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3 years ago

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 vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c

Ann [662]

Answer:

Amount of air left in the cylinder=m $_{2}$ =0.357 Kg

The amount of heat transfer=Q=0

Explanation:

Given

Initial pressure=P1=300 KPa

Initial volume=V1=0.2 $m^{3}$

Initial temperature=T $_{1}$ =20 C

Final Volume= $V_{2}$ =0.1 $m^{3}$

Using gas equation

$m_{1}=((P_{1}*V_{1})/(R*T_{1}))$

m1==(300*0.2)/(.287*293)

m1=0.714 Kg

Similarly

m2=(P2*V2)/R*T2

m2=(300*0.1)/(0.287*293)

m2=0.357 Kg

Now calculate mass of air left,where me is the mass of air left.

me=m2-m1

me=0.715-0.357

mass of air left=me=0.357 Kg

To find heat transfer we need to apply energy balance equation.

$Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})$

Where me=m1-m2

And as the temperature remains constant,hence the enthalpy also remains constant.

h1=h2=he=h

Q=(me-(m1-m2))*h

me=m1-me

Thus heat transfer=Q=0

6 0

3 years ago

You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch

djyliett [7]

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample = 24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at 96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × $10^{9}$ bits

as 1 Gib bit = $2^{30}$ bits

so required space = 2.48832 × $10^{9}$ bits ÷ $2^{30}$ bits

required space = 2.3174 Gibibits ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × $10^{9}$ bits

space required = 0.381024 × $10^{9}$ bits ÷ $2^{30}$ bits

space required = 0.3548 Gibibits ...........2

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

8 0

3 years ago

Implement a function called myFilter that does the following: Takes a sequence of integers with values from 0 to 100 Remove all

goldenfox [79]

Answer:

using System;

using System.Collections.Generic;

using System.Linq;

public class Problem1 {

public static IEnumerable < int > myFilter(IEnumerable < int > input) {

return input.Where((val, index) => {

//First remove all multiples of 6 less than 42

return !(val % 6 == 0 && val < 42);

}).Select((val, index) => {

//Then square each number

return val * val;

}).Where((val, index) => {

// Finally, remove any resulting integer that is odd

return !(val % 1 == 0);

});

}

public static void Main(string[] args) {

Random rnd = new Random(5);

//Takes a sequence of integers with values from 0 to 100

var listForProblem = Enumerable.Range(1, 100).Select(i => rnd.Next() % 101);

var answer = Problem1.myFilter(listForProblem);

foreach(int i in answer) {

Console.WriteLine(i);

}

Explanation:

The Problem 1 solution is explanatory enough.

Your problem 2 question is lacking details. However, i have written a code that sorts it.

6 0

3 years ago

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

nasty-shy [4]

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine $\rho =1050kg/m^3$

Rate at which brine is pumped $V=0.3m^3/sec$

So mass of the pumped per second

Mass = volume × density = $1050\times 0.3=315$ kg/sec

Acceleration due to gravity $g=9.8m/sec^2$

Depth h = 200 m

So work done $W=mgh=315\times 9.8\times 200=617400J$

Efficiency is given $\eta =0.9$

We have to fond the input power

So input power $=\frac{617400}{0.9}=686000J$

So input power of the geothermal power will be 686000 J

5 0

3 years ago

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+

nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

Explanation:

From the question; the equations of the velocities profile in the system are:

$u = u_{max}(Ay^2+By+C)$ ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0 ----- (a)

At y = h; u =0 -----(b)

At y = $\frac{h}{2}$ ; u = $u_{max}$ ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

$u_{max}$ = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0$

Replacing the boundary condition (b) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah \ \ \ \ \ --- (d)$

Replacing the boundary condition (c) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 = \frac{Ah^2}{4} + \frac{h}{2}(-Ah) \\ \\ 1= \frac{Ah^2}{4} - \frac{Ah^2}{2} \\ \\ 1 = \frac{Ah^2 - Ah^2}{4} \\ \\ A = -\frac{4}{h^2}$

replacing $A = -\frac{4}{h^2}$ for A in (d); we get:

$B = - ( -\frac{4}{h^2})h$ $B = \frac{4}{h}$

replacing the values of A, B and C into the velocity profile expression; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)$

To determine the volume flow rate; we have:

$Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)$

Replacing $u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u$

$\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})$