1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dimaraw [331]
3 years ago
12

In engineering, economic cost is a decision-making tangible factor. Group of answer choices True False

Engineering
2 answers:
salantis [7]3 years ago
8 0
Economic cost is a rescission making tangible factor true
polet [3.4K]3 years ago
6 0

Answer:

True

Explanation:

Economic cost is a very important factor to consider in decision making in Engineering. It is part of the four essential elements involved in decision making in engineering analysis.

Economic cost is part of the criteria to evaluate alternatives considering the time value of money by estimating a specific measure of worth of estimated commodity cost over a period of time.

Other Factors to consider in Engineering Economics as related to the Economy are:

(1) Time of Occurence of Cash flows

(2) Interest Rates

(3) Cash Flows

You might be interested in
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1
Gelneren [198K]

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }  ...........1

put here all these value and we get t2

\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }

t2 = 424.8

final temperature is 424.8 K

so correct option is e

5 0
3 years ago
I gave 15 min to finish this java program
lisov135 [29]

Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

 for (int number = 1; number <= 10; ++number) {

  int sum = 1;

  System.out.print("1");

  for (int summed = 2; summed <= number; ++summed) {

   sum += summed;

   System.out.print(" + " + Integer.toString(summed));

  }

  System.out.print(" = " + Integer.toString(sum) + '\n');

 }

}

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum  numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

4 0
3 years ago
What are the controlling LRFD load combinations for dead and floor live load?
yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

Explanation:

Load and Resistance Factor Design

there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

here load factor for L given ( * ) mean it is  permitted = 0.5 for occupancies when live load is less than or equal to 100 psf

here

D is dead load and L is live load

E is earth quake load and S is snow load

W is wind load and R is rain load

Lr is roof live load

3 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat
densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0
3 years ago
Other questions:
  • The Manufacturing sector is the only sector where Lean manufacturing philosophy can be applied. a)- True b)- False
    12·1 answer
  • A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the
    12·2 answers
  • What the Best describes the purpose of the occupational safety and health administración OSHA
    12·1 answer
  • n open feedwater heater operates at steady state with liquid water entering inlet 1 at 10 bar, 50°C. A separate stream of steam
    8·1 answer
  • Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What
    7·1 answer
  • What is a business cycle? a period of economic growth followed by economic contraction the amount of time it takes a business to
    13·2 answers
  • Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
    8·1 answer
  • Any help is appreciated.
    7·1 answer
  • 1. Using the formula above, complete this task.
    9·1 answer
  • The cylinder C is being lifted using the cable and pulley system shown.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!