A. inelastic, since the girl moves in the same direction as the thrown ball
Answer: No sure what to do
Explanation: points for points
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
![[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\](https://tex.z-dn.net/?f=%5B%2021y%20%5D_%7B0%7D%5E%7B8%7D%20-%20%5B%5Cfrac%7By%5E%7B2%7D%7D%7B2%7D%5D_%7B0%7D%5E%7B8%7D%5C%5C)
)
= 113152 = 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
