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Anit [1.1K]
3 years ago
6

In a full duplex communication by using UTP cable, Pt =140 w, Pr =10 w and NEXTdb =14,47. According to this information which an

swer is correct?
a.
Pc = 7,50 Adb = 12,97 ACRdb = 1,51

b.
Pc = 5,00 Adb = 11,46 ACRdb = 3,01

c.
Pc = 7,50 Adb = 14,47 ACRdb = 0,00

d.
Pc = 10,00 Adb = 12,97 ACRdb = 1,51

e.
Pc = 10,42 Adb = 9,97 ACRdb = 4,50
Engineering
1 answer:
Paraphin [41]3 years ago
4 0
The answer that is correct is c
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At audio frequencies (below 20kHz), electromagnetic waves have very ____ wavelengths. The wavelength is typically much larger th
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Answer:

broad, transmission or power lines, electrical circuit theory

Explanation:

Electromagnetic waves have quite broad wavelengths for audio frequencies. Normally, the wavelength is larger than any of wires used in circuit. Long transmission or power line might be an exception. If the wavelength is much smaller than length of a wire, the basic principles of electronic circuits apply and the principle of electrical circuits is not needed

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One proposed technique to reduce outage probability is to use macrodiversity, where a mobile unit’s signal is received by multip
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4 years ago
The following are related to wastewater treatment:
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Answer:

A. - Primary wastewater treatment: the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater.

- Secondary wastewater treatment: treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process.

-Tertiary wastewater treatment: Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending on the type of water contamination that must be removed.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C. - Activated Sludge, Rotating biological contractors.

Explanation:

A. Primary wastewater treatment refers to sedimentation, the physical treatment process which is used to in order to get rid of suspended solids that can settle from wastewater. The main goal of primary treatment is to remove organic and inorganic solids that settles by sedimentation, and by act of skimming removed materials end up floating.

Secondary wastewater treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process. The goal of secondary treatment is the further treatment of the effluent from primary treatment in order to get rid of the residual organics and suspended solids.

Advanced wastewater is quite in between. Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending upon the type of water contamination that must be removed. Advanced treatment processes can be combined with primary or secondary treatment or used instead of secondary treatment.

B. Primary treatment: Approximately 35% of the incoming biochemical oxygen demand (BOD).

Secondary treatment : Approximately 85% of biochemical oxygen demand (BOD).

C.

-Activated Sludge: The activated sludge process is a kind of wastewater treatment process for treating sewage/industrial wastewaters by the use of aeration and a biological floc which consists of bacteria and protozoa.

- Rotating biological contactor: Rotating biological contactors (RBCs) are fixed-film reactors that have similarities with biofilters in the sense that organisms are attached to support media.

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6 0
4 years ago
Read 2 more answers
Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points
AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component  based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure (P_{in}) = 11 MPa

Outside pressure (P_{out}) = 1 MPa

Yield strength (\sigma_y) = 1000 MPa

Therefore:

\sigma_y =\frac{\sigma_y}{N}, Substituting values,

\sigma_y =\frac{\sigma_y}{N}=\frac{1000}{3.5}=285.714 MPa

The minimum required thickness for a steel pressure vessel (t) is given by the equation:

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}. Substituting values

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}=\frac{65*10^{-3} *(11-1)10^{6} }{285.714*10^{6} } =2.275*10^{-3} =2.275 mm

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

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Answer:

43.2%

Explanation:

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The egative value of efficiency shows work is done by the engine.

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3 years ago
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