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:
22kj
Explanation:
set h = 0 at the end of slide.
final height is -12m
initial condition will be Ui = 0
Ki = 1/2mv² = 1/2 x 61 x (27)² = 22234.5J
Final condition is Ui = mgh = 61 x 9.8 x -12 = -7173J
Ki = 1/2mv²
Ki= 1/2 x 61 x (16)² = 7808J
conservation energy says that
Ui + Ki = Uf +Kf +ΔEth
so ΔEth = Ui + Ki - Uf - Kf
ΔEth = 22234.5 - 7808 + 7173
ΔEth = 21600J
ΔEth =22Kj
Explanation:
acceleration is the Changing of throut per unit time ....
and in uniform motion .....
in that the speed of an object is uniform means same
Answer:
0.05 cm
Explanation:
The compression of the original spring = 12 - 8.55 cm = 3.45 cm = 0.0345 m
By Hooke's law, F = ke
Where F is the applied force, k is the spring constant and e is the extension or compression. In the question, F is the weight of the car.
k = F/e = 1355 × 9.8 / 0.0345 = 384898.55 N/m
This is the spring constant of the original spring. The question mentions that the force constant of the new spring is 5855.00 N/m smaller. Hence, the force constant of the new spring is 384898.55 - 5855 = 379043.55 N/m
With the new spring installed, the compression will be
e = F/k = 1355 × 9.8 / 379043.55 = 0.035 m = 3.5 cm
The difference in the compressions of both springs = 3.5 cm - 3.45 cm = 0.05 cm
