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lesantik [10]
3 years ago
14

What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a

2.00 mm x 2.00 mm square cross section?
Physics
1 answer:
astra-53 [7]3 years ago
8 0

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section areaA = (2.00\times10^{-3})^2\ m

(I). We need to calculate the area of copper wire

Using formula of area

A=\pi r^2

A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}

R=0.0742\ \Omega

(II). We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}

R=1.75\ \Omega

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

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You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive
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f_{police}=1268.7 Hz    

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f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})      (1)

Now, when the police car is moving in front of the car, the frequency detected by the car will be:

f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})        (2)            

By solving equation (1) and equation (2) for v_{police} we have:

v_{police} = 44.67 m/s

Knowing that:

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Finally, we just need to put this value into the first equation to find frequency of the police car.

f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})    

f_{police}=1310(\frac{343-44.7}{343-35})  

f_{police}=1268.7 Hz    

I hope it helps you!

                                                                                                                                       

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