The power of the antenna is

and since the power is the emitted energy divided by the time, the energy emitted per second is

The frequency of the emitted photons is

So the energy of a single photon is

Therefore, to find the number of emitted photons per second, we should divide the total energy emitted by the antenna in 1 second by the energy of a single photon:
Answer:
616.3 rad/s²
Explanation:
Given that
t= 1.46 s
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 27000 rev/min
Angular speed in the rad/s given as

Now by putting the values

ωf=900 rad/s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
900 = 0 + α x 1.46

Therefore the acceleration will be 616.3 rad/s²
Because of the position on the equator, the change in rotation of the Earth on its axis throughout the year doesn't affect it much. Unlike the poles, Quito is almost constantly in direct view of the sun. So, because of lack of change in rotation, the daylight hours are hardly varied as Quito is almost constantly in more or less the same spot in relation to the sun.