Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem

Put the value into the formula


We need to calculate the magnitude of the charge q₃
Using formula of net force

Put the value into the formula






Hence, The value of charge q₃ is 40.46 μC.
Answer:
a)
, b) 
Explanation:
a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

In this case, the force is equal to the weight of the object:



The spring constant is:



b) The length of the spring is:




Resistance reduces the current. If there is more resistance, there is less current.
Answer:
Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.
The conserved form of the equation is given by
E = K + Ug + Wother.
In this case Wother = work done by the tension force.
In that form the total mechanical energy is conserved.