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miskamm [114]
4 years ago
8

Two charged particles are placed 2.0 meters apart. the first charge is +2.0 e-6 c and the second charge is +4.0 e-6

Physics
1 answer:
Yakvenalex [24]4 years ago
7 0
The Coulomb force between two charges is defined as: Fc=(k*Q₁*Q₂)/r² where k=9*10^9 N m² C⁻², Q₁ and Q₂ are charges and r is the distance between those charges. 
In our case:

Q₁=2*10^-6 C
Q₂=4*10^-6 C
r=2 m

Now we simply plug in the numbers into the equation:

Fc={(9*10^9)*(2*10^-6)*(4*10^-6)}/2^2=0.018 N

The Coulombs force between the two positive charges is Fc=0.018 N and it is a repulsive force because both charges are positive.
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Answer:

 

Explanation:

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3 years ago
Read 2 more answers
Two cars A and B are 100m apart moving towards each other with
maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

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ale4655 [162]
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