Answer: The correct answer is B
Explanation: The string is pulling right and the string is unraveling causing it to accelerate left
Answer: William Thomson, better known as Lord Kelvin
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:
Battery will run for t = 90 s
Explanation:
As we know that rate of flow of charge is known as electric current
So we will have
now we have
Answer:
x = 4 m
Explanation:
For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive
as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system
∑τ = 0
W 3 - w x = 0
x = 3W / w
x = 3 Mg / mg
x = 3 M / m
let's calculate
x = 3 60/45
x = 4 m
