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OverLord2011 [107]
4 years ago
15

Consider six proposed properties of electromagnetic radiation: wave speeds of 3.00×108 km/s3.00×108 km/s and 3.00×108m/s3.00×108

m/s , wavelengths of 673 nm673 nm and 0.221 nm0.221 nm , and frequencies of 2.39×1018 Hz2.39×1018 Hz and 4.85×1014 Hz4.85×1014 Hz . Place these according to whether they apply only to the X-ray band, only to the visible light band, to both bands, or to neither band. X-ray band only Visible light band only Both bands Neither band wavelength of 0.221 nm0.221 nm frequency of 2.39×1018 Hz2.39×1018 Hz speed of 3.00×108 m/s3.00×108 m/s wavelength of 673 nm673 nm frequency of 4.85×1014 Hz4.85×1014 Hz speed of 3.00×108 km/s3.00×108 km/s Answer Bank
Physics
1 answer:
-Dominant- [34]4 years ago
7 0

Answer:

1) Speed ​​of light (c)       visible, X-rays

2) λ = 0.221 nm                         X-rays

3) λ = 673 nm               Visible

4) .f = 2.39 10¹⁸ Hz                   X-rays

5) .f = 4.85 10¹⁴ Hz        Visible

Explanation:

Electromagnetic radiation has several characteristics

The speed of the wave regardless of its wavelength or frequency is

        c = 3 10⁸ m / s²

There is a relationship with speed

       c = λ f

The speed c is for all waves (x-rays, visible)

Visible light has a wavelength between 400 nm to 700 nm,

X-ray light corresponds to wavelengths less than 10 nm

Light gamma rays for wavelength less than 0.1 nm

Let's look for the frequencies for the two ranges

Visible

      f = c /λ

       f = 3 10⁸/400 10⁻⁹ = 7.5 10¹⁴ Hz

       f = 3 10⁸ (/ 700 10⁻⁹ = 4.29 10¹⁴ Hz

Visible range of the order of 10¹⁴ Hz

X-rays

       f = 3 10⁸/10 10⁻⁹ = 3 10¹⁶ Hz

       f = 3 10⁸ / 0.1 10⁻⁹ = 3 10¹⁸ Hz

 

X-ray higher frequency 10¹⁶ Hz

1) Speed ​​of light (c)       visible, X-rays

2) λ = 0.221 nm                         X-rays

3) λ = 673 nm               Visible

4) .f = 2.39 10¹⁸ Hz                   X-rays

5) .f = 4.85 10¹⁴ Hz        Visible

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alekssr [168]

If a 3 Ω and a 1.5 Ω resistor are wired in parallel and the combination is wired in series to a 4 Ω and a 10 V emf device then the current in the 3 Q resistor is 0.667 Ampere

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material.

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

As given in the problem 3 Ω and a 1.5 Ω resistor are wired in parallel

Their equivalent resistance would be

1/Re= 1/R1 + 1/R2

1/Re = 1/3 +1/1.5

Re= 1 Ω

Now this parallel equivalent is connected with a series combination with a 4Ω resistor

For calculating equivalent resistance in series combination.

R = R1  + R2

R = 1 + 4

R = 5 Ω

As the 10 V emf device is connected then the current flowing can be calculated by using Ohms law

V=IR

10 = 5×I

I = 2 ampere

as the voltage is divided between parallel resistor combination and series resistor

The voltage drop across parallel combination = 10×(1/1+4) volt

                                                                            = 2 volt

Now the current flows through a 3 Ω resistor by using Ohms law

I = V/R

I= 2/3

I = 0.667 Ampere

Thus, the current flowing through the 3 Ω resistor comes out to be 0.667 Ampere

Learn more about resistance from here

brainly.com/question/14547003

#SPJ1

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