Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
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
#SPJ4
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
Answer:
A
Explanation:
The compressor receive hot refrigerant and raises the pressure and temperature even further as it is send to the condenser.
