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3 years ago

What is electrons ? explain it function

Physics

2 answers:

Neko [114]3 years ago

7 0

Answer:

The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure

Explanation:

functions of electrons

and electrons being the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus

coldgirl [10]3 years ago

4 0

Answer:

A stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.

Explanation:

Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus.

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Light is shown through air on a diamond (n=2.42) and it partially reflects and refracts.

NikAS [45]

Answer:

Light is shown through air on a diamond n=2.42) and it partially reflects and reflects.

Based on the incident angle of 62.5 shown what is the angle of reflection of the light

question of options

21.5

6 0

2 years ago

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

3 0

2 years ago

Need help on physics/momentum

koban [17]

Answer:

vf=11.67m/s

Explanation:

vf:

Δp=m(vf-vi)

80=12(vf-5)

80=12vf-60

140=12vf

vf=11.67m/s

pf:

p=m*v

p=12*11.67

p=140.04 N*s

Hope this helps

5 0

3 years ago

A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. Part A) At the instant

Gre4nikov [31]

A) $2.2\cdot 10^{-19} N$

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

In this situation,

I = 5.20 A

r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

$B=\frac{(4\pi \cdot 10^7)(5.20)}{2\pi (0.046}=2.26\cdot 10^{-5} T$

The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

$F=qvB$

where

q is the magnitude of the charge of the particle

v is its velocity

B is the magnetic flux density

For the electron of this problem,

$q=1.6\cdot 10^{-19} C$ is the charge

$v=6.10\cdot 10^4 m/s$ is the speed

$B=2.26\cdot 10^{-5} T$ is the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(6.10\cdot 10^4)(2.26\cdot 10^{-5})=2.2\cdot 10^{-19} N$

B) Same direction as the current in the wire

First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:

- The thumb gives the direction of the current -> upward

- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)

Now the direction of the force can be found by using the right-hand rule. We have:

- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)

- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

8 0

3 years ago

compute the electric field experienced by a test a charge q = +0.08µC from a source charge q = +15µC in a vacuum when the test c

Fudgin [204]

here is your answer hope it's helpful for you thankyou I

3 0

2 years ago

What is electrons ? explain it function​

What is electrons ? explain it function