eturpin1
  • 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.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
eturpin1
  • eturpin1
@Michele_Laino
Michele_Laino
  • 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?
eturpin1
  • eturpin1
q=3.75?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

eturpin1
  • 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.
Michele_Laino
  • Michele_Laino
I got q=10.9 cm
Michele_Laino
  • Michele_Laino
Second question: what is the initial potential energy?
Michele_Laino
  • Michele_Laino
@eturpin1
eturpin1
  • eturpin1
IPE=1/2mv^2, is that correct?
Michele_Laino
  • Michele_Laino
no, initial potential energy is: \[\large PE = mgh = 800 \times 9.81 \times 95 = ...\]
eturpin1
  • eturpin1
PE at the top of the first hill = 745560
Michele_Laino
  • Michele_Laino
ok! Now what is the initial kinetic energy?
eturpin1
  • eturpin1
kinetic energy is zero?
Michele_Laino
  • Michele_Laino
that's right!
eturpin1
  • eturpin1
Great, so for the PE at the bottom of the first hill, would it be 800*9.81*31?
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
whe it reaches the height of 31 meters, what is the potential energy?
Michele_Laino
  • Michele_Laino
@eturpin1
eturpin1
  • eturpin1
do i still use the same formula?
Michele_Laino
  • Michele_Laino
yes!: \[PE = mgh = 800 \times 9.81 \times 31 = ...\]
eturpin1
  • eturpin1
equals 243288 at the bottom of the hill?
Michele_Laino
  • Michele_Laino
ok! so the new kinetic energy is: \[KE = 745560 - 243288 = ...Joules\]
eturpin1
  • eturpin1
equals 502272 J?
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
, when we are at the top of the second hill, what is the kinetic energy?
Michele_Laino
  • Michele_Laino
\[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 800 \times {28^2}\]
eturpin1
  • eturpin1
equals 313600 J
Michele_Laino
  • Michele_Laino
ok! so the corresponding potential energy is: \[PE = 745560 - 313600 = ...Joules\]
eturpin1
  • eturpin1
431960 J
Michele_Laino
  • Michele_Laino
correct!
eturpin1
  • eturpin1
and after the coaster comes to a rest at the end, do KE and PE both equal zero?
Michele_Laino
  • Michele_Laino
when we are at ground level, the enrgy of the roller coaster is all kinetic energy and it's value is: \[KE = 745560Joules\]
eturpin1
  • eturpin1
okay, thanks so much! Do you think you could look at that earlier question with me?
eturpin1
  • 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.
Michele_Laino
  • Michele_Laino
as I written the requested distance q is 10.9 cm, here is the computation: \[\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{8} - \frac{1}{{30}} = \frac{{22}}{{240}}\]
eturpin1
  • eturpin1
so this 10.9 equals the distance of the image?
Michele_Laino
  • Michele_Laino
so we have: \[q = \frac{{240}}{{22}} = \frac{{120}}{{11}}cm\] |dw:1433353685174:dw|
Michele_Laino
  • Michele_Laino
yes! that's right!
eturpin1
  • eturpin1
okay! so the image would be inverted since it is a convex lens, correct?
Michele_Laino
  • Michele_Laino
yes! correct!
eturpin1
  • eturpin1
wonderful! thank you a billion, you are always such a great help! Best wishes with everything! (:
Michele_Laino
  • Michele_Laino
thanks!!!!! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.