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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?
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.)
Thanks guys appreciate it to the Kingdom Come
Thanks @hartnn ;)
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 :)
ops I guess I had to change 20 ohms to 18 ohms since time wasn't used at this timeXD
yes that would be 18 ohms
It says "using the scientific notation" whereas 1.25kV doesn't do as instructed
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....
hmm...then I would suggest 1.25 * 10^0 kV
I think that would be more accurate.
What's your taken on rounding up till 4000C instead of 3900C? I think it's valid.
When a number with only one sig fig is used I can't assign the final answer any more than intermediate numbers had
Actually the least of all
Your reasoning looks good. Keep it as 4000C :)
Calculations regarding other units are correct?
Wasn't entirely sure of the coulombs unit and other units stuff.
yep, all others are correct. and for the last one, have you considered answering it to 3.9 kC ?
Yeah. I have. But I am not sure if sig figs of 2 would be appropriate once even the zeros go beyond decimal
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
So maybe I should say 4kC
or perhaps write it 3.9kC first and then round it to 4 kC yes
rounding 3900 to 4000 looks odd, rounding 3.9 to 4 seems ok, though you're doing the same thing :P
That would come off more intuitive;)
Marker-friendly I guessXD
good luck! :)
Thanks so much!