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Andreyy89
3 years ago
11

What’s another name for a service overcurrent device?

Engineering
1 answer:
alexandr402 [8]3 years ago
4 0
Overcurrent protective devices, or OCPDs
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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
Whats a meat trimmer
baherus [9]

Answer:

trime the meat

Explanation:

4 0
3 years ago
If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b
zalisa [80]

Answer:

h=32.1 km

Explanation:

<em>solution:</em>

using newton law of gravitational attraction and newton second law:

W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\

m_{E}= mass of earth

r= distance between two masses

at sea level

a=g

r=R_{E}

a=\frac{Gm_{E}}{r^{2}}.............................(1)

Gm_{E} =gR_{E}^2.........................(2)

by substituting (2) and (1) a=g\frac{R_{E}^2 }{r^{2} } acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0
3 years ago
A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much
Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0
3 years ago
Kris and James are working at a construction site that has a significant amount of stagnant water. Which type of hazard are they
Mashutka [201]
The answer is most likely Biological because insects and other organisms thrive in stagnant water.
7 0
4 years ago
Read 2 more answers
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