Question

Would you proof read due to my absentminded mind

Answer 1

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?

Answer 2

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.)

Answer 3

Thanks guys appreciate it to the Kingdom Come

Answer 4

Thanks @hartnn ;)

Answer 5

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 :)

Answer 6

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

Answer 7

yes that would be 18 ohms

Answer 8

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

Answer 9

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....

Answer 10

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

Answer 11

I think that would be more accurate.

Answer 12

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

Answer 13

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

Answer 14

Actually the least of all

Answer 15

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

Answer 16

Great! Thanks;)

Answer 17

welcome ^_^

Answer 18

Calculations regarding other units are correct?

Answer 19

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

Answer 20

yep, all others are correct.

and for the last one, have you considered answering it to 3.9 kC ?

Answer 21

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

Answer 22

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

Answer 23

So maybe I should say 4kC

Answer 24

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

Answer 25

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

Answer 26

That would come off more intuitive;)

Answer 27

Marker-friendly I guessXD

Answer 28

good luck! :)

Answer 29

Thanks so much!