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tensa zangetsu [6.8K]
3 years ago
8

You must yield the right-of-way to all of the following EXCEPT:

Engineering
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

You must yield the right-of-way to a police car, fire engine, ambulance, or other emergency vehicle that uses a siren and flashing lights. Pull as close to the right of the road as possible and stop until the emergency vehicle(s) has passed. However, don't stop in an intersection because that would cause a wreck.

Explanation:

please have a good day. and also mark as brainllest bye

[o]-[o]

\___/

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It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?
vovangra [49]

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

5 0
4 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%

3 0
3 years ago
What steel type and ASTM designation is preferred for W-shapes?
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Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

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