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attashe74 [19]
3 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]3 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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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
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<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

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