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
A. 4% you divide 2000 to 500 and you get 4 which is your answer
Without air resistance, both balls reach the ground at the same instant.
Neither horizontal motion nor weight affects vertical motion.
On the map red bands represent the earthquake
The "blue" object would look the same. Say that this blue object is a square. That "square" is every color but blue. The only reason that you see it as "blue" is because what we call "blue", it the only color of light that bounces back at our eyes. Under all colors of light, especially simultaneously. This square or any shape for that matter would absorb the other colors of light, but the blue will be rejected. Therefore, we can only see what bounces back at our eyes, which is the color blue in this case.
Hope this helps, WyattMarine501
