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matrenka [14]
3 years ago
5

According to Coulomb's law, when the distance between two point charges doubles, what happens to the electric force acting betwe

en the charges?
Physics
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

If the distance between charges is doubled, then the Force will become one-fourth.

F' = F/4

Explanation:

The Coulomb's Law gives the relation for the attractive or repulsive forces between two charges. The relation or formula is given as follows:

F = kq₁q₂/r²

where,

F = Attractive or Repulsive forces between the charges

k = Coulomb's Constant

q₁ = magnitude of 1st charge

q₂ = magnitude of 2nd charge

r = distance between charges

If we keep the magnitude of charges constant then the relation between Force an Distance between charges will be:

F α 1/r²

<u>So, it is clear from this relation that if the distance between charges is doubled, then the Force will become one-fourth.</u>

<u>F' = F/4</u>

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3 years ago
) what is kinetic energy, and how does it differ from potential energy?
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3 years ago
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto
lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}

Solving for D_{2} we get

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What are the steps to extract metal from the earths crust
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4 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

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a=  3.63 cm

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distance ,d= 0.473 mm

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C=\dfrac{\varepsilon _oA}{d}

By putting the values

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V=26.27 V

V= E d

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7 0
3 years ago
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