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3 years ago

What is productivity as it relates to labor?

Engineering

1 answer:

Anastasy [175]3 years ago

6 0

Answer:

7th A good time for you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you

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Please help with my economics problem

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Answer:

You first get a new job, and make a new company and then by amazon to traumatize Jeff Bezos after his divorce

Explanation:

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3 years ago

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3 years ago

un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti

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3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j

Anna [14]

Answer:

the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

Explanation:

Given that;

volume of cut = 25,100 m³

Volume of dry soil fill = 23,300 m³

Weight of the soil will be;

⇒ 93% × 18.3 kN/m³ × 23,300 m³

= 0.93 × 426390 kN 3

= 396,542.7 kN

Optimum moisture content = 12.9 %

Required amount of moisture = (12.9 - 8.3)% = 4.6 %

So,

Weight of water required = 4.6% × 396,542.7 = 18241 kN

Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

6 0

3 years ago