Answer
given,
length of rod = 21.5 cm = 0.215 m
mass of rod (m) = 1.2 Kg
radius, r = 1.50
mass of ball, M = 2 Kg
radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m
considering the rod is thin
![I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B1%7D%7B3%7DM_%7Brod%7DL%5E2%20%2B%20%5B%5Cdfrac%7B2%7D%7B5%7DM_%7Bball%7DR%5E2%2BM_%7Bball%7D%28R%2BL%29%5E2%5D)
![I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%201.2%20%5Ctimes%200.215%5E2%20%2B%20%5B%5Cdfrac%7B2%7D%7B5%7D%5Ctimes%202%20%5Ctimes%200.0345%5E2%2B2%5Ctimes%20%280.0345%20%2B0.215%29%5E2%5D)
I = 0.144 kg.m²
rotational kinetic energy of the rod is equal to


KE = 6.15 J
b) using conservation of energy



ω = 9.25 rad/s
c) linear speed of the ball
v = r ω
v = (L+R )ω
v = (0.215+0.0345) x 9.25
v =2.31 m/s
d) using equation of motion
v² = u² + 2 g h
v² = 0 + 2 x 9.8 x 0.248
v = √4.86
v =2.20 m/s
speed attained by the swing is more than free fall
% greater = 
= 5 %
speed of swing is 5 % more than free fall
The true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.
We have to note that from the law of conservation of mechanical energy, the sum of the kinetic and potential energy of the book at any point remains constant.
If the book is being raised at constant speed, the the true statement is that ;there is no change in the kinetic energy of the book and the potential energy of the book-Earth system increases.
Learn more about kinetic energy: brainly.com/question/15308590
Um I would help you but I don’t see the question
On foot= 1 kilometer per 5 minutes
Bike= I kilometer per 2 minutes
3 minutes fast per mile on bike