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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.
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1Second question: what is the initial potential energy?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0IPE=1/2mv^2, is that correct?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, initial potential energy is: \[\large PE = mgh = 800 \times 9.81 \times 95 = ...\]

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0PE at the top of the first hill = 745560

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! Now what is the initial kinetic energy?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0kinetic energy is zero?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0Great, so for the PE at the bottom of the first hill, would it be 800*9.81*31?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1whe it reaches the height of 31 meters, what is the potential energy?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0do i still use the same formula?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes!: \[PE = mgh = 800 \times 9.81 \times 31 = ...\]

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0equals 243288 at the bottom of the hill?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! so the new kinetic energy is: \[KE = 745560  243288 = ...Joules\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1, when we are at the top of the second hill, what is the kinetic energy?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 800 \times {28^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! so the corresponding potential energy is: \[PE = 745560  313600 = ...Joules\]

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0and after the coaster comes to a rest at the end, do KE and PE both equal zero?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1when we are at ground level, the enrgy of the roller coaster is all kinetic energy and it's value is: \[KE = 745560Joules\]

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0okay, thanks so much! Do you think you could look at that earlier question with me?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0An 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
 one year ago
Best ResponseYou've already chosen the best response.1as 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
 one year ago
Best ResponseYou've already chosen the best response.0so this 10.9 equals the distance of the image?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we have: \[q = \frac{{240}}{{22}} = \frac{{120}}{{11}}cm\] dw:1433353685174:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! that's right!

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0okay! so the image would be inverted since it is a convex lens, correct?

eturpin1
 one year ago
Best ResponseYou've already chosen the best response.0wonderful! thank you a billion, you are always such a great help! Best wishes with everything! (:
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