All categories

3 years ago

What is the critical path and time duration?

Engineering

1 answer:

kvasek [131]3 years ago

3 0

The critical path is A-B-C, with a duration of 15 minutes.

<u>Explanation</u>:

The critical path is A-B-C, with a term of 15 minutes. You don't need to be knowledgeable in computer lingo to make sense of this one (as I made sense of it absent a lot of data on that half of the condition). Here's the manner by which I made sense of this.

The main thing I saw was that "predecessor" was utilized, which implies something precedes the other thing. That provided me the primary insight that A precedes B and C, and that B must precede C. Then, I just included the term times for the three ways, which allowed me 15 minutes.

You might be interested in

What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng

skad [1K]

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

6 0

3 years ago

Marie and James are bubbling dry pure nitrogen (N2) through a tank of liquid water (H2O) containing ethane (C2H6). The vapor str

tekilochka [14]

.............................,.,.,.,.,.,.,.,.,.,.,.,.,.,.,,.,.,,,.,,.,.,

3 0

3 years ago

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

Anuta_ua [19.1K]

Answer:

$d=0.414\times 10^{-4}\ m$

Explanation:

Given that

P = 4 KPa

Contact angle = 6°

Surface tension = 1 N/m

Lets assume that atmospheric pressure = 100 KPa

Lets take that density of water = $1000\ kg/m^3$

So the capillarity rise h

$h=\dfrac{\Delta P}{\rho g}$

$h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}$

h= 9.61 m

We know that for capillarity rise h

$h=\dfrac{2\sigma cos\theta }{r\rho g}$

$r=\dfrac{2\sigma cos\theta }{h\rho g}$

$r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}$

$r=0.207\times 10^{-4}\ m$

$d=0.414\times 10^{-4}\ m$

3 0

3 years ago

Why can the human powered generator be the worst option for the rescue team

zimovet [89]

Answer:

Human powered generators are the best energy source for the rescue team as it doesn't require any lengthy time consuming resource but only the power of single human being. Just like fossil fuels, the human power is a kind of renewable energy source that does no harm to the environment and can be used whenever required.

Explanation:

4 0

3 years ago

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

Ugo [173]

Answer:

Computation of the load is not possible because E(test) >E(yield)

Explanation:

We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:

Calculation of E(test is as follows)

E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm

=0.0262

Computation of E(yield) is given below:

E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027

Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).

7 0

3 years ago

Read 2 more answers