To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,



PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

At the same time we have that centripetal acceleration is given as

Replacing



One reason is that when you have been out in the cold, your hands feet and exposed features of your face will take time to recover as the blood circulation improves and supplied warm blood to capillaries. So the relatively warm room you enter will not immediately feel warm until the blood has regained its normal circulation. Other factors are that windows are cooled from the outside and condensation forms on the inside because of moisture in the air. For this condensation to evaporate requires heat, which will be extracted from the room and the air near the windows will be cooled. The cold air will descend and form a draught at floor level and this will tend to make the room cooler.
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J