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Lerok [7]
3 years ago
8

What are the2 factors thatincrease theelectric forcebetweenobjects?​

Physics
1 answer:
harkovskaia [24]3 years ago
4 0

The two factors that affect the electric force are the charges and the separation between the objects

Explanation:

The magnitude of the electrostatic force between two 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 objects

Looking at the equation, we see that the electric force between objects depends on two factors:

  • The charges of the objects: the force is directly proportional to the product of the two charges, therefore the greater the charges, the stronger the force
  • The distance between the two objects: the force is inversely proportional to the square of the distance, therefore the smaller the distance between the objects, the stronger the force

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
6 0
3 years ago
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m_a_m_a [10]

Answer:

Driving force increases, friction forces increase, the driving force is bigger than friction 12.

Explanation:

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3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe
Citrus2011 [14]
The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.
7 0
3 years ago
The current in two identical light bulbs connected in series is 0. 25 A. The voltage across both bulbs is 110 V. The resistance
konstantin123 [22]

Answer:

220 ohms

Explanation:

I = V / R

0.25 = 110 / R

R = 110 / 0.25

R = 440 ohms

Equivalent resistance = 440 ohms

Resistance of single light bulb = Equivalent resistance / number of bulbs

= 440 / 2

= 220 ohms

8 0
2 years ago
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