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4 years ago

Which of the following activities will allow an engineer to determine that a building's design is sound?

1 answer:

3 0

Please add more details because I don’t know if you are using a book or passage, therefore I cannot help you unless you add more detail

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A buzzerd

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The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of

weeeeeb [17]

Answer:

$P_{pump}= 68600Pa$

Explanation:

We understand that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

In this way,

$P_{tank} + P_{pump} = P_{nozzle}$

For the pressure we also know that it is given by

$P=\rho gh$

$\rho gh_1 + P_{pump} = \rho gh_2$

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3 years ago

The A-36 steel drill shaft of an oil well extends 12 000 ft into the ground. Assuming that the pipe used to drill the well is su

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3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

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3 years ago