Physics Questions and Answers

F= 15cm

  M = 3

  M = $\frac{v}{u}$

  V =3u

  $\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

  $\frac{1}{15}$ = $\frac{1}{u}$ - $\frac{1}{3u}$

  $\frac{1}{15}$ =   $\frac{2}{3u}$

U =10cm

  D = u + v

  = 10 +30 = 40cm

V = -6cm

M = 0.6cm

Recall m = $\frac{V}{U}$

U = $\frac{V}{M}$

U = $\frac{6}{0.6}$

 U = 10

  $\frac{1}{U}$ + $\frac{1}{V}$ = $\frac{1}{F}$

 $\frac{1}{10}$ + $\frac{1}{-6}$ = $\frac{1}{F}$

$\frac{-4}{60}$ = $\frac{1}{F}$

 F = $\frac{60}{-4}$

=  -15cm

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The primary reason that power is transmitted at high voltages is to increase efficiency. As electricity is transmitted over long distances, there are inherent energy losses along the way. High voltage transmission minimizes the amount of power lost as electricity flows from one location to the next. How? The higher the voltage, the lower the current. The lower the current, the lower the resistance losses in the conductors. And when resistance losses are low, energy losses are low also. Electrical engineers consider factors such as the power being transmitted and the distance required for transmission when determining the optimal transmission voltage

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