What happens to has when it cools is ...
Step 1) They will start to form liquids, like condensation.
Step 2) As liquids cool, they will turn into solids.
Step 3) As solids cool, they become more stable and solid.
Explanation:
It is known that relation between torque and angular acceleration is as follows.

and, I = 
So, 
= 4 


So, 
= 1 
as 
=
Hence, 

Thus, we can conclude that the new rotation is
times that of the first rotation rate.
The frequency of the wheel is given by:

where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:

And now we can find the angular speed of the wheel, which is related to the frequency by:
The overall arrangements of the atoms produce crystals
Answer:
a) Acceleration of the car is given as

b) Acceleration of the truck is given as

Explanation:
As we know that there is no external force in the direction of motion of truck and car
So here we can say that the momentum of the system before and after collision must be conserved
So here we will have

now we have


a) For acceleration of car we know that it is rate of change in velocity of car
so we have



b) For acceleration of truck we will find the rate of change in velocity of the truck
so we have


