Answer:
Part(a): The final angular velocity is 
Part(b): The ratio of the rotational energies is
,showing the the energy of th system will decrease.
Explanation:
Part(a):
If '
' be the moment of inertia of an object and '
' be its angular velocity then the angular momentum '
' of the object can be written as

If '
' and '
' be the moment of inertia of the two cylinders and '
' and '
' be the initial angular velocity of the cylinders and '
' and '
' be their respective final angular velocity, then from conservation of angular momentum,

Given,
. From the above expression

Part(b):
Initial kinetic energy
and Final kinetic energy

Substituting the value of
,

The above expression shows that the ebergy of the system will decrease.
-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.
-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.
Answer:
The current flows through the rod is 14.9 A.
Explanation:
Given that,
Magnetic field = 0.045 T
Mass of aluminum rod = 0.19 kg
Length = 1.6 m
Angle = 30.0°
We need to calculate the force
Using resolving force


Put the value into the formula


We need to calculate the current flows through the rod
Using formula of magnetic force


Put the value into the formula

Hence, The current flows through the rod is 14.9 A.
First, calculate the initial velocity of the dog given with the vertical height and the acceleration due to gravity which is calculated through the equation,
2ad = Vo²
Substituting the known values,
2(9.8 m/s²)(1.2 m) = V₀²
V₀ = 4.85 m/s
The kinetic energy is solved through the equation,
KE = 0.5mv²
Substituting the known values to the latest equation,
KE = 0.5 (7.2 kg)(4.85 m/s)²
KE = 17.46 J
Thus, the kinetic energy is 17.46 J.
The frequency of the wave is 
Explanation:
The frequency, the wavelength and the speed of a wave are related by the following equation:

where
c is the speed of the wave
f is the frequency
is the wavelength
For the radio wave in this problem,


Therefore, the frequency is:

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