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Bumek [7]
2 years ago
14

The demand schedules for Jones, Smith, and other buyers are shown in the table below:

Engineering
1 answer:
Solnce55 [7]2 years ago
4 0

The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.

<h3>What is a demand schedule?</h3>

A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.

A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.

Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.

Note that the complete information wasn't found and an overview was given.

Learn more about demand on:

brainly.com/question/1245771

#SPJ1

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A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0
4 years ago
A strong base (caustic or alkali) is added to oils to test for: b. entrained air a)- alkalinity b)- acidity c)- contamination. d
fredd [130]

Answer:

A strong base(caustic or alkali) is added to oils test for stability

option (d) is correct

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7 0
3 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

7 0
4 years ago
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