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Mrac [35]
3 years ago
15

The distance from the center of a lens to the location where parallel rays converge or appear to converge is called the _____ le

ngth.
Physics
1 answer:
Morgarella [4.7K]3 years ago
7 0

Answer:

<h2>FOCAL</h2>

Explanation:

<em>The center of a lens is known as its optical center. </em><em>All light rays incident on a particular lens converges at a points a point known as the principal focus or the focal point after reflecting</em><em>. Note that all light incident on a reflecting surface must all converge at this focal point after reflection. </em>

The distance measured from the center of this lens to its principal focus (otherwise known as focal point) is known as the <em>focal length of the lens. </em>

<em>Based on the explanation above, it cam be concluded that the distance from the center of a lens to the location where parallel rays converge or appear to converge is called the</em><em> FOCAL</em><em> length.</em>

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An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
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Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

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