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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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PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST
stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0
3 years ago
We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient
zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicul
zepelin [54]

Answer:

b) there must be a component of force parallel to the motion of the object.

Explanation:

We know that work done on a body by an external force is calculated by the formula given below:

W = F.d = Fd Cos θ

where,

W = Work Done by the force on the body

F = Magnitude of force

d = displacement of the body

θ = The angle between the direction of motion of the body and the force applied

It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.

Therefore, the correct option will be:

b) <u>there must be a component of force parallel to the motion of object.</u>

6 0
3 years ago
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