Answer: 
Explanation:
Given
Magnitude of charge is 
Force experienced is 
Electric field intensity is the electrostatic force per unit charge

Thus, the electric field intensity is 
A..........................................
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
Answer:
(a) Acceleration of the bag will be a=16.214m/sec^2
(B) Weight of the bag will be 137.2 N
Explanation:
We have given mass of the bag m = 14 kg
Force with which bag is lifted = 227 N
(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration
So 

(b) Acceleration due to gravity 
We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity
So weight 
So weight of the bag will be 137.2 N
At time t = 0 the velocity is v1. Therefore C1 = v1 and C2 = x1. Equations (1), (2), (3), and (4) fully describe the motion of particles, or bodies experiencing rectilinear (straight-line) motion, where acceleration a is constant.