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Brut [27]
3 years ago
11

The systems engineering method applies to the advanced development phase in a similar set of four steps, as it does to the prece

ding concept definition phase. For each step in the method, compare the activities in the two phases with one another, stating in your own words (a) how they are similar and (b) how they are different.
Engineering
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

Similarities and variations between the functional device specification as well as the concept specification classification are given below.

Explanation:

  • The specification of the product requires market specifications, customer requirements, receiving goods and services, operating requirements, system eligibility criteria.
  • The paper, and on the other hand, explains the sub-systems as well as the configurations between certain subsystems. It is the perception of both the developer. The timetable as well as the testing Plan are the only other papers that you'll have to plan.
  • The scheduling paper becomes easy maintenance to be doing if you already have carefully planned out technical requirements and concept documents, then you will have protected all of the bases.
  • Because while the functional specification is evaluated by decomposing higher-level components found towards lower-level mechanisms by requirement analysis phase. Lower roles are given the output potential contributing factor with either the greater position.
  • Addressing the complete system features is one part of the system specification. Certain functionality has been mostly carried out in hardware and some in applications. The specification including its framework is to describe the maximum functionality but to carry out trade-offs including preliminary design analyses to assign these specifications to the various disciplines, like that of the performance specifications.
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Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

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Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

 A=\pi r^2\\A=3.142 (0.05)^2

 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

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A certain robot can perform only 4 types of movement. It can move either up or down or left or right. These movements are repres
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Answer:

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Answer:

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6 0
3 years ago
Read 2 more answers
A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87
Inessa [10]

Answer:

atomic radius  R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a =  0.413 nm

parameters b = 0.665 nm

parameters c =  0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and  atomic weight

solution

we apply  here atomic packing factor (x) that is

atomic packing factor (x) = \frac{volume(sphere)}{volume(unit\ cell)}  ..................1

put here value we get

atomic packing factor = \frac{8*(4/3)*\pi R^3}{3*a*b*c}

R = (\frac{3(x)(abc)}{32\pi })^{1/3}

R =  (\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}

atomic radius  R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = \frac{\rho (abc)(N_A)}{no\ of\ atom}   ....................2

metal atomic weight = \frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}  

metal atomic weight = 72.27 g/mol

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3 years ago
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