Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.
All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.
If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.
Answer:
For left = 0 N/C
For right = 0 N/C
At middle = 
N/C
Explanation:
Given data :-
б =
C/ m²
Considering the two thin metal plates to be non conducting sheets of charges.
Electric field is given by
1) To the left of the plate
= 0 N/C.
2) To the right of them.
= 0 N/C.
3) Between them.
= 
= 
= N/C
The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions
