Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² = 
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k = 
we calculate
k = 
k = 3.769 10² m
we substitute in the first equation
w² = 
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f = 
f = 19.415 / 2pi
f = 3.09 Hz
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
40m/s
Explanation:
The horizontal component of velocity remains constant because there are no external forces in that direction
By applying motion equations, V= U+ at
where ,
- v - final velocity
- u - initial velocity
- a-acceleration
- t - time
v = u +at
As no force act on the ball ( we neglect air resistance here) no acceleration is seen,
So v = u = 40m/s
Yellow and red hope that helped
