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Calculating Uncertainties of Measured quantities (Physics)

Joystar1977

Active member
Jul 24, 2013
119
d1 = 2.53 cm +/- .05 cm

d2 = 1.753 m +/- .001 m

0 = 23.5 degrees +/- .5 degrees

v1 = 1.55 m/s +/- .15 m/s

Using the measured quantities above, calculate the following. Express the uncertainty calculated value.

a = 4 v1^2 / d2

a = 4 (1.55 m/s +/-.15 m/s)^2 / 1.753 m +/- .001 m

a = 6.8 m/s ^2 / 1.754 m

a = 13.6 m/s / 1.754 m

a = 7.753705815

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d3 = 4 (d1 + d2)


d3 = 4 (2.53 cm +/- .05 cm) + (1.753 m +/- .001 m)


d3 = 10.12 cm +/- .2 cm + 7.012 m +/- .004 m


d3 = 10.32 cm + 7.016 m

I tried to work this problem out, but I don't understand it and think it's not right. Someone please help me with this problem.
 
Last edited by a moderator:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
d1 = 2.53 cm +/- .05 cm

d2 = 1.753 m +/- .001 m

0 = 23.5 degrees +/- .5 degrees

v1 = 1.55 m/s +/- .15 m/s

Using the measured quantities above, calculate the following. Express the uncertainty calculated value.

d3 = 4 (d1 + d2)

d3 = 4 (2.53 cm +/- .05 cm) + (1.753 m +/- .001 m)

d3 = 10.12 cm +/- .2 cm + 7.012 m +/- .004 m

d3 = 10.32 cm + 7.016 m

I tried to work this out, but I don't think it's right so someone please help me.
There is no single formula that you can use to get errors. Which you use depends on what kind of experiment you are doing and what data you have. One of the typical ones in use is this:

Given x, y and their respective errors \(\displaystyle \Delta x,~\Delta y\) and the equation z = x + y you can calculate
\(\displaystyle \frac{\Delta z}{z} = \sqrt{ \left ( \frac{\Delta x}{x} \right ) ^2 + \left ( \frac{\Delta y}{y} \right ) ^2}\)

You can use the same formula for z = xy or z = x/y as well. If you have more variables, such as z = x + y + w just add a term for w under the square root.

-Dan