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attashe74 [19]
4 years ago
15

During a test, a NATO surveillance radar system, operating at 12 GHz at 190 kW of power, attempts to detect an incoming stealth

aircraft at 80 km. Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity (in ?W/m2) of the beam when the beam reaches the aircraft's location? The aircraft reflects radar waves as though it has a cross-sectional area of only 0.52 m2. (b) What is the power (in mW) of the aircraft's reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value (in ?T) of the magnetic field of the reflected radar beam?
Physics
1 answer:
ollegr [7]4 years ago
5 0

Answer:

4.72491\ \mu W/m^2

2\times 10^{-3}\ mW

6.10993\times 10^{-13}\ W/m^2

2.1454\times 10^{-5}\ N/C

7.15133\times 10^{-14}\ T

Explanation:

A = Area of hemispher = 2\pi r^2

r = Distance

P = Power

I = Intensity

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

Intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{190\times 10^3}{2\pi 80000^2}\\\Rightarrow I=4.72491\times 10^{-6}\ W/m^2=4.72491\ \mu W/m^2

The intensity is 4.72491\ \mu W/m^2

Power is given by

P=IA\\\Rightarrow P=4.72491\times 10^{-6}\times 0.52\\\Rightarrow P=2.45695\times 10^{-6}=2\times 10^{-3}\ mW

The power is 2\times 10^{-3}\ mW

I=\frac{P}{A}\\\Rightarrow I=\frac{2.45695\times 10^{-6}}{2\pi 80000^2}\\\Rightarrow I=6.10993\times 10^{-13}\ W/m^2

The intensity is 6.10993\times 10^{-13}\ W/m^2

Maximum electric field is given by

E_0=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\frac{2\times 6.10993\times 10^{-13}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=2.1454\times 10^{-5}\ N/C

Maximum value of electric field is 2.1454\times 10^{-5}\ N/C

Magnetic field is given by

B=\frac{E_0}{c}\\\Rightarrow B=\frac{2.1454\times 10^{-5}}{3\times 10^8}\\\Rightarrow B=7.15133\times 10^{-14}\ T

The rms value of magnetic field is 7.15133\times 10^{-14}\ T

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sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
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Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

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      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

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for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

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3 years ago
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