Aplane flies directly between two cities, a and b, which are separated by 2300 mi. from a to b, the plane flies into a 65 mi/h headwind. on the return trip from b to a, the wind velocity is unchanged. the trip from b to a takes "65 min less" than the trip from a to b. what is the airspeed of the plane, assuming it is the same in both directions?

The speed of air is  526.22 mi/h                

Explanation:

Given that

Distance = 2300 mi

Velocity of wind ,U=65 m/h

Lets take velocity of plane = V

When plane is moving from A to B then resultant velocity of pane = V-65 m/h

Lets take time taken to plane when plane is moving from A to B = t hr

We know that

Distance = Velocity x time

2300 = (V-65) x t        

 t = 2300/ (V-65)                                                         ------------1

When plane is moving from A to B then resultant velocity of pane = V+65 m/h

Time taken by plane from B to A = t -1.08 hr               ( 65 min = 1.08 hr)

2300 = (V+65) x (t - 1.08)        

     -------------2

  Now from equation 1 and 2

V=526.22 mi/h        

The speed of air is  526.22 mi/h                                                                                      

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.