## 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

@Michele_Laino

2. Michele_Laino

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

q=3.75?

4. eturpin1

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

I got q=10.9 cm

6. Michele_Laino

Second question: what is the initial potential energy?

7. Michele_Laino

@eturpin1

8. eturpin1

IPE=1/2mv^2, is that correct?

9. Michele_Laino

no, initial potential energy is: $\large PE = mgh = 800 \times 9.81 \times 95 = ...$

10. eturpin1

PE at the top of the first hill = 745560

11. Michele_Laino

ok! Now what is the initial kinetic energy?

12. eturpin1

kinetic energy is zero?

13. Michele_Laino

that's right!

14. eturpin1

Great, so for the PE at the bottom of the first hill, would it be 800*9.81*31?

15. Michele_Laino

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

whe it reaches the height of 31 meters, what is the potential energy?

17. Michele_Laino

@eturpin1

18. eturpin1

do i still use the same formula?

19. Michele_Laino

yes!: $PE = mgh = 800 \times 9.81 \times 31 = ...$

20. eturpin1

equals 243288 at the bottom of the hill?

21. Michele_Laino

ok! so the new kinetic energy is: $KE = 745560 - 243288 = ...Joules$

22. eturpin1

equals 502272 J?

23. Michele_Laino

correct!

24. Michele_Laino

, when we are at the top of the second hill, what is the kinetic energy?

25. Michele_Laino

$KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 800 \times {28^2}$

26. eturpin1

equals 313600 J

27. Michele_Laino

ok! so the corresponding potential energy is: $PE = 745560 - 313600 = ...Joules$

28. eturpin1

431960 J

29. Michele_Laino

correct!

30. eturpin1

and after the coaster comes to a rest at the end, do KE and PE both equal zero?

31. Michele_Laino

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

okay, thanks so much! Do you think you could look at that earlier question with me?

33. eturpin1

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

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

so this 10.9 equals the distance of the image?

36. Michele_Laino

so we have: $q = \frac{{240}}{{22}} = \frac{{120}}{{11}}cm$ |dw:1433353685174:dw|

37. Michele_Laino

yes! that's right!

38. eturpin1

okay! so the image would be inverted since it is a convex lens, correct?

39. Michele_Laino

yes! correct!

40. eturpin1

wonderful! thank you a billion, you are always such a great help! Best wishes with everything! (:

41. Michele_Laino

thanks!!!!! :)