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
zimovet [89]
3 years ago
10

In unguided medium (free space), the electromagnetic (EM) signal wave spreads as it leaves the transmit antenna. Since the power

of the EM signal resides in the surface area of the wave front, signal power is described as signal power density (i.e., watts per area).
True

False

The spreading of an electromagnetic (EM) signal in an unguided medium such as air, weakens the EM signal therefore causing signal attenuation.

True

False
Engineering
1 answer:
natka813 [3]3 years ago
4 0
The real answer - for 1- true
you have to pay attention in class to actually learn this subject if you want to work in this workforce


second - false
there is no way that the Elecrtomagnetic signal can just cause awful signal attenuation.
You might be interested in
A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

NTU = 0.661

          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0
3 years ago
The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.
emmasim [6.3K]

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature =  130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

<u>Applying Amontons's Law as:</u>

\frac {P_1}{T_1}=\frac {P_2}{T_2}

So,

P_2=\frac {P_1}{T_1} \times {T_2}

P_2=\frac {1 atm}{373.15 K} \times {403.15 K}

P_2=1.0804 atm

<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>

6 0
3 years ago
A motor vehicle engine operating with a diesel engine takes in atmospherics air at a temperature of 30°C and pressure of 1 bar.
nydimaria [60]

Answer:

η=0.568

Explanation:

At inlet condition

temperature = 30 C and pressure P=1 bar

The maximum temperature = 13456 C

Compression ratio r= 12

We know that for process 1-2

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{303}=12^{1.4 -1}

T_2=818.68 K

Now for process 2-3

\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}

\dfrac{273+1345}{818.86}=\dfrac{V_3}{V_2}

\dfrac{V_3}{V_2}=1.97

So the cut off ratio ρ=1.97

Efficiency of diesel engine

\eta =1-\dfrac{\rho ^{\gamma}-1}{r^{\gamma -1}\gamma \left (\rho -1\right )}

Now put the values

\eta =1-\dfrac{1.97 ^{1.4}-1}{12^{1.4-1}\times 1.4\times \left (1.97 -1\right )}

   ⇒η=0.568

So the efficiency is 56.8%.

4 0
3 years ago
What ratio between differential gain and common-mode gain is called​
dlinn [17]

Answer:

CMRR(Common Mode Rejection Ratio) is the ratio of differential gain and the common mode gain.

Explanation:

8 0
3 years ago
An engineering student claims that a country road can be safely negotiated at 65 mi/h in rainy weather. Because of the winding n
igomit [66]

Answer:

Negotiated speed should be lower. Perception/reaction time is too less than design values.

Explanation:

Given:

- The claimed safe speed V_1 = 65 mi/h

- Sight distance D = 510 ft

- The practical deceleration a = 11.2 ft/s   ... according to standards

Find:

Assuming practical stopping distance, comment on the student whether the claim is correct or not

Solution:

- Calculate the practical stopping distance:

                      d = V_1^2 / 2*a

                      d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft

- Solve for reaction distance d_r is as follows:

                     d_r = D - d = 510 - 402.054 = 107.945 ft

- The perception/time reaction is:

                   t_r = d_r/V_1 = 107.945 / 94.9

                  t_r = 1.17 sec

Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.

Hence, the safe speed should be lower.

4 0
3 years ago
Other questions:
  • Saturated steam coming off the turbine of a steam power plant at 30 °C condenses on the outside of a 3-cm-outer-diameter, 35-m-l
    15·1 answer
  • Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra
    10·1 answer
  • If a worksite includes more than one set of management and workers, who should have access to the information, training, and con
    12·2 answers
  • 2.18 The net potential energy between two adjacent ions, EN, may be represented by the following equation: (1) Calculate the bon
    5·1 answer
  • What is arduino and for what it is used​
    12·1 answer
  • How can you use the IPDE Process to help protect motorcyclists while driving?
    6·1 answer
  • A detailed image of a brain scan with height, width, and depth is an example of a(n) 3D _________ model.
    15·1 answer
  • One reason the shuttle turns on its back after liftoff is to give the pilot a view of the horizon. Why might this be useful?
    6·2 answers
  • Need help, I will give cake :))<br><br> + branliest
    14·2 answers
  • A bridge a mass of 800 kg and is able to support up to 4 560 kg. What is its structural efficiency?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!