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3 years ago

The amount of time an activity can be delayed and yet not delay the project is termed:_________

Engineering

1 answer:

Shalnov [3]3 years ago

6 0

The correct answer is A. Total slack

Explanation:

Projects include multiple tasks or activities; moreover, each activity and the project itself have a specific due date. Besides this, activities can be delayed and this might affect or not the due date of the project. Indeed, some activities can have extra time without any effect on the date of the project, which is known as total slack or total float (delay in activities that does not delay the project.) Additionally, total slack is possible if activities or tasks are flexible, for example, if each task in a project should take 1 day and one of the tasks takes one day and a half, this can be compensated if another task takes less time. According to this, the correct answer is A.

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Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive

Kazeer [188]

The answer for the ideal mechanical advantage and actual mechanical advantage for the different scenarios are;

A) Ideal Mechanical Advantage = 18.67

B) Actual Mechanical Advantage = 4.1067

We are given;

Input distance; The diameter of the wheel; d_w = 14 inches

Output distance; The diameter of the axle that drives the wheel; d_a = 3/4 inches

The force needed to cut a one-inch-thick softwood board; F = 1.75 pounds

The efficiency of the band saw; η = 22% = 0.22

A) Formula for Mechanical advantage is;

M.A = Force output/Force input = (Input distance)/(Output distance)

Thus;

Ideal mechanical advantage = 14/(3/4)

Ideal mechanical advantage = 18.67

B) Now, we are given that efficiency of the band saw is η = 22% = 0.22.

Thus using the mechanical advantage formula above;

Actual mechanical advantage = 0.22 × Expected output

Actual mechanical advantage = 0.22 × 18.67

Actual mechanical advantage ≈ 4.1067

Read more about Mechanical Advantage at; brainly.com/question/18345299

5 0

2 years ago

A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i

Gre4nikov [31]

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F ) = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

<em>attached below is a detailed solution</em>

3 0

2 years ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

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3 years ago

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Kobotan [32]

Explanation:

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2 years ago

3) Explain how dc machines Can work as motor and generator

weeeeeb [17]

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1 year ago