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anonymous
 one year ago
Would you proof read due to my absentminded mind
anonymous
 one year ago
Would you proof read due to my absentminded mind

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Convert 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01. 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.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks guys appreciate it to the Kingdom Come

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1would 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 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ops I guess I had to change 20 ohms to 18 ohms since time wasn't used at this timeXD

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1yes that would be 18 ohms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It says "using the scientific notation" whereas 1.25kV doesn't do as instructed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03900 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....

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1hmm...then I would suggest 1.25 * 10^0 kV

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think that would be more accurate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's your taken on rounding up till 4000C instead of 3900C? I think it's valid.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When a number with only one sig fig is used I can't assign the final answer any more than intermediate numbers had

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually the least of all

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1Your reasoning looks good. Keep it as 4000C :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Calculations regarding other units are correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wasn't entirely sure of the coulombs unit and other units stuff.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1yep, all others are correct. and for the last one, have you considered answering it to 3.9 kC ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. I have. But I am not sure if sig figs of 2 would be appropriate once even the zeros go beyond decimal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For instance I could have used 4.0kC but doing so would imply I consider 0 as another sig fig which is not very good

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So maybe I should say 4kC

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1or perhaps write it 3.9kC first and then round it to 4 kC yes

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1rounding 3900 to 4000 looks odd, rounding 3.9 to 4 seems ok, though you're doing the same thing :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would come off more intuitive;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Markerfriendly I guessXD
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