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Amiraneli [1.4K]
3 years ago
7

Which option identifies the concept represented in the following scenario?

Engineering
1 answer:
dlinn [17]3 years ago
3 0

Answer:

project object

Explanation:

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HfrrghhJbfrfefefft nbgyjjiutrwdgju
Tasya [4]

Answer:

hshdhriwjajaldh skshdjdywuusg

Explanation:

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4 0
2 years ago
List two things that technological systems have in common.​
Sphinxa [80]

They all share the way that they are fundamentally designed: if they are quite complex, they will share the same basic logic foundations, like the way that the programming languages work. They also all share the method of construction and common and fundamental electronic components, like resistors, capacitors and transistors. As we humans design them, they make logical sense to at least someone, and probably only discounting the internet, you can probably draw logic diagrams and whatever to represent how they work.

Because they are designed by Humans, in a way they all mimic how our brains and society work. Also, as yet there are no truly intelligent technological systems, and are only able to react to a situation how they have been programmed to do so.

3 0
2 years ago
Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?
zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0
1 year ago
Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
makkiz [27]

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

7 0
3 years ago
Read 2 more answers
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
3 years ago
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