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Alenkinab [10]
2 years ago
11

What is one of the principles of charges in a conductor?

Physics
1 answer:
nevsk [136]2 years ago
3 0

Answer:

A conductor allows free charges to move about within it. The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. Any excess charge will collect along the surface of a conductor. Conductors with sharp corners or points will collect more charge at those points.

Explanation:

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Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 re
alexgriva [62]

Answer:

\frac{Ie}{lm} = 1.10*10^{-3}

Explanation:

GIVEN DATA:

Engine operating speed nf = 8325 rev/min

engine angular speed ni= 12125 rev/min

motorcycle angular speed N_m= - 4.2 rev/min

ratio of moment of inertia of engine to motorcycle is given as

\frac{Ie}{lm} = \frac{-N}{(nf-ni)}

\frac{Ie}{lm} = \frac{-(-4.2)}{(12125 - (8325))}

\frac{Ie}{lm} = 1.10*10^{-3}

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3 years ago
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A ball is thrown downward with an initial speed of 7 m / s. The ball's velocity after 3 seconds is m / s. (g = -9.8m / s²)
yan [13]

Answer:

-36.4 m/s

Explanation:

final velocity= initial velocity + acceleration x time

7 + (-9.8)(3)= -36.4 m/s

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3 years ago
What are some ways houses along the coastlines can protect themselves from storm surges?
BARSIC [14]
Build walls around the coast
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3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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Why won’t a magnet help you separate a mixture of salt and water
Gennadij [26K]

Answer: nor are magnetic

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