Recent content by davidepalmer

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    Magnification due to a Diverging Lens

    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.
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    Magnification due to a Diverging Lens

    Alright, so I redid the problem with a negative focal length and got a height of .4cm.
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    Magnification due to a Diverging Lens

    Is it negative since it is to the left of the lens?
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    Magnification due to a Diverging 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...
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    Resistor currect and voltage drop in a circuit.

    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!
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    Resistor currect and voltage drop in a circuit.

    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?
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    Resistor currect and voltage drop in a circuit.

    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...
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