Asmall lightbulb is 1.06 m from a screen. a) if you have a convex lens with 20 cm focal length, where are the two lens locations that will project an image of the lightbulb onto the screen?
b) what’s the magnification in the first case (smaller distance from the lightbulb)?
c) what’s the magnification in the second case (larger distance from the lightbulb)?

A)Explanation:

Let object distance be u.

Image distance v = 1.06 - u

Focal length = 20 cm = .2 m

Applying lens formula

1 / v - 1 / u = 1 / f

1 /( 1.06-u)  - 1 / u = 1 / .2

5 u² -3.3u - 1.06 =0

It will have two roots

u₁ = .1636 m, u₂ = .8964 m

B ) Magnification

= u = .1636m , v = .8964m

m = .8964 / .1636 = 5.48

C ) When u = .8964m ,v = .1634m

m = .1634 / .8964

=.182

the answer is 46 mls because when you add the speed per second it increses the speed

! 55 !  

write a paragraph or story to explain the transformation between potential and kinetic energy as shown in your picture or model.  

you should have at least 5 complete sentences with a main idea and supporting details.  

you must include the following words:  

potential energy

kinetic energy

mechanical energy

conservation of energy

dude, this is hard lol : ))