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3 years ago

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A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction.Part AIf at a particular instant and at a certa

in point in space the electric field is in the +x-direction and has a magnitude of 4.90 V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?Part B
What is the direction of the magnetic field?

1 answer:

stira [4]3 years ago

8 0

Explanation:

Given that,

Magnitude of electric field in +x direction, E = 4.9 V/m

Part A :

The relation between electric and magnetic field is given by :

$B=\dfrac{E}{c}$

c is the speed of light

$B=\dfrac{4.9\ V/m}{3\times 10^8\ m/s}$

$B=1.63\times 10^{-8}\ T$

So, the magnetic field of the wave at this same point in space and instant in time is $1.63\times 10^{-8}\ T$ .\

Part B :

The direction of electric and magnetic field is same i.e. +x direction.

Hence, this is the required solution.

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Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

Rina8888 [55]

<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>

8 0

3 years ago

Read 2 more answers

Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0

2 years ago

Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?

Lina20 [59]

To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²= (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps

3 0

3 years ago