<span>The work done is 3.0 Nm.
We can us the equation Work = Force * Distance, where Force = 75.0 N, and distance is xf – xi = 3.00 cm - -1.00 cm = 4.00 cm. Convert centimeters to meters by moving the decimal place to the left by two places to get 0.04 m. Plug these values into the Work equation:
Work = Force * Distance
Work = 75.0 N * 0.04 m
Work = 3.0 Nm</span>
The angular frequency of the cyclotron is 0.07 x 
Hz.
<h3>What is angular frequency?</h3>
- Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
- It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.
<h3>What is cyclotron?</h3>
The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.
<h3>Calculation of angular frequency:</h3>
Given,
B = 0.47 T
r = 0.68
mass of proton = 1.6x
q = 1.6 x 
so, the frequency is:
f = qB/2
m
f = 1.6 x x 0.47/2x3.14x1.6x
f = 0.07 x
Hence, the angular frequency of the cyclotron is 0.07 x Hz.
Learn more about angular frequency here:
brainly.com/question/14244057
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Protons do not move out of the nucleus of atoms although they repel each other.
Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.
Answer:
a. Wavelength = λ = 20 cm
b. Next distance of maximum intensity will be 40 cm
Explanation:
a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.
Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.
This we get the equation:
(nλ + λ/2) - nλ = 30-20
λ/2 = 10
λ = 20 cm
b. at what distance, sound intensity will be maximum again.
For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.
= 30 + λ/2
= 30 + 20/2
=30+10
=40 cm
