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Deffense [45]
4 years ago
15

On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distanc

e to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×10^8 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 40.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?
Physics
1 answer:
Arisa [49]4 years ago
4 0

Answer:

Explanation:

The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .

difference = \frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( 1-\mu  \right )

Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.

difference =\frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( \mu -1  \right )\\

=\frac{ 2\times 40\times 10^3}{3\times10^3 }\left ( 1.000293-1 \right )\\

=7.81\times 10^{-3} s

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Answer:

A) μ^ = - IA•k

B) Bx = 2D/IA

C) By = 4D/IA

D) Bz= -14D/(IA)

Step-by-step explanation:

We are given;

Torque; τ = D(2i^ − 4j^) Nm

Potential energy; U =− μ•B

Magnitude of magnetic field;

Bo = 15D/IA

a. The vector magnetic moment of the current loop is given as

μ^ = - μ•k

μ^ = -IA •k

b. Now, let's find the component of the magnetic field B.

If we assume B = Bx•i + By•j + Bz•k

Then, torque is given as

τ = μ^ ×B

τ = - IA •k × (Bx•i + By•j + Bz•k)

Note that;

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

Then,

τ = - IA •k × (Bx •i + By •j + Bz •k)

τ= -IABx•(k×i) - IABy•(k×j) - IABz•(k×k)

τ= -IABx•j + IABy•i

τ= IABy•i - IABx•j

The given torque is τ = D(2i^ − 4j^)

Comparing coefficients;

Then,

-IABx = -4D

Bx = -4D/-IA

Bx = 4D/IA

c. Also,

IABy = 2D

Then, By= 2D/IA

d. To get Bz, let's use the magnitude of magnetic field Bo

Bo² = Bx² + By² + Bz²

(15D/IA)²=(4D/IA)²+(2D/IA)² + Bz²

Bz² = (15D/IA)²- (4D/IA)²- (2D/IA)²

Bz² = 225D²/(I²A²) - 16D²/(I²A²) - 4D²/(I²A²)

Bz² = (225D² - 16D²- 4D²)/I²A²

Bz² = 205D²/I²A²

Bz = √(205D²/(I²A²))

Bz = ± 14D/(IA)

So we want to determine if Bz is positive or negative

From the electric potential,

U=− μ•B

U= -(- IA k•(Bx i+By j+Bz k)

Note, -×- = +, i.i=j.j=k.k=1

i.j=j.k=k.i=0

Then,

U= IA k•(Bx i+By j+Bz k)

U = IABz

Since we are told that U is negative, then this implies that Bz is negative

Then, Bz= -14D/(IA)

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