Answer:
heterogeneous and homogeneous
Explanation:
Hetero you can separate whilst homo is combined and cant be separated
Kinetic energy<span> increases with the square of the velocity (KE=1/2*m*v^2). If the velocity is doubled, the KE quadruples. Therefore, the </span>stopping distance<span> should increase by a factor of four, assuming that the driver is </span>can<span> apply the brakes with sufficient precision to almost lock the brakes.</span>
Answer:
A. 
B. 
C. ΔK
Explanation:
From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

A. Since the two vehicles become entangled the final mass is:

From linear momentum we got that:




B. The change in velocity of both vehicles are:
For the car

For the truck

C. The change in kinetic energy is:
ΔK=
ΔK=
ΔK
Answer:

Explanation:
Moment of inertia of given shell
where
M represent sphere mass
R -sphere radius
we know linear speed is given as 
translational 
rotational 
total kinetic energy will be


fraction of rotaional to total K.E
