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3 years ago

In which type of wave do air particles move together or apart parallel to the direction of the wave?

1 answer:

8 0

The best and most correct answer among the choices provided by the question is B.sound waves.


Particles move together or apart parallel to the direction of the sound wave.

Hope my answer would be a great help for you.
If you have more questions feel free to ask here at Brainly.

You might be interested in

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

Zarrin [17]

first object gors highest

second object goes lowest.

5 0

2 years ago

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

$F_{C} = F_{G}$

$\frac{mv^{2}}{R} = mg$

where

v = velocity

g = acceleration due to gravity

$v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s$

4 0

2 years ago

Why are there two fluoride ions in magnesium fluoride but only one fluoride ion in lithium fluoride

postnew [5]

This would be more of a chemistry question. Remember magnesium has a charge of 2+, and would need to hand off its two extra electrons. Fluorine can only take one electron at a time, so there needs to be two fluorines to take one magnesium's 2 electrons.

With lithium, it has a +1 charge, so it has one extra electron, which it can hand off to just 1 fluorine atom.

Another way of looking at this is: $Mg^{2+}$ + 2 $F^{-}$ = MgF2 (the charges must balance out to zero)

$Li^{+}$ + $F^{-}$ = LiF (the charges balance out to zero)

5 0

3 years ago

Read 2 more answers

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour

Gemiola [76]

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

Average speed = 14 blocks/hour

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity. I'm so sure of this that I'm gonna give you the solution for that too. If there's no more question, then you won't need this, and you can just discard it. I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

Displacement = 4 blocks west

Average velocity = (4 blocks west) / (1 hour)

Average velocity = 4 blocks/hour West

4 0

2 years ago

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