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eturpin1

  • one year ago

An object is placed 30.0 cm to the left of a convex lens with a focal length of +8.0 cm. Draw a ray diagram showing the location of the image, and calculate the distance of the image from the lens.

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  1. eturpin1
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    we have the subsequent drawing: |dw:1433350605642:dw| we have to apply this formula: \[\Large \frac{1}{p} + \frac{1}{q} = \frac{1}{f}\] where p= 30 cm, f= 8 cm so what is q?

  3. eturpin1
    • one year ago
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    q=3.75?

  4. eturpin1
    • one year ago
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    Oh, I've posted the wrong question, do you mind looking at this one? An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level. Draw a picture of the roller coaster drop and calculate the kinetic and potential energy at the top and bottom of each hill.

  5. Michele_Laino
    • one year ago
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    I got q=10.9 cm

  6. Michele_Laino
    • one year ago
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    Second question: what is the initial potential energy?

  7. Michele_Laino
    • one year ago
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    @eturpin1

  8. eturpin1
    • one year ago
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    IPE=1/2mv^2, is that correct?

  9. Michele_Laino
    • one year ago
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    no, initial potential energy is: \[\large PE = mgh = 800 \times 9.81 \times 95 = ...\]

  10. eturpin1
    • one year ago
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    PE at the top of the first hill = 745560

  11. Michele_Laino
    • one year ago
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    ok! Now what is the initial kinetic energy?

  12. eturpin1
    • one year ago
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    kinetic energy is zero?

  13. Michele_Laino
    • one year ago
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    that's right!

  14. eturpin1
    • one year ago
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    Great, so for the PE at the bottom of the first hill, would it be 800*9.81*31?

  15. Michele_Laino
    • one year ago
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    so the total mechanical energy, which is a constant quantity of the motion, is: \[\large E = PE + KE = 745,560Joules\] please wait a moment

  16. Michele_Laino
    • one year ago
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    whe it reaches the height of 31 meters, what is the potential energy?

  17. Michele_Laino
    • one year ago
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    @eturpin1

  18. eturpin1
    • one year ago
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    do i still use the same formula?

  19. Michele_Laino
    • one year ago
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    yes!: \[PE = mgh = 800 \times 9.81 \times 31 = ...\]

  20. eturpin1
    • one year ago
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    equals 243288 at the bottom of the hill?

  21. Michele_Laino
    • one year ago
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    ok! so the new kinetic energy is: \[KE = 745560 - 243288 = ...Joules\]

  22. eturpin1
    • one year ago
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    equals 502272 J?

  23. Michele_Laino
    • one year ago
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    correct!

  24. Michele_Laino
    • one year ago
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    , when we are at the top of the second hill, what is the kinetic energy?

  25. Michele_Laino
    • one year ago
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    \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 800 \times {28^2}\]

  26. eturpin1
    • one year ago
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    equals 313600 J

  27. Michele_Laino
    • one year ago
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    ok! so the corresponding potential energy is: \[PE = 745560 - 313600 = ...Joules\]

  28. eturpin1
    • one year ago
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    431960 J

  29. Michele_Laino
    • one year ago
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    correct!

  30. eturpin1
    • one year ago
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    and after the coaster comes to a rest at the end, do KE and PE both equal zero?

  31. Michele_Laino
    • one year ago
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    when we are at ground level, the enrgy of the roller coaster is all kinetic energy and it's value is: \[KE = 745560Joules\]

  32. eturpin1
    • one year ago
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    okay, thanks so much! Do you think you could look at that earlier question with me?

  33. eturpin1
    • one year ago
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    An object is placed 30.0 cm to the left of a convex lens with a focal length of +8.0 cm. Draw a ray diagram showing the location of the image, and calculate the distance of the image from the lens.

  34. Michele_Laino
    • one year ago
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    as I written the requested distance q is 10.9 cm, here is the computation: \[\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{8} - \frac{1}{{30}} = \frac{{22}}{{240}}\]

  35. eturpin1
    • one year ago
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    so this 10.9 equals the distance of the image?

  36. Michele_Laino
    • one year ago
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    so we have: \[q = \frac{{240}}{{22}} = \frac{{120}}{{11}}cm\] |dw:1433353685174:dw|

  37. Michele_Laino
    • one year ago
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    yes! that's right!

  38. eturpin1
    • one year ago
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    okay! so the image would be inverted since it is a convex lens, correct?

  39. Michele_Laino
    • one year ago
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    yes! correct!

  40. eturpin1
    • one year ago
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    wonderful! thank you a billion, you are always such a great help! Best wishes with everything! (:

  41. Michele_Laino
    • one year ago
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    thanks!!!!! :)

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