Over recent recents, e-Commerce has relied the following to stay successfully and competitive*
A. Logistics function
B. Make function
C. All SCOR model function
D. Non above

The pyramids are important to the study of engineering because they were some of the first structures built using engineering principles. The pyramids are an example of how engineering can be used to create amazing structures.

What is pyramid?

A pyramid is a structure with triangular exterior surfaces that converge to a single step at the summit, giving the shape the shape of a pyramid in the geometric sense. A pyramid's base can be trilateral, quadrilateral, or any polygon shape. A pyramid, as a result, has at least three exterior triangle surfaces. A typical variant is the square pyramid, which has a square base and four triangular exterior surfaces. The design of a pyramid, with the majority of the weight closer to the bottom and the pyramidion at the apex, means that less material will be pushed down from above. This weight distribution enabled early civilizations to build stable massive constructions.

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The injector supply lines must be replaced if they are removed. So, what is said by technician A is correct.

Injector refers to a device that is used to inject liquid fuel into an internal-combustion engine. The injector can also used for describing an apparatus to injcet feed water into a boiler. The injector supply lines must be replaced if they are removed. The injector can be used as a spray nozzle to break up the fuel into a fine spray - it doesn't always control the fuel flow. There are three general kind of injection systems on the market:

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Answer:

The answer is "shear".

Explanation:

Every transformation could be shown by some kind of conventional matrix.

Its standard matrices of shape k here could be any real value enabling shear transformation to parallel to the y axis.

\(\left[\begin{array}{cc}1&0\\k&1\\\end{array}\right]\)

This coefficient matrix A is the standard matrix during transformation. With the use of a the transformation could be entered into T(x)=Ax

And standard transfer function is standard.

\(\left[\begin{array}{cc}1&0\\6&1\\\end{array}\right]\)

The matrix in the form of the shear matrix.

The percent crystallinity of the specimen with a density of 1.382 g/cm^3 is approximately 85.5%.

To compute the density of totally crystalline poly(ethylene terephthalate) (PET), we can use the density and percent crystallinity values given for the material.

(a) Compute the density of totally crystalline PET:

Density of totally crystalline PET = Density / (1 - Percent Crystallinity / 100)

Density = 1.408 g/cm^3

Percent Crystallinity = 74.3%

Density of totally crystalline PET = 1.408 / (1 - 74.3 / 100)

Density of totally crystalline PET = 1.408 / (1 - 0.743)

Density of totally crystalline PET = 1.408 / 0.257

Density of totally crystalline PET ≈ 5.484 g/cm^3

(b) Compute the density of totally amorphous PET:

Density of totally amorphous PET = Density / (1 + Percent Crystallinity / 100)

Density = 1.343 g/cm^3

Percent Crystallinity = 31.2%

Density of totally amorphous PET = 1.343 / (1 + 31.2 / 100)

Density of totally amorphous PET = 1.343 / (1 + 0.312)

Density of totally amorphous PET = 1.343 / 1.312

Density of totally amorphous PET ≈ 1.023 g/cm^3

(c) Determine the percent crystallinity of a specimen having a density of 1.382 g/cm^3:

Percent Crystallinity = (Density of totally crystalline PET - Density of the specimen) / (Density of totally crystalline PET - Density of totally amorphous PET) * 100

Density of the specimen = 1.382 g/cm^3

Density of totally crystalline PET = 5.484 g/cm^3

Density of totally amorphous PET = 1.023 g/cm^3

Percent Crystallinity = (5.484 - 1.382) / (5.484 - 1.023) * 100

Percent Crystallinity ≈ 85.5%

Therefore, the percent crystallinity of the specimen with a density of 1.382 g/cm^3 is approximately 85.5%.

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Answer:

115.2 W

Explanation:

The computation is shown below:

As we know that

Power = F . v

\(F_H = F cos \theta\)

\(F_H = 30 \frac{4}{5}\)

\(F_H = 24N\)

Now we solve for V

\(V = V_0 + at\)            a = 24N ÷ 20Kg

But V_0 = 0          a = 1.2 m/s^2

F_H = ma             V = 0 + (1.2) (4)

a = F_H ÷ m        V = 4.8 m/s

Therefore

Power = F_Hv

= (24) (4.8)

= 115.2 W

By applying the above formuals we can get the power

Answer:

c

Explanation:

:)

Answer:

Explanation:

