Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Carbon belong to Group IV of elements; meaning it can bond with 4 Group I atoms such as Hydrogen to form CH4.
The answer is One atom of carbon can bond with up to four other atoms.
Explanation:
It is given that,
Mas of the object, m = 6 kg
It is lifted through a distance, h = 5.25 m
Tension in the string, T = 80 N
(a) By considering the free body diagram of the object, the forces can be equated as :
Work done by tension, 
(b) Work done by gravity, 
(c) Let v is the final speed of the object and u = 0
v = 5.91 m/s
Hence, this is the required solution.
Answer:
distance and length are the same quantity
