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anonymous

  • one year ago

A 6.0 cm object is located 20 cm in front of a convex mirror with a radius of curvature equal to 120 cm. What is the size of the image produced by the mirror?

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  1. anonymous
    • one year ago
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    @IrishBoy123

  2. anonymous
    • one year ago
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    i get 0.45 and u ?

  3. anonymous
    • one year ago
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    @IrishBoy123

  4. IrishBoy123
    • one year ago
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    4.5

  5. anonymous
    • one year ago
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    how did u do it?

  6. IrishBoy123
    • one year ago
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    how did you do it? you first....

  7. anonymous
    • one year ago
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    k i found -1.5 for di

  8. anonymous
    • one year ago
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    in the end it did (1.5)(6)/20

  9. anonymous
    • one year ago
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    to get 0.45

  10. IrishBoy123
    • one year ago
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    what is f in this question??!!

  11. IrishBoy123
    • one year ago
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    \(R_c = 120\)

  12. IrishBoy123
    • one year ago
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    so f = ??

  13. anonymous
    • one year ago
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    60

  14. anonymous
    • one year ago
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    -60 since it convex

  15. IrishBoy123
    • one year ago
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    seems we are in agreement - ish (___), cool. for me: \(\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) \(f = -60\) \(d_o = 20 \implies d_i = -15\) \( \large \frac{h_i}{h_0} = -\frac{d_i}{d_o} \) \(h_o = 6\) follows that \(h_i = 4.5\)

  16. anonymous
    • one year ago
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    -1/60-1/20=1/di???

  17. anonymous
    • one year ago
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    @IrishBoy123

  18. IrishBoy123
    • one year ago
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    -1/60 = 1/20 +1/di

  19. anonymous
    • one year ago
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    yea so when the 1/20 come to the other side it becomes negative

  20. anonymous
    • one year ago
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    ohh wait its 1200 not 120

  21. anonymous
    • one year ago
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    omg

  22. anonymous
    • one year ago
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    these question are really easy just these silly mistakes ruin them

  23. anonymous
    • one year ago
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    btw do u come to site often ? i will be needing lots of help in the coming days since i have excam

  24. IrishBoy123
    • one year ago
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    \(\large \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) \(f = - 60\) \(\large -\frac{1}{60} = \frac{1}{20} + \frac{1}{d_i}\) \(\large \frac{1}{d_i} = -\frac{1}{60} - \frac{1}{20} = -\frac{1}{20} (\frac{1}{3} + 1) = -\frac{1}{15}\) \(\large d_i = -15\) \(\large \frac{h_i}{h_o} = -\frac{d_i}{d_o}\) etc etc

  25. IrishBoy123
    • one year ago
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    plus, for "lots of help" on Physics, be sure to tag @Michele_Laino Qualified Helper, and brilliant in every sense of the word

  26. Michele_Laino
    • one year ago
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    thanks!! @IrishBoy123

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