The correct field line would be A.
To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.
So things our data are given by



PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

So the voltage for transformer 2 would be given by,

PART B) To express the number value we proceed to replace with the previously given values, that is to say



A.
According to the equation: v=λ*f
λ stands for wavelength. wavelength increases, the frequency will decrease. lower frequency deserves low energy.
The question is incomplete. You dis not provide values for A and B. Here is the complete question
Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.
A = 12
B = 18
Answer:
18.5⁰
Explanation:
Angle of incidence i = 12.0 + A
A = 12
= 12.0 + 12
= 14
Refractive index u = 1.10 + B/100
= 1.10 + 18/100
= 1.10 + 0.18
= 1.28
We then find the angle of refraction index u
u = sine i / sin r
u = sine24/sinr
1.28 = sine 24 / sine r
1.28Sine r = sin24
1.28 sine r = 0.4067
Sine r = 0.4067/1.28
r = sine^-1(0.317)
r = 18.481
= 18.5⁰
Answer: f=150cm in water and f=60cm in air.
Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:
= (nglass - ni)(
-
).
nglass is the index of refraction of the glass;
ni is the index of refraction of the medium you want, water in this case;
R1 is the curvature through which light enters the lens;
R2 is the curvature of the surface which it exits the lens;
Substituting and calculating for water (nwater = 1.3):
= (1.5 - 1.3)(
-
)
= 0.2(
)
f =
= 150
For air (nair = 1):
= (1.5 - 1)(
-
)
f =
= 60
In water, the focal length of the lens is f = 150cm.
In air, f = 60cm.