Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.
I hope my answer has come to your help. God bless and have a nice day ahead!</span>
Answer: 117.8 nm
Explanation:
Given,
Nonreflective coating refractive index : n = 1.21
Index of refraction:
= 1.52
Wave length of light = λ = 570 nm = 


Hence, the minimum thickness of the coating that will accomplish= 117.8 nm
Answer:
b) the alpha particles were found to be attracted to the nucleus
Explanation:
Answer:
hold up nvm Reaction with oxygen
Explanation: