All categories

3 years ago

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,

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

$\lambda_m$ = $\frac{1}{3}$ × 10⁻⁶

Also,

$m\lambda_m = 2L$

Therefore,

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

m = 2 × 10⁶

You might be interested in

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0

3 years ago

What is the mass of a 735-N weight on Earth?

Allisa [31]

W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.

735 N = m* 9.8

735/9.8 = m

75 = m

Mass , m = 75 kg. B.

3 0

3 years ago

Breaking a pencil in half is an example of: options:

Licemer1 [7]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

$\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0$

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that $f_{max}=\mu N$

Now, we find the acceleration:

$\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}$

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

$x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s$

8 0

3 years ago