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nataly862011 [7]
2 years ago
15

A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 a. Determine the time average power rad

iated by the dipole if the oscillating frequency is 150 mhz
Physics
2 answers:
Mnenie [13.5K]2 years ago
8 0

For A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 , the time average power radiated  is mathematically given as

P=0.1577w

<h3>What is the time average power radiated by the dipole if the oscillating frequency is 150 mhz?</h3>

Generally, the equation for the   is mathematically given as

\lambda =\frac{c}{f}

Therefore

\lambda=\frac{3\times 10^{8}}{10^{6}}

lambda=300m

In conclusion, for the power

P=40\pi^{2}(I_{0})^{2}(\frac{l}{\lambda})^{2}\\\\P=40* (3.14)^{2}\times6^{2} (\frac{1}{300})^{2}

P=0.1577w

Read more about Power

brainly.com/question/10203153

USPshnik [31]2 years ago
3 0

The time-averaged power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.

<h3>What is wavelength?</h3>

The distance between identical points (adjacent crests) in adjacent cycles of a waveform signal carried in space or down a wire is defined as the wavelength.

Given that the current is 2 amp, while the frequency is 150 MHz. Therefore, the wavelength can be written as,

\lambda = \dfrac cf = \dfrac{3 \times 10^8}{10^6} = 300\rm\ m

Now, the power can be written as,

P=40 \pi^2 \times (I_o)^2 \times (\dfrac{l}{\lambda})^2\\\\P = 40 \times \pi ^2 \times 6^2 \times (\dfrac{1}{300})^2\\\\P = 0.1577\rm\ w

Hence,  the time-average power radiated by the dipole if the oscillating frequency is 150 MHz is 0.1577 w.

Learn more about Wavelength:

brainly.com/question/13533093

#SPJ4

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A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
2 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
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5 0
3 years ago
An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
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Answer:

  about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

  y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

  t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

  = (22.8 ± √823.64)/(-9.8)

The positive solution is ...

  t ≈ 0.60195193

It takes about 602 milliseconds for the arrow to reach the ground.

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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen
Marta_Voda [28]

Answer:

a) that laser 1 has the first interference closer to the central maximum

c) Δy = 0.64 m

Explanation:

The interference phenomenon is described by the expression

         d sin θ = m λ

Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference

For the separation of the lines we use trigonometry

        tan θ = sin θ / cos θ = y / x

In interference experiments the angle is very small

          tan θ = sin θ = y / x

         d y / x = m λ

a) and b) We apply the equation to the first laser

          λ = d / 20

          d y / x = m d / 20

          y = m x / 20

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The second laser

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          d y / x = m d / 15

          y = m x / 15

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We can see that laser 1 has the first interference closer to the central maximum

c) laser 1

They ask us for the second maximum m = 2

            y₂ = 2 4.8 / 20

            y₂ = 0.48 m

For laser 2 they ask us for the third minimum m = 3

In this case to have a minimum we must add half wavelength

         y₃ = (m + ½) x / 15

         m = 3

         y₃ = (3 + ½) 4.8 / 15

         y₃ = 1.12 m

        Δy = 1.12 - 0.48

        Δy = 0.64 m

4 0
2 years ago
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