Step one:

given data

T_{wi} = 20^{\circ}C

T_{Ai}=1000K

T_{Ae}= 400kPa

P_{Wi}=200kPa

P_{Ai}=125kPa

P_{We}=200kPa

P_{Ae}=100kPa

m_A=2kg/s

m_W=0.5kg/s

We know that the energy equation is

\(m_Ah_{Ai}+m_Wh_W=m_Ah_{Ae}+m_Wh_{We}\)

making \(h_{We}\) the subject of formula we have

\(h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})\)

from the saturated water table B.1.1 , corresponding to  \(T_{wi}= 20c\)

\(h_{Wi}=83.94kJ/kg\)

from the ideal gas properties of air table B.7.1 , corresponding to T=1000K

the enthalpy is:

\(h_{Ai}=1046.22kJ/kg\)

from the ideal gas properties of air table B.7.1 corresponding to T=400K

\(h_{Ae}=401.30kJ/kg\)

Step two:

substituting into the equation we have

\(h_{We}=h_{Wi}+\frac{m_A}{mW}(h_A-h_{Ae})\)

\(h_{We}=83.94+\frac{2}{0.5}(2046.22-401.30)\\\\h_{We}=2663.62kJ/kg\)

from saturated water table B.1.2 at \(P_{We}=200kPa\)  we can obtain the specific enthalpy:

\(h_g=2706.63kJ/kg\)

we can see that \(h_g>h_{Wi}\), hence there are two phases

from saturated water table B.1.2 at \(P_{We}=200kPa\)

\(T_{We}=120 ^{\circ} C\)

The correct option for the implication of low variety is B) Flexibility needed. Variety refers to the number of different products, services, or activities that an organization offers. High variety means that a company offers a wide variety of products or services, while low variety means that a company provides a limited range of products or services.

One of the implications of low variety is that customers have limited options. They are restricted to purchasing only those goods and services that the company offers, which may not match their requirements or preferences. This may lead to losing customers to rivals who offer a greater variety of goods and services.

In the case of low variety, it is critical to have the required flexibility. Customers may ask for certain specific goods or services that are not provided by the company. To meet customer requirements, the company should be adaptable and able to rapidly respond to customer requests by expanding its product line. As a result, low variety necessitates the need for flexibility.

The other given options are not valid. Low unit cost is not related to low variety. Unit cost may be affected by economies of scale and other factors, but it is not directly related to variety. High complexity is not related to low variety as well. As the range of goods and services offered by the company increases, the complexity of the company's operations and management also increases. Matching customers' specific needs is partially correct because the company cannot match customer specific needs in low variety. In low variety, the company has a limited range of goods and services, so it is not possible to fulfill all the customers’ specific needs.

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The true power is obtained as 70.7 VA.

We define the power a the rate of doing work, we know that the power in a circuit is the product of the current and the voltage. In this case, we want to find the true power thus we have to involve the use of the phase shift in degrees.

Thus;

True power = PcosΦ

P =  100 VA

Φ = 45 degrees

True power =  100 VA * cos 45 degrees

True power = 70.7 VA

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Answer:

Stop Sign

Explanation:

It's red and shaped like an octogon

Answer:

15psig

Explanation:

Answer:

1: the negative battery terminal

Explanation:

When we design and develop an electrical circuit in electrical and electronics engineering, it is important that we define a reference point for all electronic signals from which we can measure voltage. Although, the voltage passing through a reference point in an electrical circuit is typically 0 Volts.

Also, a ground is a common return path in an electrical circuit for the current flowing through the circuit.

When working on a direct current (DC) circuit that is mainly operated using a battery, the negative terminal of the battery would be the ground (reference point) for all the components connected to the electrical circuit.

Hence, in the context of electronics, the term "ground" is often used to refer to the negative battery terminal.

Answer: the answer is B and D

Explanation:

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

The work input required to inflate the bubble to a diameter of 2.7- in is 9.12 x 10⁻⁷.

When an object is moved over a specific distance by an external force, the quantity of energy delivered to it is known as work.

Change in surface area

4π [2.7² - 2.4]

4π [7.29 -5.76]

19.22 m² = 19.22 x 0.083 = 0.132

Work done = surface tension x change into area

T x 2A

0.0027 x 2 x 0.132 lbf

0.00071 lbt. H

0.0007 x 0.00128 Bt

9.12 x 10⁻⁷

Therefore, the work input required to inflate the bubble to a diameter of 2.7- in is 9.12 x 10⁻⁷.

