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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

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

Please help, and show steps. Thank you very much!

Vikentia [17]

V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)

6 0

3 years ago

6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

$v_{f}^{2}=v_{i}^{2} +2*a*x$

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

$v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]$

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3 years ago

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NikAS [45]

Answer:

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Explanation:

1. Find the equation of eht maximal friction force:

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2. Replace values in the equation to find the answer:

$Fr=0.8*1600kg*9.8\frac{m}{s^{2}}$

$Fr=12544N$

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