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Zielflug [23.3K]
3 years ago
13

A gas laser has a cavity length of 1/3 m and a single oscillation frequency of 9.0 x 1014 Hz. What is the cavity mode number?

Physics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer:

2 × 10⁶

Explanation:

Data provided in the question:

Cavity length, L = \frac{1}{3}m

Oscillation frequency, f_m = 9.0 × 10¹⁴ Hz

Now,

we know,

f_m=\frac{c}{\lambda_m}

here,

c is the speed of light = 3 × 10⁸ m/s

\lambda_m = Wavelength of mode m inside the laser cavity

m is the cavity mode number

Thus,

9.0\times10^{14}=\frac{3\times10^8}{\lambda_m}

or

\lambda_m = \frac{1}{3} × 10⁻⁶

Also,

m\lambda_m = 2L

Therefore,

m × \frac{1}{3} × 10⁻⁶ = 2 × \frac{1}{3}

or

m = 2 × 10⁶

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Answer:

C) amplitude

Explanation:

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Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is
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Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

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2 years ago
A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

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Initially, it is given by

C=\frac{\epsilon_0 A}{d}

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A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

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We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

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- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

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4 0
3 years ago
Suppose the three resistors in the picture represents a 240 W
Airida [17]

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2 years ago
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Can I get help and an explanation on C?
Diano4ka-milaya [45]

Answer:

1.67 m/s

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