Trey is testing a properly working resistor in a circuit and finds its value to be far lower than he expects what’s a likely err
or
A. The resistor is faulty
B. The circuit is energized
C. He didn’t cut a lead on the resistor
D. The resistor is installed backwards
A. The resistor is faulty
B. The circuit is energized
C. He didn’t cut a lead on the resistor
D. The resistor is installed backwards
1 answer:
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Answer:
Hooke's law, given by is denoted with a negative sign to regard that the direction of restoring force that occurs opposite in direction of the force that caused the displacement
. F represents the force required to create a spring displacement of
.
Explanation:
As with many things in physics, the negative sign represents direction. In this case, after a spring is displaced a distance of from equilibrium, the restoring force created by the spring will be in opposite direction of the force applied to move the spring there, represented by F.
Answer:
a) 60 V
b) 125 V
c) 125 V
Explanation:
<u>Given</u>
We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.
<u>Required </u>
We are asked to calculate the potential at the distances
(a) r = 10.0 cm
(b) r = 20.0 cm
(c) r = 40.0 cm
<u>Solution</u>
(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by
V = q/4∈_o (1)
r is the distance where the potential is measured and the term 1/4∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m
V= 1*q/4∈_o*r
=60 V
(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by
V = 1*q/4∈_o*R (2)
Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m
V = 1*q/4∈_o*R
= 125 V
(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)
V= 125 V