Afighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. on the ground, the plane’s pilot weighs (53 kg) (9.8 m/s 2 ) = 519.4 n. what is his/her apparent weight during the plane’s turn? answer in units of n

"Apparent weight during the "plan's turn" is  519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector

In magnitude,

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

Which is quite heavier than his/her true weigh of 519.4 N

the assestment will probley apele with cause and effect