Answer:
I'm not sure it is c I'm sure it is d
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:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
Answer:
The answer to the questions is;
In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)
The distance 4.1 cm is equivalent to λ/4
Explanation:
For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point
When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound
The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ
Therefore 4.1 cm is λ/4
The mass of an object always stays the same since it is really just the amount of matter in an object so no matter the force applied, as long as the object does not lose or gain matter, the object stays the same
