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Simora [160]
3 years ago
11

Suppose you are observing the interference pattern formed by a Michelson interferometer in a laboratory and a joking colleague h

olds a lit match in the light path of one arm of the interferometer. Will this have an effect on the interference pattern
Physics
1 answer:
Troyanec [42]3 years ago
7 0

Answer:

Explanation:

In Michelson interferometer , two light waves from different directions are made to overlap so that fringes are formed on the screen due to interference . In it,  two monochromatic and coherent light are made to overlap which have some path difference or phase difference. They form dark and bright fringes .

Now when a match stick is lit in the path of a wave , the fringes will disappear  and an general illumination will be observed on the screen as the light from the lit match stick will not be coherent . Incoherent light can not form stable fringes.

You might be interested in
Which of the following statements are not true regarding sound?
Likurg_2 [28]

Answer:

All  are  not true.

Explanation:

Sound needs medium to travel.

sound speed is fastest in solids.

5 0
3 years ago
Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends
lesya692 [45]

Answer:

0.00384 kg/m

Explanation:

The fundamental frequency of string waves is given by

f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}

For some tension (F) and length (L)

f\propto\frac{1}{\mu}

Fundamental frequency of G string

f_G=196\ Hz

Fundamental frequency of E string

f_E=659.3\ Hz

Linear mass density of E string is

\mu_E=3.4\times 10^{-4}\ kg/m

So,

\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m

The linear density of the G string is 0.00384 kg/m

4 0
3 years ago
The acceleration due to gravity of the earth is 9.81 m/s2 and the acceleration due to gravity of the moon is 1.625 m/s2. If your
Viktor [21]

Answer:

686 N

113.75 N

Explanation:

On Earth:

W= mg

= 70 x 9.8

= 686 N

On Moon:

W= mg

= 70 x 1.625

= 113.75 N

3 0
3 years ago
A 9 µF capacitor and a 16 H inductor are connected in series with a 60 Hz source whose rms output is 116 V. Find the rms current
AfilCa [17]

Answer:

19.2 mA

Explanation:

C = 9 micro farad = 9 x 10^-6 F

L = 16 H

f = 60 Hz

Vrms = 116 V

X_{L}=2\times \pi \times f\times L

X_{L}=2\times 3.14 \times  60 \times 16

XL = 6028.8 ohm

X_{C}=\frac{1}{2 \times \pi  \times f \times C}

X_{C}=\frac{1}{2\times 3.14  \times 60 \times 9 \times 10^{-6}}

Xc = 294.88 ohm

Let Z be the impedance of the circuit.

Z=\sqrt{X_{L}^{2}+X_{C}^{2}}

Z=\sqrt{6028.8^{2}+294.88^{2}}

Z = 6036 ohm

I_{rms}=\frac{V_{rms}}{Z}

I_{rms}=\frac{116}{6036}

Irms = 0.0192 A

Irms = 19.2 mA

6 0
4 years ago
• what is the typical distance between two adjacent pins on a 14-pin dual-in-line ic package?
muminat

A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.

The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.

The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.

we are asked to calculate the typical distance between two adjacent pins.

The typical distance between two adjacent pins is calculated as-

                                                                 Typical\ distance =\frac{dimensional\ length}{number\ of\ pins\ in\ each\ row}

                                    =\frac{18 mm}{7}

                                    = 2.5714 mm    [ans]                  

7 0
3 years ago
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