Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
Sorry I'm so late, but I just took this test and the answer is white (for people who didn't study well ;) )
Answer: 321 J
Explanation:
Given
Mass of the box 
Force applied is 
Displacement of the box is 
Velocity acquired by the box is 
acceleration associated with it is 

Work done by force is 

change in kinetic energy is 

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
![\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%2BW_f%3D%5CDelta%20K%5Cquad%20%5BW_f%3D%5Ctext%7BWork%20done%20by%20friction%7D%5D%5C%5C%5C%5C%5CRightarrow%20375%2BW_f%3D54%5C%5C%5CRightarrow%20W_f%3D-321%5C%20J)
Therefore, the magnitude of work done by friction is 
My guess is A. I'm not 100% positive but i'm pretty sure.
An animal is a lion, it may compete for food and it’s territory. The animal may have to fight other animals to get these things