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:
The diameter of the axle is 5.08 cm.
Explanation:
Given that,
Force = 800 N
Distance = 78.0 m
Suppose we need to find the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm.
We need to calculate the radius of axle
Using formula of linear velocity
Where, v =velocity
r = radius
=angular velocity
Put the value into the formula
We need to calculate the diameter of axle
Using formula of diameter
Hence, The diameter of the axle is 5.08 cm.
acceleration= change in velocity/ time taken
initial velocity=(36×1000m)/3600
= 10m/s
final velocity= (54×1000m)/3600
=15m/s
change in velocity= final - initial
=15 - 10
=5m/s
time taken = 5s
acceleration=15/5
=3metre per second square
Answer:5m/s² well I am not pretty sure but hope it's help
Explanation:
u=0m
final velocity ,v=50m/s
t=10s
(v-u)/t=(50-0)/10=50/10=5m/s²
