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Luden [163]
3 years ago
5

A shift of one fringe in the michelson-morley experiment corresponds to a change in the round-trip travel time along one arm of

the interferometer by one period of vibration of light when the apparatus is rotated by 90 degrees. a particular interferometer has arm lengths of 8.50 meters, and is using light with a wavelength of 658 nm. what velocity through the ether would be deduced from a shift of one fringe?
Physics
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer : [tex]v = 59.02\ km/s[/tex]

Explanation :  given that,

Arm length = 8.50 meters

Wavelength = 658 nm

Number of fringe = 1

We know that,

Formula of the speed is

v = c\sqrt\dfrac{N\lambda}{2L}

Now, put the values

v = 3\times10^{8}\ m/s\sqrt\dfrac{1\times658\times10^{-9}\ m}{2\times8.50\ m}

v = 590.21\times10^{2}\ m/s

v = 59.02\ km/s

Hence, this is the required solution.







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3 years ago
A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to
maksim [4K]

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius r=\frac{d}{2}=\frac{15.3}{2}=7.65cm

So area A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt

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Discuss the difference between waveform graphs and vibration graphs​
Kitty [74]

Difference exists mainly in the label for x axis.

Explanation:

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Irregular galaxies are very large. True or False?
xenn [34]
It's false i hope this helps :)

5 0
3 years ago
Read 2 more answers
PART ONE
stira [4]

Answer:

3.64×10⁸ m

3.34×10⁻³ m/s²

Explanation:

Let's define some variables:

M₁ = mass of the Earth

r₁ = r = distance from the Earth's center

M₂ = mass of the moon

r₂ = d − r = distance from the moon's center

d = distance between the Earth and the moon

When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

M₁ / r² = M₂ / (d − r)²

M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0

d² − 2d r + (1 − M₂/M₁) r² = 0

Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

g = GM / r²

g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²

g = 3.34×10⁻³ m/s²

4 0
2 years ago
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