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3 years ago

Why does laser beam might appear blue and another laser beam might appear red?

2 answers:

6 0

A laser emits light at one single frequency (one single wavelength).
But not all lasers emit the same one.

The frequency (wavelength) that comes out of a laser depends on the
material that's doing the 'lasing' ... different material fabricated into a
laser diode ==> different wavelength in the output.

Many lasers, but not all, emit a frequency (wavelength) that's in the visible
range, and a single given laser may emit red, orange, yellow, green, blue,
or violet light. Lasers of other designs emit at frequencies outside the visible
range, in the microwave, infrared, or ultraviolet.

Alenkasestr [34]3 years ago

4 0

The red laser will have a hotter temperature than the blue, so it is red because it is hot
<span>hope this helps!!!</span>

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UkoKoshka [18]

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7 0

3 years ago

With the aid of a string, a gyroscope is accelerated from rest to 16 rad/s in 0.40 s. what is its angular acceleration in rad/s2

vovikov84 [41]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (ending speed) minus (starting speed)

Change in angular speed = (16 rad/s) - (zero) = 16 rad/s .

Angular acceleration = (16 rad/s) / (0.4 s)

(Average) angular acceleration = 40 rad/s²

4 0

2 years ago

What is the electric field at the point x=6.5 m? positive e-fields point to the right.?

Mademuasel [1]

6,5 6,4 6,3 6,2 6
if you know,1

5 0

2 years ago

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

4 0

3 years ago

Read 2 more answers

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago