The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs
<h3>What is angular velocity?</h3>
Angular velocity is the ratio of the angle turned to the time taken.
The kinematic equation for angular velocity are presented as follows;
ω = ω₀ + α·t
θ = θ₀ + ω₀·t + 0.5·α·t²
Where;
θ₀ = The initial angle turned = 0
ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise
α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²
t = The duration of the motion;
When the angular velocity, ω is zero, we get;
0 = 115 - 46/17·t
t = 85/2
Which indicates;
θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25
θ = 7331.25 radians
θ = 7331.25/(2×π) ≈ 1166.8 rev
The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero
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Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
<h3>Average power produced by the rocket</h3>
The average power produced by the rocket is calculated as follows;
P = FV
where;
- P is the average power
- F is the force exerted by the rocket
- V is the velocity of the rocket
Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
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As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.
Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.
As per Planck's quantum theory -

Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.
The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.
We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

[
is the wavelength of the wave ]
It is given that wavelength =21 cm=0.21 m
The frequency is calculated as -




Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz
Here Hertz[ Hz] is the unit of frequency.