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Dmitry_Shevchenko [17]
3 years ago
5

Which of this following have the minimum frequency in their sound?

Physics
1 answer:
Verdich [7]3 years ago
5 0

Answer:A MAN

Explanation:

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Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 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.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

4 0
4 years ago
The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information  

number of half-lives past is

=  15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

4 0
3 years ago
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months
o-na [289]
Mem me e m even have. Jags. Shah. Shiv side esicjm is n meh dish so do indbbd
4 0
2 years ago
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is trans- formed i
ra1l [238]

Answer:

a) v = 4.4 m/s

b) F = 400 N

Explanation:

a) ½kx² = ½mv²

v = √(kx²/m)

F = kx

v = √(Fx/m)

v = √(800(0.012) / 0.5) = √19.2 = 4.3817...

b) Fd = ½mv²

F = mv²/2d

F = 0.5(19.2) / (2(0.012) = 400 N

3 0
3 years ago
A cyclist moves at a constant speed of 5 m/s if the cyclist does not accelerate during the next 20 seconds he will travel at?
Verizon [17]
5 m/s because the speed is constant 
8 0
3 years ago
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