Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

Now the total force required is:
0.0702N+0.803N=0.873N
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
The energy carried by a single photon of frequency f is given by:

where

is the Planck constant. In our problem, the frequency of the photon is

, and by using these numbers we can find the energy of the photon:
Physical changes: melting, evaporating, and condensation. This is a physical change.