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anonymous

  • one year ago

Would you proof read due to my absentminded mind

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  1. anonymous
    • one year ago
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    Convert the following measurements as indicated using the scientific notation: 750 mV = ? V 1250 µV = ? V 2.5 V = ? mV 0.15 V = ? µV 4.5 kV = ? V 1250 V = ? kV 2. An electric charge Q passes through a section of an electric circuit in three minutes. If the magnitude of the current is 120 mA, calculate the amount of electrical charge Q. 3. What is the resistance of a flashlight bulb if 250 mA flows through it when its 1.5 V is connected across it? 4. A hair dryer draws 6.5 A when plugged into a 120 V DC line. What is its resistance and how much charge passes through it in 10 min?

  2. anonymous
    • one year ago
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    1. a. 750mV=7.5*10^-1V b. 1250 µV=1.25*10^-3V c. 2.5V=2.5*10^3mV d. 0.15V=1.5*10^5 µV e. 4.5kV=4.5*10^3V f 1250V=0.125*10^1kV 2. Since the charge Q passes the electric circuit in 3 minutes, and given the magnitude of the current 120mV, Using the formula I=Q/T Q=TI 3min=180s Q=0.12A*180s=21.6C After adjusting for sig figs Therefore the electric charge is 22C 3. Since 250mA flows through the bulb when 1.5 V is connected across it, Since R=V/I where R is the resistance V is the voltage I is the ampere R=1.5V/0.25A=6.0 Therefore the resistance of this particular circuit is 6.0 Ω 4. The resistance of this circuit is calculated as follows Since electric current=6.5A Voltage=120V R=120V/6.5A=18.461 Adjusting for sig fig of 1, The resistance in this particular circuit is 20 Ω (Since time is not used in calculation) To calculate the charge that passed through in 10 minutes, 10min=600s 6.5A*600s=3900C The amount of charge passed through is 4000 C (because of the 10 min sig fig is 1.)

  3. anonymous
    • one year ago
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    Thanks guys appreciate it to the Kingdom Come

  4. anonymous
    • one year ago
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    Thanks @hartnn ;)

  5. hartnn
    • one year ago
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    would it be better to write f 1250V=0.125*10^1kV as f 1250V=1.25 kV ? Also, why is the 10 min significant figure considered in last problem? whats the problem with 3900C ?? Everything else is correct :)

  6. anonymous
    • one year ago
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    ops I guess I had to change 20 ohms to 18 ohms since time wasn't used at this timeXD

  7. hartnn
    • one year ago
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    yes that would be 18 ohms

  8. anonymous
    • one year ago
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    It says "using the scientific notation" whereas 1.25kV doesn't do as instructed

  9. anonymous
    • one year ago
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    3900 has two sig figs where "10min" was used to calculate so. Since there is one sig fig in 10 min the final answer has to have only 1 sig fig....

  10. hartnn
    • one year ago
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    hmm...then I would suggest 1.25 * 10^0 kV

  11. anonymous
    • one year ago
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    I think that would be more accurate.

  12. anonymous
    • one year ago
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    What's your taken on rounding up till 4000C instead of 3900C? I think it's valid.

  13. anonymous
    • one year ago
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    When a number with only one sig fig is used I can't assign the final answer any more than intermediate numbers had

  14. anonymous
    • one year ago
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    Actually the least of all

  15. hartnn
    • one year ago
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    Your reasoning looks good. Keep it as 4000C :)

  16. anonymous
    • one year ago
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    Great! Thanks;)

  17. hartnn
    • one year ago
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    welcome ^_^

  18. anonymous
    • one year ago
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    Calculations regarding other units are correct?

  19. anonymous
    • one year ago
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    Wasn't entirely sure of the coulombs unit and other units stuff.

  20. hartnn
    • one year ago
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    yep, all others are correct. and for the last one, have you considered answering it to 3.9 kC ?

  21. anonymous
    • one year ago
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    Yeah. I have. But I am not sure if sig figs of 2 would be appropriate once even the zeros go beyond decimal

  22. anonymous
    • one year ago
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    For instance I could have used 4.0kC but doing so would imply I consider 0 as another sig fig which is not very good

  23. anonymous
    • one year ago
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    So maybe I should say 4kC

  24. hartnn
    • one year ago
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    or perhaps write it 3.9kC first and then round it to 4 kC yes

  25. hartnn
    • one year ago
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    rounding 3900 to 4000 looks odd, rounding 3.9 to 4 seems ok, though you're doing the same thing :P

  26. anonymous
    • one year ago
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    That would come off more intuitive;)

  27. anonymous
    • one year ago
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    Marker-friendly I guessXD

  28. hartnn
    • one year ago
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    good luck! :)

  29. anonymous
    • one year ago
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    Thanks so much!

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