Answer:
U = 102.8 J (100 J to two significant digits)
Explanation:
potential energy converted = 20(9.8)(1.8) = 352.8 J
kinetic energy at base of track = ½(20)5.0² = 250 J
energy (work) of friction 352.8 - 250 = 102.8 J
4 seconds
bullet flies 200m/s. thus 800m/200m=4sec
B v1= 12.5m/s
As the acceleration is constant and time taken is also same and distance covered is halfed so the speed will also be halfed.
D
I hope these is correct
