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3 years ago

How does a rubber rod become negatively charged through friction?

2 answers:

8 0

Thank you for posting your question here and Brainly!~

Even though you have not provided answer choices, I believe the answer is whatever letter corresponds with the answer: "It is rubbed with another object, and electrons move onto the rod."

Hope I helped!~ Brainliest appreciated.

kakasveta [241]3 years ago

6 0

Answer:Explained

Explanation:

When a rubber rod is rubbed against, let's say animal fur the electrons from the animal is being transferred to rubber rod as rubber rod has greater attraction for electron i.e. rubber rod has higher electron negativity .That's why it become negatively charged.

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Which of the following represents an image that is located in front of a lens?

Lostsunrise [7]

-di represents an image in front of a lens

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3 years ago

Read 2 more answers

Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

nexus9112 [7]

Answer:

Explanation:

KE = 1/2 mv^2

=1/2(30kg)( 3.4 m/s)^2

=173.4 joules

=1.7×10^2 joules

6 0

3 years ago

Galileo's observational contributionsGalileo Galilei was the first scientist to perform experiments in order to test his ideas.

ozzi

<h2>Answer: </h2><h2>- Jupiter has orbiting moons.</h2><h2>- The Sun has sunspots and rotates on its axis.</h2><h2>- The Moon has mountains, valleys, and craters.</h2><h2>- Venus goes through a full set of phases.</h2>

Explanation:

In 1609 Galileo built a telescope, with which he observed mountains and craters on the Moon, discovered Jupiter’s major satellites and the next year he published these discoveries in his book <em>The Sidereal Messenger</em>.

In addition, Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth. This is because <u>Venus presented its smaller size when it was in full phase and the largest size when it was in the new one, when it is between the Sun and the Earth. </u>

<u />

On the other hand, <u>although Galileo was not the first to observe sunspots</u>, he gave the correct explanation of their existence, which supported the idea that planets revolve around the Sun.

These observations and discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.

4 0

3 years ago

What are the advantages and disadvantages of renewable and non-renewable energy?

bogdanovich [222]

Renewable energy

<u>Advantages :-</u>

1. Easily regenerate

2. Boost economic growth

3. Easily available

4. Support environment

5. Low maintenance cost

<u>Disadvantages :-</u>

1. Weather dependency

2. High installation cost

3. Noise caused by wind energy

4. Fluctuation problem (solar)

5. Intermittency issue (wind)

Non-renewable energy

<u>Advantages :-</u>

1. Concentrated energy source

2. Reliable energy source

3. Can be built anywhere

4. No radioactive waste

<u>Disadvantages :-</u>

1. Produces greenhouse gases

2. Contributes to global warming

3. Produces acid rain

4. Harmful to environment when they are burnt

<em>I hope this helps.....</em>

5 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).

6 0

3 years ago