Answer:
D) Electric power distribution.
Explanation:
Electric power distribution requires high voltages to efficiently transmit electric power. This requires use of a transformer which uses electromagnetic induction.
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
Part A ) Replacing our values,
Part B ) To find the torque we apply the equation as follow,
