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2 years ago

A neutral metal sphere is attracted to a positively-charged amber rod. Why does this occur?.

Physics

1 answer:

aniked [119]2 years ago

3 0

Hi there!

A neutral metal sphere is comprised of both ELECTRONS and PROTONS.

Recall that electrons have a NEGATIVE (-) charge, and protons have a POSITIVE (+) charge.

Opposite charges attract, and like charges repel.

In this instance, the positively-charged rod will cause the electrons in the neutral metal sphere to move towards the side facing the rod (opposite charges attract), which causes the two to be attracted to one another.

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If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

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Explanation:

the vehicles displacement, since displacement deals with position

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3 years ago

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P

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Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure $P_{in}= 1.90\times10^{5}\ Pa$

Area $A=3.90\times10^{-4}\ m^2$

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

$F_{net}=F_{in}-F_{out}$

$F_{net}=(P_{in}-P_{out})A$

Where, $P_{in}$ =inside Pressure

$P_{out}$ =outside Pressure

A = area

Put the value into the formula

$F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}$

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Hence, The magnitude of the force is 34.59 N.

6 0

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:

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:

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

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)

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.

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.

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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Don’t trust those links they usually pull up your IP

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