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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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2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
Viefleur [7K]

Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

<u>Determine the force transmitted by the coupling between the nozzle and hose </u>

attached below is the remaining part of the  detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2  

= 32 ( 25 / 35 )^2

= 16.33 m/s

4 0
2 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
2 years ago
Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67
defon

Answer:

#include <iostream>

#include <string>

using namespace std;

bool isPalindrome(string str)

{

   int length = str.length();

   for (int i = 0; i < length / 2; i++)

   {

       if (tolower(str[i]) != tolower(str[length - 1 - i]))

           return false;

   }

   return true;

}

int main()

{

   string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};

   int i;

   for(i=0; i<6; i++)

   {

       //Testing function

       if(isPalindrome(s[i]))

       {

           cout << "\n " << s[i] << " is a palindrome... \n";

       }

       else

       {

           cout << "\n " << s[i] << " is not a palindrome... \n";

       }

   }    

       

   return 0;

}

5 0
3 years ago
ILL GIVE BRAINLIEST!!!
Sedbober [7]
Get the app socratic I saw the answer to your question on the app but I ran out of screen time to show you
6 0
2 years ago
A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

3 0
3 years ago
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