Recent content by davidepalmer

yay! thank you very much!

So, I redid my equation: 1/[(1/-74cm)-(1/37cm)]. This gave me: -24.667. Then I redid my magnification equation: -(-24.667)/(37). This gave me .667 for my magnification. Then when multiplied by the height i got .4 cm.

Alright, so I redid the problem with a negative focal length and got a height of .4cm.

Is it negative since it is to the left of the lens?

Homework Statement The focal length of the diverging lens is 74 cm. If an object with height h = 0.6 cm is located at d = 37 cm, what is the height of the image? Homework Equations Thin-Lens Equation: (1/do)+(1/di)=1/f Magnification Equation: m=-di/do The Attempt at a...

I think I've finally figured out what you've been explaining to me. I was able to work it backwards and get the answer now. Thank you so so much for your help!

Therefore if I'm able to find the voltage after R1, that voltage is the same through R4 and R23. So with this information I can find the current through R23 and since R2 and R3 are in series they have the same current as R23?

Consider this circuit with R1 = 30 ohm, R2 = 10 ohm, R3 = 10 ohm, R4 = 20 ohm, R5 = 60 ohm. (the circuit is attached) A. What is the current across the resistor R2, if a voltage of 110 Volt is applied to the two terminals at the bottom of the figure? B. What is the voltage drop across R2...