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slava [35]
3 years ago
7

What is a fundamental difference between gravitational forces and electrostatic forces?

Physics
1 answer:
Kryger [21]3 years ago
8 0

Answer:

The differences are: Electrostatic force is force between two charges separated by distance in space while gravitational force is the force between two masses separated by a distance. Electrostatic force is either repulsive or attractive while gravitational force is always attractive.

Explanation:

hope it helps u stay safe

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Mars was formed by the __________ of smaller objects.
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Answer: debris, dust???
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A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗
Masteriza [31]

The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

  • k is coulomb's constant
  • q₁ and q₂ are the charges
  • r is the distance between the charges
<h3>Distance between the charges</h3>

|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units

F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \  N

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

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3 0
2 years ago
A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on
svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}

Put the value into the formula

\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}

\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}

\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}

\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}

\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1

x=0.32\ m

Hence, The position on the x axis is 0.32 m.

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