Answer:
ω' = 0.815 rad/s
Explanation:
Given,
R = 1.20 m
Inertia of merry-go- round= 240 kg.m²
Rotating speed = 9 rpm = 
=0.9424 rad/s
mass of the child, m = 26 kg
angular speed of the merry-go-round=?
we know
Angular momentum, L = I ω
Moment of inertia of the child
I' = m r² = 26 x 1.2² = 37.44 kgm²
Conservation of angular momentum
initial angular momentum = Final angular momentum
I ω = (I+I')ω'
240 x 0.9424 = (240+37.44) ω'
226.176= 277.44 ω'
ω' = 0.815 rad/s
new angular speed of the merry-go- round is equal to 0.815 rad/s
Answer:
The scientific defenision of work reveals its relationship to energy; whenever work is done, energy is transfered.
Answer:
the initial velocity of bullet is u = 330.335 m /s
Explanation:
given,
mass of bullet(m) = 7 g
Mass of pendulum ( M ) = 1.5- kg
The bullet emerges from the block with a speed (V) = 200 m/s
The maximum height ( h ) = 12 cm.
Let the initial speed of the bullet be ' u '
Apply, Law of conservation of momentum :
m u = ( m + M ) V ...............................(1)
Apply, Law of conservation of Energy :
...............(2)
From equation (1) and equation (2) , we have
But V = √2gh
u = 330.335 m /s
hence, the initial velocity of bullet is u = 330.335 m /s
Answer:
The three forms of water are gas, liquid, and ice, I'm not sure if these were mentioned in the passage since you did not attach a picture.
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