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jasenka [17]
3 years ago
7

Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the

water inside the kettle heats up and begins to boil. What is happening? A) Light energy is transforming into heat. Eliminate B) Heat energy is transforming into electricity. C) Electrical energy is transforming into heat energy. D) Electrical energy is transforming into light and heat energy.
Physics
2 answers:
Elodia [21]3 years ago
5 0

Answer: The correct answer is option C.

Explanation:

Electrical energy is an energy possessed by the electrical charges in motion.

Heat energy is an energy due to the motion of particles like atoms, ions etc in solid , liquid and gases .

In an electric kettle:

Electric energy → Heat energy   (Law of conservation Energy)

In an electric kettle electric energy is changing into heat energy which results in heating up of water present in side it.The electric energy from the power source is transforming into heat energy which gets transferred to water molecules.

Hence, the correct answer is option C.

Setler [38]3 years ago
4 0
C) electrical energy is transformed into heat energy
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Answer:

Wavelength of the incident wave in air = 1 m

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Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

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\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

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n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

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The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

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n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

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