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Nikitich [7]
3 years ago
8

Your new mobile phone business is now approaching its first anniversary and you are able to step back and finally take a deep br

eath and consider positive changes that could be implemented. One of these changes is to transform you from a reactive maniac running around putting out fires, to a role model, achieving personal excellence. What specific steps will you use to make this transition? What specific programs will you put in place to help your key personnel do the same? Submit your paper to Drop Box 10.3.

Engineering
1 answer:
Leya [2.2K]3 years ago
7 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the
adelina 88 [10]

Answer:

Please look at attachment carefully.

7 0
4 years ago
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype
ella [17]

Answer:

Explanation:

Given

scale i.e. L_r=1:15

Using Reynolds number similarity

(Re)_m=(Re)_p

(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p

Properties of air

\nu _{air}=1.57\times 10^{-4} ft/s

Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )

V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )

V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}

V_p=0.963\ ft/s

8 0
4 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
3 years ago
According to the rules of dimension how should written notes appear
Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular  views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with  horizontal but may be made at any convenient angle except vertical or horizontal.

7 0
3 years ago
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