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Oliga [24]
3 years ago
5

A chamber 5.0 cm long with flat, parallel windows at the ends is placed in one arm of a Michelson interferometer (see below). Th

e light used has a wavelength of 500 nm in a vacuum. While all the air is being pumped out of the chamber, 29 fringes pass by a point on the observation screen. What is the refractive index of the air?
Physics
1 answer:
Ivahew [28]3 years ago
6 0

Answer:

1.00029

Explanation:

A wave is a disturbance which travels through a medium and transfers energy without displacing the medium itself.

interference is one of the characteristics of a wave motion

The Michelson interferometer produces interference fringes by splitting a beam of monochromatic light so that one beam strikes a fixed mirror and the other a movable mirror. When the reflected beams are reflected, an interference pattern results.

∝Δm=Δud

∝=wavelength  500nm

Δm=number of fringes

d=length 9f the  Michelson interferometer chamber

Δu= change in refractive index

500*10^-9*29=0.05*Δu

Δu=0.00029

REFRACTIVE INDEX=1+0.00029

refractive index=

1.00029

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A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
Q 28.7: A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T. What is the maximum Hall
Zinaida [17]

Answer:the  maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x  w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/  velocity of the strip, = 25 cm/s  to m/s becomes 25/100=0.25m/s

and w is the width of the strip=  1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

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Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the  maximum Hall voltage across the strip=0.00168V

3 0
2 years ago
Which statement best compares the momentum of a 5-kilogram fish swimming at a speed of 10 meters/second and a 2-kilogram fish sw
ruslelena [56]
M = mass of the larger fish =5kg
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<span>m = mass of the smaller fish =2kg</span>
<span>v = velocity of the smaller fish =10m/s
</span>formula=
<span>MV = mv 
5kg*10m/s=2kg*10m/s
biggern mass fish has more momentum
hope this helps

</span>
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3 years ago
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Whats the answer to 10a² - 6ab + 10 -2a² - 4ab +15b?​
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8a2-10ab+15b+10 Explaintion:

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