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.
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Answer:
First one, third one, and fourth one
The thermal energy that is generated due to friction is 344J.
<h3>What is the thermal energy?</h3>
Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
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The true scientific way is the last: using the water displacement method
Explanation:
We define force as the product of mass and acceleration.
F = ma
It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.
Given Data:
Width of the pool = w = 50 ft
length of the pool = l= 100 ft
Depth of the shallow end = h(s) = 4 ft
Depth of the deep end = h(d) = 10 ft.
weight density = ρg = 62.5 lb/ft
Solution:
a) Force on a shallow end:
b) Force on deep end:
c) Force on one of the sides:
As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.
1) Force on the Rectangular part:
2) Force on the triangular part:
here
h = h(d) - h(s)
h = 10-4
h = 6ft
now add both of these forces,
F = 25000lb + 150000lb
F = 175000lb
d) Force on the bottom:
