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For (a) I know the image is virtual and upright.
For (b) I calculated the focal length to be: -0.2 m
1/f = 1/do + 1/di
do (object) : 0.20 m
di (image) : -0.1 m ---> 0.20 m (1/2) = 0.1 m (Did I do thatright? I also made it negative.)
1/f = 1/0.20m + 1/-0.1 m ---> f = 1/-5 = -0.2 m
For (c) I need help on! Any help would be appreciated!Thanks!
I understood all the steps that were used to integrate andsolvethis problem. However, my initial idea was to solve thisusingθf = θi + vt+1/2at2. This shouldwork outto θf = 0 + 0 + 1/2(10 +6t)t2. Icalculated this to be 272, which is clearlydifferent than theintegrated result. Am I missing somethingobvious?
54. Line 2 (A) for (int i = 1; i>=height/2; i+=10) { (B) for (int i = height; i>0; i--) { (C) for (int i = 1; i
