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Genrish500 [490]
3 years ago
15

An injection-molding machine has a first cost of $1,050,000 and a salvage value of $240,000 in any year. The maintenance and ope

rating cost is $235,000 with an annual gradient of $75,000. The MARR is 10%. What is the most economic life?
a. years (EUAC - $759.376)
b. 7 years (EUAC - $672,616)
c. 3 years (EUAC - $904,384)
d. 5 years (EUAC - $608,428)
Engineering
1 answer:
Ilya [14]3 years ago
4 0

Explanation:

First cost =$ 1,050,000

Salvage Value =$ 225,000 Maintenance & Operating cost =$ 235,000 Maintenance & Operating Gradient =$ 75,000 MARR =10 %

EUAB - EAUC $=\$ 1,050,000(\mathrm{~A} / \mathrm{P}, 10 \%, \mathrm{n})+\$ 225,000(\mathrm{~A} / \mathrm{F}, 10 \%, \mathrm{n})$

$$

-\$ 235,000-\$ 75,000(\mathrm{~A} / \mathrm{G}, 10 \%, \mathrm{n})

$$

Try $\mathrm{n}=4$ years:

$$

\text { EUAB - EAUC }=\$ 331,275+\$ 48,488-\$ 235,000-\$ 103,575=-\$ 621,362

$$

Try $\mathrm{n}=5$ years:

EUAB - EUAC=-\$ 276,990+\$ 36,855-\$ 235,000-\$ 135,750=-$ 610,885

Try n=6 years

{ EUAB - EUAC }=-$ 241,080+$ 29,160-$ 235,000-$ 166,800=-$ 613,720

Thus, year 5 has the minimum EUAB - EUAC, hence the most economic life is 5 years.

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Answer:

metals, composite, ceramics and polymers.

Explanation:

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i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

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Explanation:

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Integrating both sides we get

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x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

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v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

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