Answer:
The maximum induced emf is 15.08 V
Explanation:
Given;
number of turns of the generator, N = 6 turns
area of the loop, A = 0.04 m
magnetic field of the loop, B = 0.2 T
frequency of loop, f = 50 Hz
The maximum induced emf is given by;
E = NBAω
Where;
ω is the angular speed = 2πf
E = NBA(2πf)
E = 6 x 0.2 x 0.04 x (2 x 3.142 x 50)
E = 15.08 V
Therefore, the maximum induced emf is 15.08 V
Answer:
57300 N
Explanation:
The container has a mass of 5300 kg, the weight of the container is:
f = m * a
w = m * g
w = 5300 * 9.81 = 52000 N
However this container was moving with more acceleration, so dynamic loads appear.
w' = m * (g + a)
w' = 5300 * (9.81 + 1) = 57300 N
The rating for the cable was 50000 N
The maximum load was exceeded by:
57300 / 50000 - 1 = 14.6%
Answer:
E. Some charges in the region are positive, and some are negative.
Explanation:
Electric potential is given as;

where;
W is the work done in moving a charge between two points which have a difference in potential
Q is quantity of charge in the given region
If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.
Therefore, the correct statement in the given options is "E"
E. Some charges in the region are positive, and some are negative.
the answer your looking for is Optical instrument.