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Contact [7]
3 years ago
8

Calculate the value of two equal charges if they repel one another with a force of 10 N having 10 m distance apart in a vacuum.

What would be the size of the charges if they were situated in an insulting liquid whose permittivity was ten times that of a vacuum.
​
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Explanation:

a) From Coulombs law,

F = (kq1q2)/r²...............1

F = 10 N

r = 10m

k = 9 × 10^9 m/F

q1 = q2

from equation 1, make q the subject

q = (Fr²)/2k

q = (10×10²)/(2×9×10^9)

q = 1000/1.8×10^10

q = 5.556×10^-8 C

b) K = k/ko

k = Kko = 10times permittivity of free-space

Goodluck

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
Pleaseeee help I really need to pass to graduate
Rus_ich [418]

Answer:

kWh I think

Explanation:

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Which of the following units are the Standard International units for
umka21 [38]
The answer is A. meters (m)
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Which of the following statements is TRUE about updating the exposure control plan?
iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

<em>Updates must have the  reflection of changes in tasks as well in procedures.</em>

<em>Updates must reflect changes in positions that affect occupational exposure.</em>

<em>Updates must have the cost of PPE that is needed and  necessary to reduce exposure</em>

An exposure control plan can be regarded as  the framework for compliance between the employer and the workers.

  • This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

  • This plan gives hope to workers in term of protection when working with their Employer.

  • There are some elements that is associated with  Exposure Control Plan, and theses are;
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Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0
2 years ago
What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?
choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

3 0
3 years ago
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