Answer
Please see the work below.
Work Step by Step
(a) We know that the flux at $ t=0.25s $ is about $8Wb $ and at $0.55s $ it is about $-3Wb $. Thus, the flux is greater at $0.25s $ than $ t=0.55s $.
(b) The rate of change of flux is given as
$\frac{\Delta \phi}{\Delta t}=\frac{-5Wb-10Wb}{0.6s-0.2s}$
$\frac{\Delta \phi}{\Delta t}=-37.5V $
The two emfs are the same because the change in flux is the same at both times.
(c) We have: $\epsilon=-N[\frac{\Delta \phi}{\Delta t}]$
We plug in the known values to obtain:
$\epsilon=-(1)[\frac{-5Wb-10Wb}{0.6s-0.2s}]$
$\epsilon=0.04KV $
