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3 years ago

How many electrons make up a charge of 3.5kC?

2 answers:

vladimir1956 [14]3 years ago

7 0

Well now, I don't know !
Let's figure it out.

The charge on one electron is 1.60 x 10⁻¹⁹ Coulomb .

To get 3,500 Coulombs of charge, you'd need to
go around and collect

   (3,500) / (1.60 x 10⁻¹⁹) electrons.

That's 2.19 x 10²² of the little fellas.

That's a big number. But what's REALLY amazing is the
mass and weight of that many electrons:

   Mass: 1.989 x 10⁻⁵ gram

   Weight: about 0.0000007 ounce !

That's not one electron. That's the whole <span>2.19 x 10²² uvvum
that it takes to hold 3,500 Coulombs of charge.</span>

ollegr [7]3 years ago

6 0

21.8452851 * 10^21 e. Which is 21.845.285.100.000.000.000.000 electrons. A lot.

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POR FAVOR AYUDENME A COMLETARLO

Zigmanuir [339]

Answer:

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Explanation:

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3 years ago

14. Which one of the following pictures shows the object that is the most dense? *

rewona [7]

Answer:

Explanation:

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2 years ago

A ray of light incident in air strikes a rectangular glass block of refractive index 1.5 at angle of incidence of 45.

Gwar [14]

Refraction is said to occur when there is a change in the speed of light.

<h3>What is the angle of refraction?</h3>

We know that refraction is said to occur when there is a change in the speed of light as it travels form one medium to another.

Given that the refractive index of the rectangular glass block is 1.5. The angle of refraction can be obtained by the use of the Snell's law;

n = sin i /sinr

n = refractive index

sini = angle of incidence

sin r = angle of refraction

sinr = sini/n

sinr = sin 45/1.5

= 0.471

r = 28 degrees

b) Now;

sinr =sin 45/1.2

sinr = 0.589

r = 36 degrees

For the glass

sinr = sin 36/1.5

sin r = 0.392

r = 23 degrees

Learn more about angle of refraction:brainly.com/question/2660868

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4 0

2 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago