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3 years ago

Please help me with my question

Physics

1 answer:

Mashutka [201]3 years ago

3 0

Answer:

Brownian motion and diffusion

Explanation:

Brownian motion: taking into consideration, SMOKE. when a burning fire is kindled, smoke is seen to move in the way or direction that is not straight but keep changing continuously ( in haphazard manner). Haphazard manner of smoke is an evidence that matter are made of molecules and are constantly in motion

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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

Which of the following is a chemical change?

klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
a lot of new compounds when it burns.

3 0

3 years ago

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

Ahat [919]

<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = ?

Time, t = 6.8 s

Displacement, s = 1/4 mi = 400 meters

Substituting

s = ut + 0.5 at²

400 = 0 x 6.8 + 0.5 x a x 6.8²

a = 17.30 m/s²

Now we have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = ?

Time, t = 6.8 s

Acceleration, a = 17.30 m/s²

Substituting

v = u + at

v = 0 + 17.30 x 6.8

v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0

3 years ago

A football punter accelerates a .55 kg football

Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the football

$\Delta t=0.25 s$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.55 kg is the mass of the football

u = 0 is the initial velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we find the force exerted on the ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N$

5 0

4 years ago

The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3

nika2105 [10]

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

$F=\dfrac{Gm_em_m}{r^2}$