Height of the image of my friend on the film

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The discussion revolves around verifying the calculations related to the height of an image in a film. The original poster doubts their answer, specifically questioning the conversion of a 29 m distance into centimeters and subsequently into millimeters. Participants emphasize the importance of accurate conversions and suggest re-evaluating the calculation of the image distance (d_i). It's noted that a crucial step may have been overlooked in the calculations. Accurate conversions and careful checking of all steps are essential for obtaining the correct height of the image.
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Homework Statement
You have a movie camera with a single lens of focal length 98 mm. You would like to take a picture of a friend who is standing 29 m away. Your friend is 155 cm tall, what is the height of the image of your friend on the film in mm?
Relevant Equations
1/f+1/di+1/do
1744710442452.png

Hi, I just want to check if my working is correct cuz I seems to get the wrong answer in my first attempt by inserting 0.054mm as the heigh.... the height seems a bit small to me.
 
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Check again your conversion of the 29 m distance into cm.
And make sure you get the distance image in the last step.
 
nasu said:
Check again your conversion of the 29 m distance into cm.
All conversions were to mm.
syllll_213 said:
View attachment 359892
Check your calculation of ##d_i##. There is something you forgot to do.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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