If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the turn?

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

answeralert drivers who drift off the road;

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2