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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
1 answer:
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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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:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
in the second case, the horizontal velocity is increased to
And so the new distance travelled will be
So, the distance increases linearly with the horizontal velocity.