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nataly862011 [7]

3 years ago

8

Calculate the wavelength of the electromagentic waves with the given frequencies, and determine the type of electromagnetic radi

ation for each combination. frequency: 4.38×1014 Hz wavelength : m type frequency: 4.14×1020 Hz wavelength : m type frequency: 3.24×1012 Hz wavelength : m type

1 answer:

ololo11 [35]3 years ago

5 0

To develop this problem we require the concepts related to wavelength and its expression to calculate it.

The wavelength is given by

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

Where,

$c=3*10^8 m/s$ light velocity

f = frequency.

Our values are given by,

$f_1=4.38*10^{14}Hz$

$f_2= 4.14*10^{20}Hz$

$f_3 = 3.24*10^{12}Hz$

Then,

$\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m$ Visible

$\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m$ Gamma Ray

$\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m$ Infrared

<em>*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.</em>

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where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
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So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

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2 years ago

Compare blue and red light from the visible spectrum. which has: the longer wavelength? the greater frequency? the greater energ

Hitman42 [59]

1) By looking at the table of the visible spectrum, we see that blue light has a wavelength in the range [450-490 nm], while red light has wavelength in the range [620-750 nm]. Therefore, red light has longer wavelength than blue light.

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$f= \frac{c}{\lambda}$
where c is the speed of light. We see that the frequency is inversely proportional to the wavelength, so the shorter the wavelength, the greater the frequency. In this case, blue light has shorter wavelength than red light, so blue light has greater frequency than red light.

3) The energy of the photons of an electromagnetic wave is given by
$E=hf$
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2 years ago

The wavelength of visible light is about 5500 times longer than the wavelength of x-rays. If you represent the wavelenth of an x

Pie

Answer:

458.33 ft

Explanation:

We are given that

Wavelength of of an x- ray photon=1 in

Wavelength of of an x- ray photon== $\frac{1}{12}$ ft

1 in= $\frac{1}{12}$ feet

We have to find the length of line (in feet) drawn by you to represent the wavelength of visible light.

According to question

Wavelength of visible light= $5500\times \frac{1}{12}$ ft

Wavelength of visible light $=\frac{5500}{12}$ ft

Wavelength of visible light=458.33 ft

Hence, the 458.33 ft line must drawn by you to represent the wavelength of visible light.

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3 years ago

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Answer:

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<u>Answer</u>

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5 0

3 years ago

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