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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

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The depletion in the Ozone layer is caused by:
andrew11 [14]
D. Chlorofluorocarbon

This is a man-made carbon that causes the gradual thinning(deception) in the earth’s Ozone layer.

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~ Luna
3 0
2 years ago
Read 2 more answers
An engineering drawing shows the: (A) dimensions, tolerances, cost, and sales or use volume of a component.(B) dimensions, toler
Leto [7]

Answer:

(B) dimensions, tolerances, materials, and finishes of a component.

Explanation:

An engineering drawing :

  An  engineering drawing is a technical drawing which draws the actual component .

An engineering drawing shows

1. Materials

2.Dimensions

3.Tolerance

4.Finishes of a component

Engineering drawing does not shows any information about the cost of component.

So the option B is correct.

3 0
3 years ago
Natalie solved the problem below incorrectly. Identify his error and justify your reasoning. −19+3x−11+2x=2 5x−19−11=2 5x−30=2 5
arsen [322]
5x-30=2
5x=2+30 (not -28) when the -30 is brought over to the RHS, 30 should be added to 2 instead of subtracted

hence, 5x=32
x = 6.4
5 0
2 years ago
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
saul85 [17]

Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0
2 years ago
Wright Company deposits all cash receipts on the day when they are received and it makes all cash payments by check. At the clos
alina1380 [7]

Answer:

                                              Wright Company

                                             Bank Reconciliation

                                                 May 31, 2013

Credit side                                                                                   Debit side

Bank statement $26200                 |                          Book balance $27900

<em>Add;                                                    </em>

Deposit on May 31 $6400

Bank error $420

Sub-total=$33020

Deductions;                                        |                       Deduct

ions

Outstanding checks $5800              |                 Bank service charge $120

Adjusted bank balance $27220       |                  NSF check $560

                                                                             Total deduction $680

                                                      Adjusted book balance $27220

8 0
2 years ago
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