Coulomb's law states<span> that: The magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distance between them.
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Answer:
, repulsive
Explanation:
The magnitude of the electric force between two charged particles is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges of the two particles
r is the separation between the two charges
The force is:
- repulsive if the two charges have same sign
- Attractive if the two charges have opposite signs
In this problem, we have two electrons, so:
is the magnitude of the two electrons
is their separation
Substituting into the formula, we find the electric force between them:

And the force is repulsive, since the two electrons have same sign charge.
Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :




So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer:

Explanation:
Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(
) and inversely proportional to the square of the distance(d) that separates them.
Replacing the given values, where k is the Coulomb constant:

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