C. A 1200kg car is going 15m/s
Answer:
The angle between the emergent blue and red light is 
Explanation:
We have according to Snell's law

Since medium from which light enter's is air thus 
Thus for blue incident light we have

Similarly using the same procedure for red light we have

Thus the absolute value of angle between the refracted blue and red light is

Im pretty sure A-10 b-0 c-40
The moon is 230,100 miles from planet earth.
Given:
Height of tank = 8 ft
and we need to pump fuel weighing 52 lb/
to a height of 13 ft above the tank top
Solution:
Total height = 8+13 =21 ft
pumping dist = 21 - y
Area of cross-section =
=
=16

Now,
Work done required = 
= 
= 832
)
= 113152
= 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb