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2 years ago

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

2 answers:

5 0

I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

   Frequency = (speed) / (wavelength)

   = (6 m/s) / (12 m)

   = 0.5 / sec

   =    0.5 Hz .

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)

Anna [14]2 years ago

4 0

Frequency(ν) = speed of infrared light( c )/wavelength( λ ) = 6/12 = 1/2 = 0.5 s^-1 .

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2 years ago

A ball rolls off a table with a horizontal velocity of 3 m/s. If it takes 0.3 seconds for the ball to reach the floor, how high

LekaFEV [45]

The height of the table above the ground is 0.45 m.

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Horizontal velocity (u) = 3 m/s
Time (t) = 0.3 s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?

<h3>How to determine the height </h3>

The height of the table can be obtained by using the following formula:

h = ½gt²

h = ½ × 10 × 0.3²

h = 5 × 0.09

h = 0.45 m

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6 0

2 years ago

HELP ;

mestny [16]

Answer:

The frequency would double.

Explanation:

Given:

Speed of wave (v) = constant.

Frequency of wave initially (f₁) = 2 Hz

Initial wavelength of the wave (λ₁) = 1 m

Final wavelength of the wave (λ₂) = 0.5 m

Final frequency of the wave (f₂) = ?

We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.

Therefore, framing in equation form, we have:

Wavelength × Frequency = Speed

$\lambda\times f=v$

It is given that speed of the wave remains the same. So, the product must always be a constant.

Therefore,

$\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2$

Now, plug in the given values and solve for 'f₂'. This gives,

$1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz$

Therefore, the final frequency is 4 Hz which is double of the initial frequency.

f₂ = 2f₁ = 2 × 2 = 4 Hz

So, the second option is correct.

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3 years ago

Read 2 more answers

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Answer:

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2 years ago