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AnnZ [28]
2 years ago
8

What happens to the force between two charged objects when you triple the magnitude of both charges?

Mathematics
1 answer:
Vika [28.1K]2 years ago
8 0

If the force between two charged objects when you triple the magnitude of both charges. The Force will become 9 times the initial force.

<h3>What is coulombs law?</h3>

According to coulombs law force between two charges is given by  

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

Here, R is the distance between both the charges Q and q.

Let the force on the charges be F1

The distance of separation = r

The magnitudes of the charges q1 and q2

K = Coulumb's constant

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

Let the force on the charges be F

The distance of separation = r

The magnitudes of the charges 3q1 and 3q2

K = Coulumb's constant

So,

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

F = \dfrac{1}{4\pi \epsilon }\dfrac{3Q\times 3q}{r^2}\\\\F = \dfrac{1}{4\pi \epsilon }\dfrac{9Qq}{r^2}\\\\F =9  \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}\\\\F = 9F_1

Learn more about coulombs law;

brainly.com/question/13106909

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To simplify this fraction, multiply the entire fraction by the conjugate of the denominator. The conjugate of a square root and a number being added to it would be the number <em>subtracted </em>from the square root. In other words, the conjugate of \sqrt{a} + b would be\sqrt{a} - b.


Applying that information to our fraction shown here, the conjugate of the denominator would be \sqrt{3} - 4. We will multiply both the numerator and denominator of our original fraction by this expression to obtain our answer, as shown below.


\Big(\dfrac{1}{\sqrt{3} + 4}\Big)\Big(\dfrac{\sqrt{3} - 4}{\sqrt{3} - 4}\Big)

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Our answer is \boxed{\dfrac{4 - \sqrt{3}}{13}}.


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3 years ago
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Using

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Given collinear points E, F and G such that point F is the midpoint of segment EG. Find the new length of EG given that EF =5x+9
Veseljchak [2.6K]

Answer:

The length of EG is 58 units

Step-by-step explanation:

The midpoint of a segment divides it into two equal part

Let us use this rule to solve our question

∵ E, F, and G are collinear points

∵ Point F is the midpoint of segment EG

→ That means F divides EG into 2 equal segments EF and FG

∴ EF = FG

∵ EF = 5x + 9

∵ FG = 3x + 17

→ Equate them

∴ 5x + 9 = 3x + 17

→ Subtract 3x from both sides

∵ 5x - 3x + 9 = 3x - 3x + 17

∴ 2x + 9 = 17

→ Subtract 9 from both sides

∴ 2x + 9 - 9 = 17 - 9

∴ 2x = 8

→ Divide both sides by 2 to find x

∵ \frac{2x}{2}=\frac{8}{2}

∴ x = 4

→ Substitute x by 4 in EF and FG to find their lengths

∵ EF = 5(4) + 9 = 20 + 9 = 29

∵ FG = 3(4) + 17 = 12 = 17 = 29

∵ EG = EF + FG

∴ EG = 29 + 29 = 58

∴ The length of EG is 58 units

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