Answer:
Part a)

Part b)

Explanation:
Time period of sun is given as



Now the radius of the orbit of sun is given as



Part a)
centripetal acceleration is given as




Part b)
orbital speed is given as



Answer:
energy of motion decrease
Explanation:
yes
Answer:
Explanation:
Even though the minimum temperature is more than the freezing point, Frost was observed on the ground because the ground will cool rapidly as cool air tends to move towards the ground, the temperature of the ground is lower than the atmosphere a few feet above it.
As the thermometer is kept some feet above the ground so ground temperature may be lower than the minimum recorded temperature.