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3 years ago

7

What is the technique used during the activities definition process that breaks down the "work units" in a work package into sma

ller and smaller chunks to identify what activities are required to complete that "work chunk"?

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

In project management this technique is called "Logical Dependency Technique"

Explanation:

A dependency highlights the logical relationship between two or more activities or task to be carried out in project management

There are basically four types of dependencies

1. Start to start

2. Finish to start

3. Start to finish

4. Finish to finish

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A private plane pilot is what kind of individual transportation position? professional level mid-level entry-level EPA-certified

gayaneshka [121]

Answer:

A private plane pilot is a mid-level position.

Explanation:

The six pilot certifications in the US are as follows:

Sport Pilot
Recreational Pilot
Private Plane Pilot
Commercial Plane Pilot
Flight Instructor
Airline Transport Pilot

From this listing, it is evident that the Private Plane Pilot is in the mid of the line so it is a mid-level position.

3 0

3 years ago

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

$E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}$

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet $P_1$ = 96 kPa

the pressure at the exit $P_2$ = ???

the pressure $\rho$ = 1000 kg/m³

∴

$0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}$

$0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$\dfrac{800000}{2} = P_2-96000$

400000 = P₂ - 96000

400000 + 96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0

3 years ago

Georgia Tech is committed to creating solutions to some of the world’s most pressing challenges. Tell us how you have improved o

Kaylis [27]

Answer:

Georgia Tech is committed to WGAR 53566 THE ANSWER IS JELLY IS KING AND THE JELLY IS KING AND hope to improve the human condition in your community.

Explanation:

6 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

4 years ago