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worty [1.4K]
3 years ago
12

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.36 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the intensity of the radio signal at that point?
a. 240 µW/m^2
b. 340 µW/m^2
c. 86 µW/m^2
d. 170 µW/m^2
e. 120 µW/m^2
Physics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

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VashaNatasha [74]
Work = force × distance
= 35 N × 200 m
= 7000 J
4 0
3 years ago
A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle
dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

\phi = BAcos\theta where:

\phi is the magnetic flux

B is the magnetic field

A is the cross sectional area

\theta is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

\phi = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0
2 years ago
Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
andre [41]
<h3>Answer:</h3>

1.5 m/s²

<h3>Explanation:</h3>

We are given;

Force as 60 N

Mass of the Cart as 40 kg

We are required to calculate the acceleration of the cart.

  • From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
  • That is, F = ma , where m is the mass and a is the acceleration

Rearranging the formula we can calculate acceleration, a

a = F ÷ m

  = 60 N ÷ 40 kg

  = 1.5 m/s²

Therefore, the acceleration of the cart is 1.5 m/s²

3 0
3 years ago
g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
sdas [7]

Answer:

The vertical distance is  d = \frac{2}{k} *[mg + f]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    E  =  P +  W_f

Here E the the energy of the spring which is increasing and this is mathematically represented as

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

        P is the potential energy of the cylinder which is mathematically represented as

     P  = mgd

And

     W_f  is the workdone by friction which is mathematically represented as

      W_f  =  f *  d

So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

=>    \frac{1}{2} * k  *  d^2 =  d[mg +  f    ]

=>  \frac{1}{2} * k  *  d =  [mg +  f    ]

=> d = \frac{2}{k} *[mg + f]

5 0
3 years ago
A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a
Aleks04 [339]

Answer:

E= -3.166 cosωt   V

Explanation:

Given that

I = Imax sinωt

L= 8.4 m H

Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π  rad/s

We know that self induce E given as

E=-L\dfrac{dI}{dt}

\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t

E=-L\times Imax \ \omega\ cos\omega t

E=-8.4\times 120\times \pi \ cos\omega t

E= -3166.72 cosωt  m V

E= -3.166 cosωt   V

This is the induce emf.

3 0
3 years ago
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