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Oksi-84 [34.3K]
3 years ago
14

PLEASE HELP!! PHYSICS QUESTION-DOES ANYONE KNOW HOW TO DO THIS WITH THE WORKING OUT?!?

Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

1.5 \Omega

Explanation:

The resistance of a piece of wire is given by

R=\frac{\rho L}{A}

where

\rho is the resistivity of the material

L is the length of the wire

A is the cross-sectional area

In this problem, we have

\rho = 3.3\cdot 10^{7} m

L = 7.0 m

The diameter of the wire is 0.14 cm, so the radius is 0.07 cm, therefore the cross sectional area is

A=\pi r^2 = \pi (0.07)^2=0.0154 cm^2 = 0.0154\cdot 10^{-4} m^2

Therefore, the resistance is

R=\frac{(3.3\cdot 10^{-7})(7.0)}{(0.0154\cdot 10^{-4})}=1.5 \Omega

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P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

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hf-32=57.29

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North Dakota Electric Company estimates its demand trend line​ (in millions of kilowatt​ hours) to​ be: D​ = 75.0 ​+ 0.45​Q, whe
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Answer:

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The demand forecast for spring is 145.08 millions KWH

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Explanation:

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D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

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D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

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We need to calculate the demand forecast for spring

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

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We need to calculate the demand forecast for summer

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D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

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We need to calculate the demand forecast for fall

Using given formula

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The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

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