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Basile [38]
2 years ago
9

A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

wer?
Physics
1 answer:
Zanzabum2 years ago
5 0

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

where

\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

P_{radiated} = P_{t}

P_{radiated} = \frac{R}{I^{2}}

Now, the current I is given by:

I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A

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According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car
Leni [432]

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

W=\dfrac{1}{2}m(v^2-u^2)

Also, W = F.d

Fd=\dfrac{1}{2}m(v^2-u^2)

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

6 0
2 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current
sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  U = 0.1355 \ m

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  I_1 =  (27.5i) A

    The force per unit length on each wire is  Z =  295 \mu N/m = 295*10^{-6}  N/m

Generally the force per unit length is mathematically represented as

         Z = \frac{F}{l}  =  \frac{\mu_o I_1I_2}{2\pi y}

=>      \frac{\mu_o I_1I_2}{2\pi y}  =  295

Where  \mu_o is the permeability of free space with a constant value of  \mu_o  =  4\pi *10^{-7} \ N/A2

substituting values

       \frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340}  =  295 *10^{-6}

=>    I_2 =  18.23 \ A

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

   So  

         \frac{\mu_o  I_2  }{2 \pi U} =  \frac{\mu_o  I_1  }{2 \pi(y -  U) }

substituting values

          \frac{  18.23  }{ U} =  \frac{ 27.5 }{(0.34 -  U) }

         6.198 -18.23U = 27.5U

          6.198=45.73U

          U = 0.1355 \ m              

5 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
2 years ago
An earth satellite in an elliptical orbit travels fastest when it is
Serggg [28]
ANY orbiting object travels fastest when it's closest to the central body.
4 0
2 years ago
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wear
Ivahew [28]

Answer:P=14.6 W

Explanation:

According to the Stefan-Boltzmann law for real radiating bodies:

P=\sigma A \epsilon T^{4} (1)

Where:

P is the energy radiated (in Watts)

\sigma=5.67(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A is the Surface area of the body  

T=30\°C + 273.15= 303.15 K is the effective temperature of the body (its surface absolute temperature) in Kelvin

\epsilon=0.6 is the body's emissivity

On the other hand, we are told the human body is roughly approximated to a cylinder of length L=2.0m and circumference C=0.8m.

The circumference of a circle is:C=0.8m=2 \pi r where r is the radius. Hence r=\frac{0.8m}{2 \pi}=0.1273 m.

Now we have to input this value for r  in the Area of a cylinder formula:

A=\pi r^{2}L

A=\pi (0.1273 m)^{2}(2 m)

A=0.0509 m^{2} (2)

Substituting (2) in (1):

P=(5.67(10)^{-8}\frac{W}{m^{2} K^{4}}) (0.0509 m^{2}) (0.6) (303.15 K)^{4} (3)

Finally:

P=14.62 W \approx 14.6 W

7 0
2 years ago
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