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3 years ago

How are project deliverables determined?

Engineering

2 answers:

Greeley [361]3 years ago

7 0

Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

It should be within the development of the work.
The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.

So that the above seems to be the right answer.

Alex17521 [72]3 years ago

7 0

It is determined by:

For any output to be classified as a “deliverable” within a project, it has to meet a few criteria: It must be within the scope of the project. Stakeholders external or internal must agree to it. It must be the result of deliberate work.

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3 years ago

Find values of the intrinsic carrier concentration n for silicon at –70° 0° 20° C, 100° C, and C. At 125° each temperature, what

Dominik [7]

Answer:

Part (i) at –70° C, intrinsic carrier concentration of silicon is 2.865 x 10⁵ carriers/cm³ and fraction of the atoms ionized is 5.37 x 10⁻¹⁸

Part (ii) at 0° C, intrinsic carrier concentration of silicon is 1.533 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 3.067 x 10⁻¹⁴

Part (iii) at 20° C, intrinsic carrier concentration of silicon is 8.652 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 1.731 x 10⁻¹³

Part (iv) at 100° C, intrinsic carrier concentration of silicon is 1.444 x 10¹² carriers/cm³ and fraction of the atoms ionized is 2.889 x 10⁻¹¹

Part (iv) at 125° C, intrinsic carrier concentration of silicon is 4.754 x 10¹² carriers/cm³ and fraction of the atoms ionized is 9.508 x 10⁻¹¹

Explanation:

$ni^2 = BT^3(e^{\frac{-E_g}{KT}})\\\\ni = \sqrt{ BT^3(e^{\frac{-E_g}{KT}})}$

where;

B = 5.4 x 10⁻³¹

Eg = 1.12 ev

K = 8.62 x 10⁻⁵ eV/K

T = (273 + ⁰C) K

Number of atoms in silicon crystal = 5 x 10²² atoms/cm³

Part (i) For –70° C, T = (273 -70 ⁰C)K = 203 K

$ni = \sqrt{ 5.4*10^{31}*203^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*203}})}} \ =2.685*10^5 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{2.685*10^5}{5 *10^{22}} = 5.370 *10^{-18}$

Part (ii) For 0° C, T = (273 +0 ⁰C)K = 273 K

$ni = \sqrt{ 5.4*10^{31}*273^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*273}})}} \ =1.533*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.533*10^9}{5 *10^{22}} = 3.067 *10^{-14}$

Part (iii) For 20° C, T = (273 + 20 ⁰C)K = 293 K

$ni = \sqrt{ 5.4*10^{31}*293^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*293}})}} \ =8.652*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{8.652*10^9}{5 *10^{22}} = 1.731 *10^{-13}$

Part (iv) For 100° C, T = (273 + 100 ⁰C)K = 373 K

$ni = \sqrt{ 5.4*10^{31}*373^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*373}})}} \ =1.444*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.444*10^{12}}{5 *10^{22}} = 2.889 *10^{-11}$

Part (v) For 125° C, T = (273 + 125 ⁰C)K = 398 K

$ni = \sqrt{ 5.4*10^{31}*398^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*398}})}} \ =4.754*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{4.754*10^{12}}{5 *10^{22}} = 9.508 *10^{-11}$

7 0

3 years ago

One reason that driving lends itself to aggressive behavior is:

Maurinko [17]

Answer:

C. Drivers feel empowered by anonymity

Explanation:

Sorry for the late answer hope it helps you and others who need it.

STay Safe and healthy!

5 0

3 years ago

Read 2 more answers