1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion?
Rom4ik [11]

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation F=k\frac{|q_{1}*q_{2}|}{r^{2} }, where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

(20.0 cm x 1m)/100cm = 0.20 m

Using the Coulomb's Law equation:

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}}(1.6x10^-7\frac{C^{2} }{m^{2} } })\\F = 1440N

5 0
3 years ago
The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h f
zmey [24]

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

8 0
3 years ago
State keplers law........​
S_A_V [24]

Answer:

There are actually three, Kepler's laws that is, of planetary motion: 1) every planet's orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its

4 0
3 years ago
Plants use sunlight to fuel photosynthesis (the process they use to produce their food). Which of the following is the best desc
Montano1993 [528]
I think it’s “light energy to chemical energy” let me know if i’m wrong
6 0
3 years ago
(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?
VashaNatasha [74]

Answer:

\boxed {3.43 x 10^{3}}

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as v=\frac {d}{t} where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

d=vt

Substituting 343 m/s for d and 10 s for t then

d= 343\times10= 3430= 3.43 x 10^{3}

Therefore, the distance between speaker and deter is \boxed {3.43 x 10^{3}}

6 0
3 years ago
Other questions:
  • A hummingbird can flutter its wings 4,800 times per minute.
    6·2 answers
  • What is the universe's estimated age
    14·1 answer
  • An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:
    8·1 answer
  • A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal acceleration if the circle has a radius of 27 m? A. 3.3
    9·1 answer
  • Which image shows both potential and kinetic energy
    5·2 answers
  • How to do gizmos for science
    15·1 answer
  • Why does an archaeologists use master tree ring patterns?
    15·1 answer
  • Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10
    11·1 answer
  • What do you know about velocity, acceleration and reference frame?
    15·1 answer
  • In every trench over 4 feet (1.2 m) deep, there must be an exit every​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!