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kati45 [8]
3 years ago
9

Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more ele

ctrical resistance it has. Who is correct?
Engineering
1 answer:
shutvik [7]3 years ago
8 0

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

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Tamiku [17]

Technician B is right say that hard water potting i usually jut a Surface problem that can be wahed off.

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The amount of dissolved calcium and magnesium in the water determines its hardness. Calcium and magnesium are the main dissolved minerals in hard water. The last time you washed your hands, you might have actually felt the effects of hard water.

What do you mean by acid rain?

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6 0
2 years ago
1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur
Anit [1.1K]

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, T_L = 36^0 F = 275.372 K

Outside temperature, T_H = 100^0F = 310.928 K

rate of insulation, Q = 100 Btu/h

To get the minimum electrical power required, use the relation below:

\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt

V = 5 V

Power = IV

W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A

If the cooler is supposed to work for 4 hours, t = 4 hours

I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr

Minimum battery size needed = 3.03 Amp-hr

6 0
3 years ago
When plotting a single AC cycle beginning at zero degrees and moving forward in time the cycles negative peak occurs at
Lana71 [14]

Answer:A 270 degrees

Explanation:

4 0
3 years ago
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0
4 years ago
1. A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions
otez555 [7]

Answer:

Maximum number of vehicle = 308

Explanation:

See the attached file for the calculation.

6 0
4 years ago
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