#### Answer

(a) The person's apparent weight is $518.5~N$
(b) The person's apparent weight is $521.5~N$
(c) The radius of the Ferris wheel is $45.2~m$

#### Work Step by Step

(a) The radial acceleration is: $~a_r = \omega^2~R$
At the top, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight.
$\sum F = m~a_r$
$mg-F_N = m~a_r$
$F_N = mg - m~\omega^2~R$
At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$
Therefore, the person's apparent weight is $518.5~N$ because the apparent weight at the top is less than the person's actual weight.
(b) At the bottom, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight.
$\sum F = m~a_r$
$F_N - mg = m~a_r$
$F_N = mg + m~\omega^2~R$
At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$
Therefore, the person's apparent weight is $521.5~N$ because the apparent weight at the bottom is greater than the person's actual weight.
(c) We can find the radius $R$:
$m~\omega^2~R = 1.5~N$
$R = \frac{1.5~N}{m~\omega^2}$
$R = \frac{1.5~N}{(520.0~N/9.80~m/s^2)~(0.025~rad/s)^2}$
$R = 45.2~m$
The radius of the Ferris wheel is $45.2~m$.