The electric potential energy of the electron depends on the potential difference applied between the two ends of the cable. Indeed, the electric potential energy of a charge is given by

where q is the magnitude of the charge, while

is the potential difference applied. So, U depends on

.
Answer:
The magnitude of the angular acceleration is 
Explanation:
From the question we are told that
The angular speed of CD is 
time taken to decelerate is 
The final angular speed is 
The angular acceleration is mathematically represented as

substituting values


The negative sign show that the CD is decelerating but the magnitude is

Answer:
Explanation:
Magnetic field near current carrying wire
= 
i is current , r is distance from wire
B = 10⁻⁷ x 
force on second wire per unit length
B I L , I is current in second wire , L is length of wire
= 10⁻⁷ x
x 33 x 1
= 3234 x 
This should balance weight of second wire per unit length
3234 x
= .075
r =
x 10⁻⁷
= .0043 m
= .43 cm .
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 answer is a inductive