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olya-2409 [2.1K]
3 years ago
10

You are the project manager assigned to construct a new 10-story office building. You are trying to estimate the costs for this

project. You start by assigning the costs associated with each of the project activity. Then you sum up all the individual costs into a final cost estimate. Which type of cost estimation technique did you use?
Engineering
1 answer:
Semmy [17]3 years ago
8 0

Answer:

Bottom-up Estimation

Explanation:

Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.

This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.

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If a front gear had 24 teeth, and a rear gear has 12 teeth:
zubka84 [21]

Answer:

  4 times around

Explanation:

The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.

If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...

  (48 teeth)/(12 teeth/turn) = 4 turns

4 0
3 years ago
In a steady flow device, the properties of the system remains constant with time. a)True b) False
Leviafan [203]

Answer:

True

Explanation:

By definition of steady flow we have

\frac{\partial f(x,y,z,t) }{\partial t}=0

where f(x,y,z,t) is any property of the system under consideration

=> f(x,y,z,t) = constant

7 0
3 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=10.3MPa

7 0
3 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)
konstantin123 [22]

Answer:

hi

Explanation:

4 0
3 years ago
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