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ruby92:
A measurement of 327.66ms-1 accurate to 3%
which of the following gives her result expressed to the appropriate no of significant figures?
a) 327.7 b)328 c)330 d)300

ruby92:
the current in a resistor is marked as (2.50+/-0.05)mA
the resistor is marked as 4.7 +/- 2% ohms
if these values were used to calculate the power dissipated what is the percentage uncertanity
a) 2% b) 4% c)6% d) 8%

A.T:

--- Quote from: ruby92 on March 24, 2010, 11:03:39 am ---the current in a resistor is marked as (2.50+/-0.05)mA
the resistor is marked as 4.7 +/- 2% ohms
if these values were used to calculate the power dissipated what is the percentage uncertanity
a) 2% b) 4% c)6% d) 8%

--- End quote ---

For resistance the percentage uncertainity is given which is 2%
For current u can find it by 0.05/2.5 * 100 = 2%
The formula is P=I^2 * R
so we'll multiply the current's uncertainity by 2 which will give 4%
now add both % uncertainities = 4+2=6%
hence answer is C

ruby92:
but current is given a mA we dont take the 'm' into account here?
and dont u have to find the fractional uncertanity for both and add them?

A.T:

--- Quote from: ruby92 on March 24, 2010, 11:13:28 am ---but current is given a mA we dont take the 'm' into account here?
and dont u have to find the fractional uncertanity for both and add them?

--- End quote ---
For uncertainity u dont have to count the milli but for findind the actual power u will need to count it.
The simplest way to find the % uncertainity is find the % unc. of all the variables and add then so there is no neeed to find fractional unc.

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