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Liono4ka [1.6K]
3 years ago
6

Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.39~\mathrm{W/m^2}1.39 W/m ​2 ​​ . If a rada

r unit leaks 10.0~\text{W}10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)
Physics
1 answer:
wolverine [178]3 years ago
3 0

Answer:

0.763 m

Explanation:

Intensity I = power P ÷ area A of exposure (spherical area of propagation)

I = P/A

A = P/I

Power = 10.0 W

Intensity = 1.39 W/m^2

A = 10/1.39 = 7.19 m^2

Area A = 4¶r^2

7.19 = 4 x 3.142 x r^2

7.19 = 12.568r^2

r^2 = 7.19/12.568 = 0.57

r = 0.753 m

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Ira Lisetskai [31]

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s = 37.81 m

Explanation:

Given that,

Mass of an apple, m = 1.5 kg

The apple falls for 2.75s before striking the ground.

We need to find how far the apple dropped before hitting the ground. It means we need to find the distance covered. Using the second equation of motion to find it :

s=ut+\dfrac{1}{2}gt^2

u is initial velocity, u = 0 (at rest)

s=\dfrac{1}{2}gt^2\\\\s=\dfrac{1}{2}\times 10\times (2.75)^2\\\\=37.81\ m

So, Apple dropped 37.81 m before hitting the ground.

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3 years ago
9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?
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Option B (1.51 m)

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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

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

Wavelength of the incident wave in air:

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

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

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

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

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

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}

E = 10r, E = 10t

3 0
3 years ago
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