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stepladder [879]
3 years ago
15

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

a proton is 1.60218×10−19 C and its mass is 1.67262 × 10−27 kg. What is the cyclotron frequency? Answer in units of rad/s.
Physics
1 answer:
klio [65]3 years ago
8 0

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, q=1.60218\times 10^{-19}\ C

Mass of a proton, m=1.67262 \times 10^{-27}\ kg

The cyclotron frequency is given by :

f=\dfrac{qB}{2\pi m}

f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}

f = 2286785.40 Hz

or

\omega=14368296.44\ rad/s

\omega=1.43\times 10^7 rad/s

Hence, this is the required solution.

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A force of 100N moves a body on a horizontal frictionless surface when......
zalisa [80]

Answer:

I. Friction force exerted on the body is less than 100N

Explanation:

For  a body to be static, the moving force must be equal to the frictional force. Since the frictional force is a force of opposition. It tends to oppose the moving force acting on an object.

Hence if the moving force is greater than the force of friction, the Force of fiction will not be able to overcome the moving hence the body will tend to move.

Therefore, for a body to move, Fm > Ff or Ff < Ff

Fm is the moving force

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Given

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Answer:

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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
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I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
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And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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