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cluponka [151]
3 years ago
10

How to build a machine that can create anything

Engineering
1 answer:
jeka943 years ago
5 0

Answer:

by getting a good machine

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What is over head line
CaHeK987 [17]

Explanation:

An overhead power line is a structure used in electric power transmission and distribution to transmit electrical energy across large distances. It consists of one or more conductors (commonly multiples of three) suspended by towers or poles. Since most of the insulation is provided by air, overhead power lines are generally the lowest-cost method of power transmission for large quantities of electric energy.

<h3><em><u>Constr</u></em><em><u>uction</u></em></h3>

Towers for support of the lines are made of wood (as-grown or laminated), steel or aluminum (either lattice structures or tubular poles), concrete, and occasionally reinforced plastics. The bare wire conductors on the line are generally made of aluminum (either plain or reinforced with steel or composite materials such as carbon and glass fiber), though some copper wires are used in medium-voltage distribution and low-voltage connections to customer premises. A major goal of overhead power line design is to maintain adequate clearance between energized conductors and the ground so as to prevent dangerous contact with the line, and to provide reliable support for the conductors, resilience to storms, ice loads, earthquakes and other potential damage causes. Today overhead lines are routinely operated at voltages exceeding 765,000 volts between conductors.

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6 0
4 years ago
20 POINTS
Kobotan [32]

Answer:

True ...................

5 0
3 years ago
Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0
4 years ago
Relationship between civil engineering and geology​
natka813 [3]

Answer:

Geological and civil engineers sometimes work together, but have very different responsibilities. While geological engineers study the earth's inner and outer surface to evaluate potential mining and infrastructure construction sites, civil engineers design the infrastructure to be built.

5 0
3 years ago
Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should w
Setler79 [48]

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

Re_{5} = Re_{12}

\sqrt{\frac{V_{5}XD_{5}  }{v_{5}}}= \sqrt{\frac{V_{12}XD_{12}  }{v_{12}}}

Re_{5} = Reynold number of water pipe

Re_{12} = Reynold number of oil pipe

V_{5} = Velocity of water 5 diameter pipe = ?

V_{12} = Velocity of oil 12 diameter pipe = 2.30

v_{5} = Kinetic Viscosity of water = 1 x 10^{-6} m^{2}/s

v_{12} = Kinetic Viscosity of oil =  4 x 10^{-6} m^{2}/s

D_{5} = Diameter of pipe used for water = 0.05 m

D_{12} = Diameter of pipe used for oil = 0.12 m

Use the formula

\sqrt{\frac{V_{5}XD_{5}  }{v_{5}}}= \sqrt{\frac{V_{12}XD_{12}  }{v_{12}}}

By Removing square rots on both sides

{\frac{V_{5}XD_{5}  }{v_{5}}}= {\frac{V_{12}XD_{12}  }{v_{12}}}

{V_{5}= {\frac{V_{12}XD_{12}  }{v_{12}XD_{5}\\}}xv_{5}

{V_{5}= [ (0.23 x 0.12m ) / (4 x 10^{-6} m^{2}/s) x 0.05 ] 1 x 10^{-6} m^{2}/s

{V_{5} = 1.38 m/s

4 0
4 years ago
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