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svlad2 [7]
2 years ago
8

A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

lar frequency of the proton?
Physics
1 answer:
Ghella [55]2 years ago
7 0

The angular frequency of the cyclotron is 0.07 x 10^{8}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 = 1.6x10^{-27}

q = 1.6 x 10^{-19}

so, the frequency is:

f = qB/2\pim

f = 1.6 x 10^{-19} x 0.47/2x3.14x1.6x10^{-27}

f = 0.07 x 10^{8}

Hence, the angular frequency of the cyclotron is  0.07 x 10^{8}Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

#SPJ4

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We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
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The moment of inertia is
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4 0
3 years ago
Which is not a common property of ionic compounds?
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Explanation:

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3 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
3 years ago
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