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3 years ago

How are mechanical waves and electromagnetic waves alike?

Physics

2 answers:

miss Akunina [59]3 years ago

6 0

I got this answer from the internet

Mila [183]3 years ago

5 0

Answer: Electromagnetic waves can travel through a vacuum; mechanical waves require a medium. Electromagnetic waves are visible; mechanical waves are invisible. Electromagnetic waves always reflect; mechanical waves always refract.

Explanation: There are two types of waves:
Mechanical waves: these waves consist of oscillations of the particles in a medium. Therefore, they can only propagates if there is a medium. Examples of mechanical waves are sound waves
Electromagnetic waves: they consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave. All electromagnetic waves travel in a vacuum always at the same speed, the speed of light: . They are the only types of waves that do not require a medium to propagate, so they can also travel through a vacuum.
We can now analyze the different statements:
Both types of waves require a medium. --> FALSE. Electromagnetic waves do not require a medium.
Both types of waves have a frequency. --> TRUE. All waves are characterized by their frequency, which is the number of complete oscillations per second.
Both types of waves transmit matter. --> FALSE. Waves transmit energy, but not matter.
Both types of waves have a pitch --> FALSE. Pitch tells how we perceive the frequency of a sound wave: but electromagnetic waves are not sound waves, so this statement is false.

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In a material for which o=5.0 S/m and e, =1 the electric field intensity is E- 250 sin 10" (V/m). Find the conduction and displa

Elodia [21]

Answer:

Explanation:

Given that:

The conductivity of the material $\sigma$ = 5.0 s/m

$The \ relative \ permittivity \ of \ the \ material$ ${\varepsilon_r} = 1$ ; &

The electric field intensity of the material E = 250 sin(10¹⁰ t) V/m

(a) The conduction current density $( J_c) = \sigma E$

$= 5.0 \times 250 \ sin ( 10^{10} \ t )$

$\mathbf{ = 1250 \ sin (10 ^{10} t ) \ A/m^2 }$

(b) Displacement current density $(J_d) = \varepsilon _d * \dfrac{\delta E}{\delta t}$

Recall that:

$\varepsilon _o = 8.854 \times 10^{-12}$

∴

$(J_d) = 8.854 \times 10^{-12} \times \dfrac{d}{dt} \times (250 \ sin \ (10^{10} \ t))$

$(J_d) = 8.854 \times 10^{-12} \times 250 \times 10^{10} \times \ cos \ (10^{10} \ t)$

$(J_d) = 22.135 \ cos \ 10^{10} \ t \ A/m^2$

$f = \dfrac{\sigma}{2 \pi \varepsilon_o \varepsilon_r}$

By substitution

$f = 18 \times 10^9 \times \dfrac{\sigma }{\varepsilon_r}$

$f = 18\times 10^9 \times \dfrac{5}{1 }$

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