation you provided:

tan(angle) = v^2 / (g x R)

where:
- angle is the angle of tilt of the wings (40 degrees in this case).
- v is the velocity of the plane (130.5556 m/s).
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- R is the radius of the circle.

To solve for R, let's rearrange the equation:

R = v^2 / (g x tan(angle))

Now we can substitute the known values into the equation:

R = (130.5556 m/s)^2 / (9.8 m/s^2 x tan(40 degrees))

Calculating the value:

R ≈ (17,000 m/s^2) / (9.8 m/s^2 x 0.839)

R ≈ 1792.65 meters

Therefore, the radius of the circle in which the plane is flying is approximately 1792.65 meters.