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kap26 [50]
3 years ago
14

What is the term RF exiciter?

Engineering
1 answer:
Sati [7]3 years ago
7 0
The exciter provides fully coherent receiver local oscillator signals at radar frequency band as well as requisite, auxiliary high frequency clock signals. The exciter function is divided into an internal frequency synthesizer and an upconverter.

Hope this helps :)))
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Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?
skad [1K]

Answer:

d

Explanation:

is the because that's the amount of work in making machine can do producing heat

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3 years ago
Describe the Optical Fiber Network( i need 2 paragraphs, but give me about 5 sentences or like 3)
frez [133]

Answer: Fiber Optic Network Fiber-optic networks have been used for decades to transmit large volumes of traffic across the country. The economics of fiber networks have only recently allowed for connecting the fiber directly to the home, creating a fiber-to-the-home (FTTH) network.

Explanation:

6 0
3 years ago
Read 2 more answers
What is capillary action?
BARSIC [14]
Capillary action occurs When the adhesion to the walls stronger than dirt cohesive forces between a liquid molecules. the head towards Capillery action will take water in a uniform circular is limited by surface tension and, of course, gravity.
4 0
2 years ago
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m K), and the wire/sheath interface i
Semmy [17]

Answer:

maximum allowable electrical power=4.51W/m

critical radius of the insulation=13mm

Explanation:

Hello!

To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)

Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and  conduction as shown in the attached image

to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient

r=\frac{k}{h} =\frac{0.13}{10}=0.013m=13mm

3 0
3 years ago
Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

7 0
3 years ago
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