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pychu [463]
3 years ago
13

A state government is trying to decide what type of energy its state should rely on for electricity in the future. Which of the

following is a benefit associated with the use of coal power that the government should take into consideration?
A.Coal power plants release toxic chemicals into the air
B.Mining for coal hurts the environment
C.Coal power can be generated both day and night
D.Burning coal releases greenhouse gases
Physics
2 answers:
Irina-Kira [14]3 years ago
7 0
Answer choice C is a benefit of coal; all the others are negatives.

Feel free to ask me more questions; I'm happy to help. (Don't forget Brainliest!)
fgiga [73]3 years ago
4 0

Answer: C. Coal power can be generated both day and night.

Explanation:

Coal is an important form of source of energy. It is a non-renewable source of energy. It is obtain by mining of the surface below the earth crust.  Coal plays an important role in generation of electricity worldwide.  

It acts as a raw material for the generation of electricity in both day and night. When the coal is turned into a gas. The gas turbine is hot enough to boil the water that can be used to make the steam. The steam spin another type of turbine so as to generate the electricity.

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From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

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Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

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Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

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=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

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