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Korolek [52]
4 years ago
6

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

0 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional). Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
Physics
1 answer:
Rus_ich [418]4 years ago
7 0

Answer:

6.99535\times 10^{-6}\ V/m

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity of Electric radiation

I=\frac{P}{A}\\\Rightarrow I=\frac{P}{4\pi r^2}\\\Rightarrow I=\frac{1000}{4\pi\times 35000000^2}\ W/m^2

Intensity of Electric radiation

I=\frac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\frac{2\times \frac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m

The amplitude of the electric field vector is 6.99535\times 10^{-6}\ V/m

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2 years ago
By comparing fossils of a species from long ago to those of today, we can conclude that *
Ratling [72]

Answer:

today's species evolved from those of the past.

Explanation:

By comparing fossils of a specie from long ago to those of today, we can conclude that today's species evolved from those of the past.

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4 0
3 years ago
Ship A is located 4.2 km north and 2.7 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a vel
kotykmax [81]

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, \vec{u_{A}} = 22 km/h towards South = - 22\hat{j}

Velocity of ship B, \vec{u_{B}} = 39 km/h Towards North east at an angle of 36^{\circ} = \vec{u_{B}} = 39sin36^{\circ} \hat{j}

Now, the velocity of ship A relative to ship B:

\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}

\vec{u_{A}} = - 22\hat{j}

\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

Now,

\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}

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3 years ago
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3 years ago
Read 2 more answers
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

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3 years ago
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