Recent content by mrlevis

Thanks goktr001, but i have just solved it, its actually 44. I summed the area under the graph, but only till the 11sec, i didnt notice that 11.5 is actually 3 quarders of triangle. Problem solved anyway. thanks.

Homework Statement A velocity of a runner is described in the graph below. The runner started from X=0 on the X axis. In what distance from X=0 (beginning) will be the runner, after 11.5 sec. Homework Equations The Attempt at a Solution Please tell me why it isn't 42m.:confused:

ok, thanks for help

oh... you right. i don't know, for some reason i didnt think it would work. but can you explain why is that? is there a simple proof to it?

you mean this?: 1/f=1/U+1/V i used this for the lens 1, and found that V, image distance, is 12, which means it is located 6 to the negative's right, 2 after the focal point. So how can i use the equation for the second lens? U is supposed to be an object, not converging rays.

Homework Statement S is a lighted object, height=1. what is the final image that is created by the system? Homework Equations The Attempt at a Solution i just don't understand how do i find the location of a virtual image created by a negative lens when the rays converge to a...