2.7549 x 10^4 is the answer I hope this helped u
A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

and substituting t = 75 seconds, we find

In degrees, it is

So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by

where we have

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find

Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
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