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Leviafan [203]
3 years ago
14

What differentiates the master builder approach prior to the Renaissance from later approaches? Projects do not depend on indivi

dual knowledge but leverage the total knowledge of the team. The engineering and construction roles dominated projects because architectural designs were so simple. The architect and engineer roles are clearly separated into separate project phases. Knowledge is confined to particular individuals and not shared or formalized into a recognizable set of standards.
Engineering
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

Design-Build, though not new as a delivery method for building projects, appears to be on the rise. Traditionally known as the Master-Builder method, it is a means of building where one party holds responsibility for both the design and the construction. The Master-Builder method was the only method before the now ubiquitous design-bid-build project structure. The Romans for example, famous for their roads, aqueducts, and amphitheaters, did not design a project, bid it out to subcontractors, and then select the low bidder to build it, but rather designed and built structures in a collaborative, somewhat simultaneous fashion. In building a house for a client during Colonial times, one party, such as a master carpenter, was responsible for delivering the general layout and exterior details, selecting structural members, and completing construction. In this way, the carpenter acted as the architect, engineer, and builder simultaneously. The concept of design, bid, and build arose out of the natural specialization of the architect, engineer, and builder in the post-1850s world, where modern structural engineering was born and separated from architecture, and architects and builders fully separated as distinct and separate entities.

Explanation:

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase
ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

m_{Fe} = 97.5\,g

If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

The volume percent of graphite is 91.906 per cent.

6 0
2 years ago
Manufacturing employees who perform assembly line work are referred to as
mamaluj [8]

Answer:

C. assembly line workers.

Explanation:

8 0
3 years ago
Read 2 more answers
"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.  

6 0
3 years ago
A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and
Talja [164]

Answer:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.

Formulate the optimization problem.

Explanation:A rectangular region ABCD is to be built inside a semicircle of radius 10 m with points A and B on the line for the diameter and points C and D on the semicircle with CD parallel to AB. The objective is to find the height h * that maximizes the area of ABCD.

Formulate the optimization problem.

7 0
3 years ago
P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

4 0
3 years ago
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