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1 year 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?

1 answer:

Ghella [55]1 year 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 = 1.6x $10^{-27}$

q = 1.6 x $10^{-19}$

so, the frequency is:

f = qB/2 $\pi$ m

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

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

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Question 14 (2 points)

sveticcg [70]

Answer:

after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

7 0

2 years ago

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

$KE = \frac{1}{2}mv^2$

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago

You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a

soldier1979 [14.2K]

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

<u>s = 12.57 m</u>

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

<u>ω = 5 RPM</u>

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

<u>v = 3.142 m/s</u>

8 0

3 years ago

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

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2 years ago

Read 2 more answers

Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$

$=289.9 \ m/s$

The time will be:

⇒ $t=\frac{d}{v}$

$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

$=0.041 \ sec$

hence,

⇒ $N=\frac{1}{t}$

$=\frac{1}{0.041}$

$=24.39 \ per \ sec$

4 0

2 years ago