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3 years ago

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A machine used to shred cardboard boxes for composting has a first cost of $10,000, an AOC of $7000 per year, a 3-year life, and

no appreciable salvage value. Assume you were told that the service provided by the machine will be needed for only 5 years, which entails a repurchase and retention for only 2 years. What salvage value, S, is required after 2 years in order to make the 2-year AW value the same as it is for its 3-year life cycle AW? Assume i = 10% per year. Solve using factors and, if assigned, using a spread- sheet and Goal Seek.

1 answer:

borishaifa [10]3 years ago

6 0

Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-

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What is stress corrosion cracking?

aksik [14]

Answer:

The growth of crack formation in a corrosive environment.

Explanation:

6 0

3 years ago

One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0

3 years ago

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0

2 years ago

Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area

dybincka [34]

Answer:153.76 MPa

Explanation:

$Initial Area\left ( A_0\right )=12.56 mm^2$

$Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2$

$Die angle=30^{\circ}$

$\alpha =\frac{30}{2}=15^{\circ}$

$\mu =0.08$

$Yield stress\left ( \sigma _y \right )=350 MPa$

$B=\mu cot\left ( \aplha\right )=0.2985$

$\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B$

$\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}$

$\sigma _{pressure}=153.76 MPa$

8 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

a)

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

b)

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

c)

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

d)

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

e)

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago