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
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
, and
Sampling rate 
The positive frequencies at the output of the sampling system are :
When n = 0,
when n = 1,
When n = 2,
Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Displacement is how far something is from its starting position
for example if X moved 10 feet, turned around and moved 30 feet, its distance would be 40, but the displacement would be 20,
and if Y Moved 2 feet, and turned around and moved 22 feet, its distance would be 24, but its displacement would also be 20
I think it has 11 but i don't really know for sure.Correct me if i'm wrong.
