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svetoff [14.1K]

3 years ago

9

A power of 100 kW (105 W) is delivered to the other side of a city by a pair of power lines, between which the voltage is 12,000

V. (a) Use the formula P 5 IV to show that the current in the lines is 8.3 A. (b) If each of the two lines has a resistance of 10 Ω, show that there is a 83-V change of voltage along each line.

1 answer:

vivado [14]3 years ago

3 0

Answer:

I = 8.3 Amp

potential drop = 83 V

Explanation:

Power = 100 KW

V = 12,000 V

R = 10 ohms

a)

Calculate current I in each wire:

P = I*V

I = P / V

I = 100 / 12 = 8.333 A

b)

Calculate potential drop in each wire:

V = I*R

V = (8.3) * (10)

V = 83 V

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Explanation:

<u>(a)</u>

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The expression for thermal conductivity is:

$q=-\kappa \bigtriangledown T$

Where,

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$\bigtriangledown T$ is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

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<u>(c)</u>

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Where,

$\alpha$ is thermal diffusivity

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3 years ago

Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type co

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Answer:

I am Providing Answer in C Language Program.

Explanation:

Please find attachment regarding code of taking two numbers input and adding them.

I would like to recommend you please use software which supports C language.

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For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

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Answer:

135 hour

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It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

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So $\frac{x_1^2}{t}=constant$

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3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

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$l = \sqrt{1\,km^{2}}$

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$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

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$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

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$\Delta t_{min} = \frac{1}{3000}\,h$

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$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

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3 years ago

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Answer:

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