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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.

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The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc

balandron [24]

Assuming the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°) + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

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2 years ago

Fast plz-The mirror check may involve ______________.

barxatty [35]

Answer:

Realigning the mirror

Explanation:

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2 years ago

For each topic, find the total number of blurts that were analyzed as being related to the topic. Order the result by topic id.

photoshop1234 [79]

Answer:

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3 years ago

1. A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions

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Answer:

Maximum number of vehicle = 308

Explanation:

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6 0

3 years ago

A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a

OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS A B C

Benzene 1 4.72583 1660.652 -1.461

Toluene 2 4.07827 1343.943 -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5% = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis for all components

So for Benzene ; p × y1=x1 × p1s ------let this be equation 1

for Toluene ; p × y2=x2 × p2s ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0

2 years ago