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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

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The ladder has a uniform weight of 80 lb and rests against the smooth wall at B. If the coefficient of static friction at A is m

natali 33 [55]

I have no idea how to answer this

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3 years ago

PLEASE I NEED HELP!

insens350 [35]

Answer:

it depends on how you do it it's mot9

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3 years ago

2- A 2-m3 insulated tank containing ammonia at -20 C, 80% quality, is connected by a valve to a line flowing ammonia at 2 MPa, 6

alekssr [168]

Answer:

The valve should be closed at 1081 kPa ⇒ 11 bar

Explanation:

$Q_{C_V} + m_ih_i = m_2u_2 - m_1u_1 + W_{C_V}$

$Q_{C_V} = W_{C_V} = 0$

$m_1 = \frac{V}{v_1} = \frac{2}{0.49927} = 4.006 kg$

$m_i = m_2 - m_1 = 15 - 4.006 = 10.994 kg$

$u_1 = 1057.5, h_i = 1509.9$

$u_2 = m_ih_i + m_1u_1$

$m_2 = \frac{[10.994*1509.9] + [4.006*1057.5]}{15} = 1389.1 kJ/kg$

$v_2 = \frac{V}{m_2} = \frac{2}{15} = 0.1333 m^3/kg$

Therefore, v₂, u₂ fix state 2.

By trial and error, P₂ = 1081 kPa & T₂ = 50.4° C

1081 kPa ⇒ 11 bar

8 0

4 years ago

An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu

Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

$T_{heater} = 120$ Degree F

$T_{room} = 60$ Degree F

A = 30 ft^2

$\eta = 100$ %

Heat loss before previous final value $= \frac{A \Delta T}{R}$

$=\frac{30\times *(120-60)}{5}$

= 360 Btu/hr

Heat loss after new value $= \frac{30\times \times (120-60)}{15} = 120 Btu/hr$

saving would be $= 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year$

= 616.1782 kw hr/yr

$cost = 616.1782 \times 0.06$ $

= 36.917 $\yr

6 0

4 years ago

What tools do you need to change a flat tire

Lyrx [107]

Answer:

Spare tire (if you’re dealing with a flat and not just rotating tires or accessing the brakes)
Carjack
Lug wrench
Wheel wedges
Work gloves
Tire repair kit (if you’re attempting to fix your flat tire)
Flare/reflective triangles (if you’re changing the tire on the side of the road or in a parking lot)

6 0

4 years ago

Read 2 more answers