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Assuming that the electrons in a material follow the Fermi-Dirac distribution function and  EF​ is 0.3eV below EC​, the temperature at which the probability of an electron occupying an energy state at E=(EC​+0.025)eV is 8×\(10^{-6}\) is 271.74 K.

A mathematical depiction of the probability distribution of the energy of the quantum states that electrons can reside in at a specific temperature is the Fermi-Dirac probability function.

It explains what happens to the electrons within metal solids as their temperature rises. The Fermi Level Formula is - k=1.38×10-23 J/K=8.62×10-5 eV/K. The detailed solution is attached below.

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The average shear stress on the pin at position A is roughly 30552 psi.

What is the formula for maximal shear stress? V*Q / I*b is the formula for computing maximum shear stress. Simply expressed, this is 12*V / 8bh. Shear force, cross-sectional breadth, and height are all denoted by the letters V, B, and h respectively.

In terms of stress (elastic modulus – strain). In terms of stress (elastic modulus – strain). A dimensional study of this connection reveals that the elastic modulus has the same physical unit as stress since strain has no dimensions.

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Answer:

1) The bending moment diagram of the shaft is shown in Figure 1. The weakest cross section A is located at the point where the bending moment is maximum.

2) The weakest point on cross section A is located at the point where the bending moment is maximum.

3) The von-Mises stress at the weakest point is given by:

σ = M/I

where M is the bending moment and I is the moment of inertia of the cross section.

4) The factor of safety n is given by:

n = Sy/σ

where Sy is the yield strength of the shaft and σ is the von-Mises stress at the weakest point.

Explanation:

Hope this helps!

Answer: The correct way to achieve proper drive pinion gear bearing preload is to torque the pinion nut to the manufacturer’s specifications and then check for preload.

Explanation: Typical starting points for torque specifications are 55 lb-ft for 3/8-inch bolts/nuts, 75 lb-in for 7/16-inch bolt/nuts, and 125 lb-in for 1/2-inch nuts/bolts. Once the nut is properly torqued, check for preload by tapping the end of the pinion gear with a punch to ensure the races are fully seated, shake it to evaluate for excessive play.

Pinion bearings are bearings that support the weight of the drive shaft and transfer power from the differential to the wheels. Preload is the tension that is put on a bearing. In the case of pinion bearings, preload takes place in the form of a collapsible spacer that stays between the bearings. The required amount of tension is obtained by adjusting the space manually.

Other ways to achieve proper drive pinion gear bearing preload include using a crush sleeve or shims. Crush sleeves are used in place of spacers and are designed to be crushed when the pinion nut is tightened. This crushes the sleeve and creates tension on the bearings. Shims can also be used to adjust the space between bearings.

Common problems with drive pinion gear bearing preload include over-tightening or under-tightening of the pinion nut, which can lead to damage or failure of the bearings. It is important to follow manufacturer specifications when torquing the pinion nut and checking for preload.

Hope this helps, and have a great day!

Answer: Both A and B

Explanation: because they are both true

9514 1404 393

Answer:

  (x, y) = (5.76, 1 5/7)

Explanation:

The location of the centroid in the x-direction is the ratio of the first moment of area about the y-axis to the total area. Similarly, the y-coordinate of the centroid is the first moment of area about the x-axis, divided by the area.

For the moment about the y-axis, we can define a differential of area as ...

  dA = (y2 -y1)dx

where y2 = √(x/k2) and y1 = k1·x^3

The distance of that area from the y-axis is simply x.

So, the x-coordinate of the centroid is ...

  \(\displaystyle c_x=\frac{a_x}{a}=\frac{\int{x\cdot dA}}{\int{dA}}\\\\a_x=\int_0^{12}{x(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{5k_2^{1/2}}\cdot12^{5/2}-\frac{k_1}{5}12^5\\\\a=\int_0^{12}{(k_2^{-1/2}\cdot x^{1/2}-k_1x^3)}\,dx=\frac{2}{3k_2^{1/2}}\cdot12^{3/2}-\frac{k_1}{4}12^4\\\)

For k1 = 4/12^3 and k2=12/4^2, these evaluate to ...

