Choice c it's confusing to understand but it's c
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
Answer: C.
250 kg-m/s
Explanation:
Given that the
Mass M = 1,500 kg
Force F = 500 N
Time t = 0.5 seconds
From Newton's second law of motion which state that the rate of change of momentum is proportional to the applied force.
F = mV/t
Ft = mV
Where ft = impulse: the product of time and applied force
Substitutes force and time into the formula
Ft = 500 × 0.5 = 250 Ns
