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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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(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in
bearhunter [10]

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

6 0
2 years ago
A solid cube of edge length r, a solid sphere of radius r, and a solid hemisphere of radius r, all made of the same material, ar
juin [17]

Answer:

Explanation:

The rate at which heat will be radiated is given by the expression

E = e Aσ ( T⁴ - T₀⁴ )

E is heat radiated , e is emissivity , A is area of surface , σ  is stephan's constant T is temperature of the object and T₀ is temperature of the surrounding .

For all the objects given , e , σ T and T₀ are same so E will solely dependent on area of the surface

surface area of cube= 6 r² ,

surface area of sphere = 4 π r²

= 12.56 r²

hemisphere = 2 π r²

= 6.28 r²

12.56 r² >6.28 r² > 6 r²

heat radiated by sphere > heat radiated by hemisphere > heat radiated by cube .

8 0
2 years ago
A gymnast dismounts the uneven parallel bars with some angular momentum about her transverse axis. Just after release, she is in
Luda [366]

Answer:

a. Her moment of inertia increases and she rotates slower.

Explanation:

As we know that initially when she starts her motion she is in piked position due to which her whole mass is concentrated near the axis of rotation

So here the rotational Inertia of her body will be smaller

Now when is comes closer to the position of landing she extends into layout position due to which her mass will move away from the axis of rotation

Due to this the rotational inertia of her body will increase

now we know that there is no external torque on the system

so here angular momentum must be conserved

So we will have

I\omega = constant

so if rotational inertia is increasing then angular speed must be slower

so correct answer will be

a. Her moment of inertia increases and she rotates slower.

7 0
2 years ago
You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g
Ray Of Light [21]

Answer:

\displaystyle t=\frac{2v_o}{g}

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

\displaystyle y=\frac{gt^2}{2}

The horizontal distance is

x=v_ot

We need to find the time when both distances are equal, thus

\displaystyle \frac{gt^2}{2}=v_ot

Simplifying by t

\displaystyle \frac{gt}{2}=v_o

Solving for t

\displaystyle \boxed{t=\frac{2v_o}{g}}

8 0
2 years ago
A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
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