Ask question

Ask question

All categories

3 years ago

9

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.

You might be interested in

Given the Earth's density as 5.5 g/cm3 and the Moon's density as 3.34 g/cm3, determine the Roche limit for the Moon orbiting the

Assoli18 [71]

Answer:

The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth

Explanation:

The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.

The relation to determine Roche limit is:

$Roche\ limit=2.423\times R_{P}\times\sqrt[3]{\frac{D_{P} }{D_{M} } }$ ....(1)

Here $R_{P}$ is radius of planet and $D_{P}\ and\ D_{M}$ are density of planet and moon respectively.

According to the problem,

Density of Earth, $D_{P}$ = 5.5 g/cm³

Density of Moon, $D_{M}$ = 3.34 g/cm³

Consider $R_{E}$ be the radius of the Earth.

Substitute the suitable values in the equation (1).

$Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }$

$Roche\ limit= 2.86R_{P}$

8 0

3 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

I need help on this which is correct? plz help me its timed.

mario62 [17]

Answer:

Watershed should be your answer!

3 0

3 years ago

Where on this diagram does the ball have the highest point of gravitational potential energy?

mixer [17]

It should be at the very top since it has more space to fall which gives it more potential energy

3 0

3 years ago

A car could move at constant speed on an icy curve which is banked for _______________ (all, one, no) speed(s) of the car.

son4ous [18]

All is the answer I believe.

4 0

3 years ago