The speed of water can be split into vertical and horizontal speed components:
Due to the force of gravity, the y component will be parabolic. The x component will be linear:
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
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
Tetraphosphorus disulfide
- R3KTFORGOOD ☕
Solve for "x"
X = speed of the 5 kg
3/15=5/x
cross multiply
3x=75
because (15)(5)=(3)(x)
divide 3 in both side
X=25
or
A=25 m/s^2
Hope this helps
