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Strike441 [17]
3 years ago
9

Briefly explain how an electric motor works.

Physics
1 answer:
NeTakaya3 years ago
3 0
You put electricity<span> into it at one end and an </span>axle<span> (metal rod) rotates at the other end giving you the power to drive a machine of some kind. 


</span>

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200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?
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That's "<em><u>insolation</u></em>" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

           (200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .

Note that this is the intensity of the <em><u>incident</u></em> radiation.  It doesn't say anything
about how much soaks in or how much bounces off.

Wait ! 
I just looked back at the choices, and realized that I didn't answer the question
at all.  I have no idea what  "1 sun"  means.  Forgive me.  I have stolen your
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Wait again !
I found it, through literally several seconds of online research.

           1 sun = 1 kW/m².

So 2/3 of a kW per m²  =  2/3 of 1 sun

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I feel better now, and plus, I learned something.


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3 years ago
Compare and contrast climate and weather
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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
Brut [27]

Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

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