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3 years ago

What are the four basic parts of process plan

Engineering

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

1. Goal setting

2. Developing the planning premises

3. Reviewing Limitations

4. Deciding the planning period

5. Formulation of policies and strategies

6. Preparing operating plans

7. Integration of plans

Explanation:

I don't have an explanation but i thru in a few extra answers for like extra credit or something if your teacher does that

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In a morphological matrix, which of the following contains the parameters that are essential to a design?

choli [55]

In a morphological matrix, the parameters that are essential for a design are in the left column.

<h3>What is a morphological matrix?</h3>

The morphological matrix is a matrix where columns and rows represent the various parameters for solving a problem. The first column is used for the characteristics relevant to the problem; the horizontal lines are filled with possibilities for each of these parameters.

With this information, we can conclude that in a morphological matrix, the parameters that are essential for a project are in the left column.

Learn more about morphological matrix in brainly.com/question/21120930

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3 years ago

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3 years ago

Cleaning the shop's floor is being discussed. Technician A says to flush the waste and dust into a floor drain. Technician B say

lawyer [7]

Answer:

Technician B

Explanation:

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3 years ago

The best way to become a better reader is to study longer and harder in every subject. engage in a variety of extracurricular ac

harina [27]

Answer:

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I just did the test on enginuity and it also is the only one that makes sence

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3 years ago

Read 2 more answers

A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The

attashe74 [19]

Answer:

$q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C$

Explanation:

$R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W$

So, heat rate can be calculated as follows:

$q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW$

Surface temperatures can be calculated as follows:

$T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C$

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4 years ago