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True [87]
3 years ago
11

Select the qualification that is best demonstrated in each example.

Engineering
2 answers:
bonufazy [111]3 years ago
7 0

Answer:

Select the qualification that is best demonstrated in each example.

Rebecca is a Plumber who trains an apprentice.

✔ leadership and teaching skills

Janet is a Construction Manager who creates a budget for a complex project.

✔ financial skills

Dennis is a Carpenter who works alone frequently.

✔ independence

Explanation:

hope it helps :)

Dmitrij [34]3 years ago
4 0

Answer:

what are the choices for the answer not ABC

Explanation:

mark me as brainiest please it also helps me thank

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What is over head line
CaHeK987 [17]

Explanation:

An overhead power line is a structure used in electric power transmission and distribution to transmit electrical energy across large distances. It consists of one or more conductors (commonly multiples of three) suspended by towers or poles. Since most of the insulation is provided by air, overhead power lines are generally the lowest-cost method of power transmission for large quantities of electric energy.

<h3><em><u>Constr</u></em><em><u>uction</u></em></h3>

Towers for support of the lines are made of wood (as-grown or laminated), steel or aluminum (either lattice structures or tubular poles), concrete, and occasionally reinforced plastics. The bare wire conductors on the line are generally made of aluminum (either plain or reinforced with steel or composite materials such as carbon and glass fiber), though some copper wires are used in medium-voltage distribution and low-voltage connections to customer premises. A major goal of overhead power line design is to maintain adequate clearance between energized conductors and the ground so as to prevent dangerous contact with the line, and to provide reliable support for the conductors, resilience to storms, ice loads, earthquakes and other potential damage causes. Today overhead lines are routinely operated at voltages exceeding 765,000 volts between conductors.

<em>Please</em><em> </em><em>mark</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em><u>brainliest</u></em><em>. </em><em>Follow</em><em> </em><em>me</em><em> </em><em>I </em><em>w</em><em>ill</em><em> </em><em>fo</em><em>llow</em><em> you</em><em> back</em><em>. </em>

6 0
4 years ago
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
Why did the dodge demon cross the road
fgiga [73]

Answer:

to get to the other side

5 0
3 years ago
Read 2 more answers
what are three things that we would need and unlimiated amount of for all of us to obtain everything that we want?
Elanso [62]
Unlimited wants is an economic term that refers to humans’ insatiable appetite for things. We never get enough because there is always something else that we need or want. The term ‘unlimited wants’ is the side of human nature that wants an infinite number of things. However, the resources we have available to get these wants are limited.

There are two halves of scarcity that have plagued us ever since we first set foot on this Earth:

Limited resources.
Unlimited wants.
The Economics of Seinfeld says the following regarding the term:

“Unlimited wants essentially mean that people never get enough, that there is always something else that they would like to have.”

“When combined with limited resources, unlimited wants result in the fundamental problem of scarcity.”

Unlimited wants – limited resources

What we want and need has no limit, i.e., it is infinite. However, what we can afford is finite, i.e., it has a limit. This is a basic condition of human existence.

We are never completely satisfied with everything we consume. We consume a variety of goods and services, but they are never enough.

In other words, there is always something else that I, you, or anybody else would want or need.

The term applies to all socioeconomic groups. Low-income groups have limited resources, and their wants always exceed those resources. However, the same happens with middle-income and upper-income groups. They never feel they have enough.

The reason is a very simple one. Every income group’s resources are finite. However, unlimited want is a feature of every human.

Put simply; our wants and needs are infinite, but our wealth is not.

The economic problem – unlimited wants

‘The economic problem‘ is a term that economists use. It states that the finite resources of an economy are not enough to satisfy all our wants and needs. We also call it ‘the central economic problem‘ or ‘the basic economic problem.’

The main question we ask when considering ‘the economic problem’ is: “How do we satisfy unlimited wants with limited resources?”

As we cannot produce everything, we have to prioritize. We must decide what to produce, how to produce it, and how much to produce. We must also determine for whom to produce.

Human wants are constant and infinite, but the resources to satisfy them are finite. The resources cannot exceed the amount of human and natural resources available.

We produce things that we know people want, as long as we have the resources to make them. How strong or weak demand is determines how much we charge for those things. It also determines how much we produce (supply).

In other words, markets fores, i.e., the forces of supply and demand, in a free market economy, determine prices.

Wants vs. needs

Needs are things without which we cannot survive. Wants are things we desire. However, we can survive without those wants.

Food, water, and housing, for example, are needs. Clothing is also a need. Without food or water, we would die. We would probably die too without housing. In cold countries, we would not survive without clothing.

A nice car, smartphone, and vacation by the beach are wants. If I don’t have a nice car, I will still live. If I don’t go to Cancun for my winter break, I won’t die. However, I want these things.

Fundamental needs are key in the function of the economy. Wants, however, are the driving forces that stimulate demand for things, i.e., demand for goods and services.

We can say either ‘unlimited wants’ or ‘unlimited wants and needs.’

3 0
3 years ago
An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow
borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

\\ E_{inflow} = E_{outflow}

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy    \\ E_{inflow} = m_{1} h_{1} .

\\

E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}

\\

Where m_{1} h_{1} are the mass flow rate and the enthalpies at the inlet  at a pressure of 3Mpa \\,

m_{2} h_{2} are the mass flow rate and the enthalpies  at the outlet 2 where we have a pressure of 500kpa respectively.\\,

and  m_{3} h_{3}   are the mass flow rate and the enthalpies  at the outlet 3 where we have a pressure of 50kpa respectively.\\,

We can now express write out the required equation by substituting the new expression for the energies \\

m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}   \\

from the above equation, the unknown are the enthalpy values and  the mass flow rate. \\

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table.  \\

It is more convenient to start from outlet 3 were we have a temperature 100^{0}C and pressure value of (50kpa or 0.05Mpa ). using double interpolation method  on the superheated water table to determine the enthalpy value with careful calculation we have  \\

h_{3}  = 2682.4 KJ/KG , at this point also from the table the entropy value ,s_{3} value is 7.6953 KJ/Kg.K. \\

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy  is constant at all inlet and outlet. \\

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of  7.6953 KJ/Kg.K with careful  interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.\\

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K  we arrive at enthalpy value of 3851.2KJ/Kg. \\

Now we determine the mass flow rate at each inlet and outlet. since  mass must also be balance, i.e  m_{1} = m_{2} + m_{3} \\

From the question the, the mass flow rate at the inlet m_{1}}  is 2Kg/s \\

Since 5% flow is delivered into the feedwater heating,  \\

m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s \\

Also for the outlet 3 the remaining 95% will flow out. Hence

m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s \\

Now, from m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}   \\ we substitute values

W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}

W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)

\\

W_{out} = 2285.19 kW.

Hence the power produced is 2285kW

7 0
3 years ago
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