  \(a_x=115.2\\a=20\\c_x=5.76\)

The y-coordinate of the centroid requires we find the distance of the differential of area from the x-axis. We can use (y2 +y1)/2 for that purpose. Then the y-coordinate is ...

  \(\displaystyle c_y=\frac{a_y}{a}\\\\a_y=\int_0^{12}{(\frac{y_2+y_1}{2}(y_2-y_1))}\,dx=\frac{1}{2}\int_0^{12}{(\frac{x}{k_2}-(k_1x^3)^2)}\,dx\\\\a_y=\frac{12^2}{4k_2}-\frac{k_1^212^7}{14}=\frac{240}{7}\\\\c_y=\frac{12}{7}\approx1.7143\)

The centroid of the shaded area is ...

  (x, y) = (5.76, 1 5/7)

Answer:

\(X_t=2.17391304*10^{-4}\)

\(X_r=2.89855072*10^{-4}\)

\(e_t=0.0026\)

\(e_r=0.0035\)

Explanation:

From the question we are told that:

Dimension \(12*12\)

Thickness \(l_t=5mm=5*10^-3\)

Normal tensile force on top side \(F_t= 15kN\)

Normal tensile force on right side  \(F_r= 20kN\)

Elastic modulus, \(E=115Gpap=>115*10^9\)

Generally the equation for Normal Strain X is mathematically given by

 \(X=\frac{Force}{Area*E}\)

Therefore

For Top

 \(X_t=\frac{Force_t}{Area*E}\)

Where

 \(Area=L*B*T\)

 \(Area=12*10^{-2}*5*10^{-3}\)

 \(Area=6*10^{-4}\)  

 \(X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}\)

 \(X_t=2.17391304*10^{-4}\)

For Right side\(X_r=\frac{Force_r}{Area*E}\)

Where

Area=L*B*T

 \(Area=12*10^{-2}*5*10^{-3}\)

 \(Area=6*10^{-4}\)  

 \(X_r=2.89855072*10^{-4}\)

 \(X_r=2.89855072*10^{-4}\)

Generally the equation for elongation is mathematically given by

 \(e=strain *12\)

For top

 \(e_t=2.17391304*10^{-4}*12\)

 \(e_t=0.0026\)

For Right

 \(e_r=2.89855072*10^{-4} *12\)

 \(e_r=0.0035\)

Answer:

We well know that water is a conductor of charges. So, when it is between capacitor plates, the charges flow from positive plate to negative plate hence discharge occurs

Explanation:

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Both Technician A and Technician B are correct. An analog oscilloscope can store the waveform for viewing later, and the trigger level must be set on most scopes to be able to view a changing waveform.

An oscilloscope is a device used to measure and analyze electrical signals in oscillatory form. It consists of a display, vertical and horizontal amplifiers, and a trigger. The trigger is used to capture a specific point of the waveform and synchronize it with the display. It is important to set the trigger level correctly in order to get an accurate reading of the waveform.

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The starting workpiece in steel hot rolling of plate and sheet stock is (d) slab, as it provides the most suitable form for this process.

The steel hot rolling process involves converting a starting workpiece into a thinner and longer sheet or plate stock. This process is widely used in industries such as automotive, construction, and manufacturing. The starting workpiece in steel hot rolling can vary depending on the desired final product. However, there is one best answer to the question of which starting workpiece is used for plate and sheet stock. The correct answer is (d) slab. Slabs are rectangular blocks of steel that have been casted in a mold and allowed to cool. They are then reheated and rolled into thinner sheets or plates using a hot rolling mill.

In summary, the starting workpiece for steel hot rolling of plate and sheet stock is a slab. This is an important step in the manufacturing process that allows for the creation of high-quality steel products used in various industries.

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Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

\(CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\\)

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in \(CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles\)

Calculating the Moles in \(HCl = 0.0200 \ moles\)

Calculating the volume of \(HCl\):

\(\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\\)

                \(= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\\)

Calculating the reaction among the acid and base:

\(CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-\)

\(0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200\)

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of \(CH_3NH_3^+\).

Total volume\(= 100 + 200 = 300\ mL = 0.300\ L\)

\([CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M\)

Using the ICE table

\(CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+\)

\(I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}\)

Calculating \(K_a\) from \(K_b\)

\(\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\\)

                           \(= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}\)

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

\(\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\\)

We have the formula to calculate pH.

\(\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91\)

The pH at the equivalent point for this titration is "5.91".

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