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WARRIOR [948]
3 years ago
8

What do radio waves and microwaves have in common?

Physics
2 answers:
Delicious77 [7]3 years ago
6 0

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡⊕

Answer:

<em>C.) Both are located to the right of visible light on the spectrum</em>

     ↓↓↓

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They travel with the speed of light.

Both radio and microwaves are electromagnetic waves which are adjacent to each other in electromagnetic spectrum towards lower frequency and higher wavelength.

The relationship between wavelength and frequency of the wave follows the equation:

where,

v = frequency of the wave

c = speed of light  

= wavelength of the wave               ↑↑↑

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡⊕

Hope I helped!!! ∞

<em>~Nuha</em>

   <3

o-na [289]3 years ago
4 0

Answer: C) Both are located to the right of visible light on the spectrum

Explanation:

BECAUSE....

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They travel with the speed of light.

Both radio and microwaves are electromagnetic waves which are adjacent to each other in electromagnetic spectrum towards lower frequency and higher wavelength.

The relationship between wavelength and frequency of the wave follows the equation:

\nu=\frac{c}{\lambda}

where,

v = frequency of the wave

c = speed of light  

{\lambda}= wavelength of the wave

<em>* Hopefully this helps: ) Mark me the brainliest:)!!</em>

<em>~234483279c20~</em>

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<u>Kinematics:-</u>

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The images show two different types of Galapagos tortoises that scientists believe descended from the same species. The first ty
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2 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

4 0
3 years ago
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