Would you proof read due to my absentminded mind

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- anonymous

Would you proof read due to my absentminded mind

- katieb

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- anonymous

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?

- anonymous

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.)

- anonymous

Thanks guys appreciate it to the Kingdom Come

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## More answers

- anonymous

Thanks @hartnn ;)

- 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 :)

- anonymous

ops I guess I had to change 20 ohms to 18 ohms since time wasn't used at this timeXD

- hartnn

yes that would be 18 ohms

- anonymous

It says "using the scientific notation" whereas 1.25kV doesn't do as instructed

- anonymous

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....

- hartnn

hmm...then I would suggest 1.25 * 10^0 kV

- anonymous

I think that would be more accurate.

- anonymous

What's your taken on rounding up till 4000C instead of 3900C? I think it's valid.

- anonymous

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

- anonymous

Actually the least of all

- hartnn

Your reasoning looks good.
Keep it as 4000C :)

- anonymous

Great! Thanks;)

- hartnn

welcome ^_^

- anonymous

Calculations regarding other units are correct?

- anonymous

Wasn't entirely sure of the coulombs unit and other units stuff.

- hartnn

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

- anonymous

Yeah. I have. But I am not sure if sig figs of 2 would be appropriate once even the zeros go beyond decimal

- anonymous

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

- anonymous

So maybe I should say 4kC

- hartnn

or perhaps write it 3.9kC first and then round it to 4 kC
yes

- hartnn

rounding 3900 to 4000 looks odd, rounding 3.9 to 4 seems ok,
though you're doing the same thing :P

- anonymous

That would come off more intuitive;)

- anonymous

Marker-friendly I guessXD

- hartnn

good luck! :)

- anonymous

Thanks so much!

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