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3 years ago

Please answer ASAP!!

Engineering

1 answer:

Mrrafil [7]3 years ago

6 0

Answer:

С.) The force acting on the pole exceeded its yield strength.

Explanation:

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Consider a finite wing with an aspect of ratio of 7; the airfoil section of the wing is a symmetric airfoil with an infinite-win

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4 years ago

The author claims that engineers should always have the final say and last word when it comes to setting safety standards. Selec

Lisa [10]

Answer:

I select false because engineers are not the only thing we have, we have scientists doctors mathematicians and much more to give safety standards

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4 years ago

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A steel with 1.0 % carbon is known as which one of the following: (a) eutectoid, (b) hypoeutectoid, (c) hypereutectoid, or (d) w

natta225 [31]

Answer:

Option c)

Explanation:

a.The carbon content of eutectoid steel, which is an carbon- iron alloy, is in between 0.80-0.83%

b. In hypoeutectoid steel, the carbon composition is less than 0.80% and is approximately 0.76%

c. In hypereutectoid steel, the carbon composition ranges from 0.80% - 2.0%

d. In wrought iron, the iron composition is quite low about 0.08%

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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

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4 years ago

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Setler [38]

Answer: ummmm wut?

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3 years ago

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