A plane monochromatic electromagnetic wave with wavelength ? = 3 cm, propagates through a vacuum. Its magnetic field is described byB? =(Bxi^+Byj^)cos(kz+?t)

where Bx = 3.3 X 10-6 T, By = 3.9 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.

1)

What is f, the frequency of this wave?

GHz

2)

What is I, the intensity of this wave?

W/m2

3)

What is Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0?

W/m2

4)

What is Ex, the x-component of the electric field at (x = 0, y = 0, z = 0) at t = 0?

V/m

5)

Compare the sign and magnitude of Sz, the z-component of the Poynting vector at (x=y=z=t=0) of the wave described above to the sign and magnitude of SIIz, the z-component of the Poynting vector at (x=y=z=t=0) of another plane monochromatic electromagnetic wave propagating through vaccum described by:

B? =(BIIxi^?BIIyj^)cos(kz??t)

where BIIx = 3.9 X 10-6 T, BIIy = 3.3 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.

SIIz < 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)

SIIz < 0 and magnitude(SIIz) = magnitude(Sz)

SIIz > 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)

SIIz > 0 and magnitude(SIIz) = magnitude(

Answers

Answer 1

Answer:

Final answer:

The question involves computation of frequency, intensity, Poynting vector and electric field of an electromagnetic wave, and comparison between two such waves. The solutions result in approximately: 10 GHz for frequency, 3.07 x 10^-12 W/m^2 for intensity, 1.3 X 10^-19 W/m^2 for the z-component of Poynting vector, and 1.43 V/m for the electric field. Moreover, the comparison yields that SIIz is less than zero and not equal to Sz in magnitude.

Explanation:

The subject of your question relates to

electromagnetic waves

and their properties such as frequency, intensity, Poynting vector, and the electric field component. These concepts belong to the realm of physics, and more specifically, are topics in the study of electromagnetic theory.

To solve your questions:

The frequency f can be found using the formula: f = c/λ where c is the speed of light in vacuum (~3x10^8 m/s). For λ = 3 cm or 0.03 m, computation yields f ≈ 10^10 Hz or 10 GHz.
The intensity I of an electromagnetic wave in a vacuum can be given by the equation I = 0.5*c*ε0*E^2, where E is the electric field amplitude. To compute I, first, we need to find E which is given by E = c*B, where B is the magnetic field amplitude. Here, B is the square root of (Bx^2 + By^2) resulting in approximately 4.77*10^-6 T. Thus, E ≈ 1.43 V/m and solving for I gives us I ≈ 3.07 x 10^-12 W/m^2.
The z-component of the Poynting vector Sz at a specified point and time is given by Sz = E x H, where H = B/μ0, μ0 represents the permeability of free space. At t = 0, Sz = Ex*Hy - Ey*Hx = Ex*By, resulting in Sz ≈ 1.3 x 10^-19 W/m^2.
The x-component of the electric field at t = 0 Ex≈1.43 V/m.
Finally, comparing Sz of both waves (magnitudes and signs), we find that SIIz < 0 and the magnitude of SIIz does not equal the magnitude of Sz.

Learn more about Electromagnetic Waves here:

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Answer 2

Answer:

Final answer:

The frequency of the wave is 10 GHz. While we can't expressly calculate the intesity, Sz, and Ex without more information, we can note that if the signs of Bx and By are swapped in a new wave, the Poynting vector would be flipped, hence SIIz would be negative and of equal magnitude to Sz.

Explanation:

An electromagnetic wave propagating through vacuum is described by certain electromagnetic fields which are associated with frequency, intensity, and Poynting vector which indicates the direction of energy flow. These can be calculated using certain formulas derived from wave equations.

Frequency can be acquired from the wavelength (λ) with the formula: f = c/λ, where c is the speed of light in vacuum. Using given λ = 3 cm, we get f = 10^10 Hz or 10 GHz.

The total Intensity (I) can be calculated as the average of the sum of the intensities in the x and y direction, given by: 1/2 ε_0 c E^2, where ε_0 is the permittivity of free-space and E is the electric field amplitude. However, more information might be needed to calculate this value. Similarly, without further information, we cannot calculate the exact values of Sz and Ex.

When comparing Sz and SIIz, if the signs of Bx and By are swapped in a new wave, this would flip the direction of the Poynting vector (since it is related to E × B), hence SIIz < 0 and its magnitude would still equal to Sz because the magnitudes of Bx and By do not change.

Learn more about Electromagnetic wave properties here:

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Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by

Answers

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^(\circ)\,\hat{i}+\sin 30^(\circ)\,\hat{j})+4\cdot (-\sin 30^(\circ)\,\hat{i}-\cos 30^(\circ)\,\hat{j})$

$\vec R = (6\cdot \cos 30^(\circ)-4\cdot \sin 30^(\circ))\,\hat{i}+(6\cdot \sin 30^(\circ)-4\cdot \cos 30^(\circ))\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Why is there so much more carbon dioxide in the atmosphere of Venus than in that of Earth? Why so much more carbon dioxide than on Mars?

Answers

Explanation:

The reason for the more concentration of carbon dioxide in the atmosphere of Venus than in the Earth -

On the Earth , most amount of the carbon dioxide is in the ocean water and in sea sediments .

Considering Venus , in the planet Venus , there is no Ocean water , hence , carbon dioxide can not get dissolved into the water , hence , it is found in the atmosphere .

So , the escape velocity for carbon dioxide on Mars is smaller than Venus .

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the