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natita [175]
3 years ago
9

Yellow light is incident upon a medium from air at an angle of 45.0°. The light bends toward the normal and refracts at 28.0°. W

hat is the index of refraction of yellow light in this medium? DOK 2
a. 1.61
b. 0.66
c. 1.51
d. 1.00
Physics
2 answers:
Sav [38]3 years ago
6 0

i would say C. This link should ex plain it well https://www.britannica.com/science/refractive-index

Flauer [41]3 years ago
3 0

Answer:

c. 1.51

Explanation:

i = angle of incidence made by yellow light at air-medium interface = 45.0°

r = angle of refraction made by yellow light at air-medium interface = 28.0°

n₁ = index of refraction of air = 1

n₂ = index of refraction of medium = ?

Using Snell's law air-medium interface

n₁ Sini = n₂ Sinr

1 Sin45 = n₂ Sin28

n₂ = 1.51

So the correct choice is

c. 1.51

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A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
Why is traveling between the stars (by creatures like us) difficult?
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Answer:

Stars are at large distances that even light takes thousands of years to reach us from there.

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Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t
wlad13 [49]

Answer:

a) DOWN direction,  b)  directed INTO THE SCREEN, c)    F = 0

Explanation:

The direction of the force is

for electric force

           F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

           F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

       F = q v  B sin 0

       F = 0

6 0
3 years ago
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