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3 years ago

Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address

Engineering

1 answer:

Arada [10]3 years ago

5 0

Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

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Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

$COP_{i} = \frac{250\,K}{300\,K-250\,K}$

$COP_{i} = 5$

The real Coefficient of Performance is:

$COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}$

$COP_{r} = 12.571$

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0

3 years ago

Illustrate the crowbar protection for silicon controlled rectifier

julsineya [31]

Answer:

The SCR over voltage crowbar or protection circuit is connected between the output of the power supply and ground. ... It also clamps the gate voltage at ground potential until the Zener turns on. The capacitor C1 is present to ensure that short spikes to not trigger the circuit.

4 0

3 years ago

I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th

SIZIF [17.4K]

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

4 0

3 years ago

What can employers do to help ensure safety in a confined space? Employers can:

Lera25 [3.4K]

Answer:

option 3 is the correct answer

5 0

3 years ago