Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q = Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]
In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]
Now we can use the first statement to get the first equation:
where:
W₁₋₂ = work from the state 1 to 2.
where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]
Answer
given,
mass of the ball = 3 kg
swing in vertical circle with radius = 2 m
work done by the gravity = ?
work done by the tension = ?
Work done by the gravity = - m g Δh
Δ h = 2 + 2 = 4 m
Work done by the gravity =
= -117.6 J
work done by gravity is equal to -117.6 J
Work done by tension will be equal to zero.
Zero because tension is always perpendicular to velocity
work done by tension is equal to 0 J
Answer:
hope this will help you
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