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complete the table below. round based on the applicable rules of accura…

Question

complete the table below. round based on the applicable rules of accuracy/precision.

measured valuepercent errorabsolute errorrange
$m_2$13.39 cm3.133 %cmcm
$m_1 - m_2$cm----cmcm
$\frac{m_1}{m_2}$cm%--------

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Explanation:

Step1: Calculate percent - error for $M_1$

Percent error formula is $\text{Percent Error}=\frac{\text{Absolute Error}}{\text{True Value}}\times100\%$. Since we don't know the true value, assume for the first row we use the given absolute error and measured value. $\text{Percent Error}=\frac{0.59}{39.83}\times 100\%\approx1.48\%$.

Step2: Calculate the range for $M_2$

The range is given by $[\text{Measured Value}-\text{Absolute Error},\text{Measured Value}+\text{Absolute Error}]$. For $M_2 = 13.39$ cm and absolute - error $e = 0.59$ cm, the range is $[13.39 - 0.59,13.39+0.59]=[12.8,13.98]$ cm.

Step3: Calculate $M_1 - M_2$

$M_1 - M_2=39.83 - 13.39 = 26.44$ cm.

Step4: Calculate the absolute error for $M_1 - M_2$

When subtracting two measured values $A$ and $B$ with absolute errors $\Delta A$ and $\Delta B$, the absolute error in $A - B$ is $\Delta(A - B)=\Delta A+\Delta B$. Here, assume the absolute error for $M_2$ is also used in the subtraction error calculation. So the absolute error for $M_1 - M_2$ is $0.59+0.59 = 1.18$ cm.

Step5: Calculate the range for $M_1 - M_2$

The range for $M_1 - M_2$ is $[(M_1 - M_2)-\text{Absolute Error},(M_1 - M_2)+\text{Absolute Error}]=[26.44 - 1.18,26.44+1.18]=[25.26,27.62]$ cm.

Step6: Calculate $\frac{M_1}{M_2}$

$\frac{M_1}{M_2}=\frac{39.83}{13.39}\approx2.9746$

Step7: Calculate the relative error in $\frac{M_1}{M_2}$

The relative error formula for $\frac{A}{B}$ is $\frac{\Delta(\frac{A}{B})}{\frac{A}{B}}\approx\frac{\Delta A}{A}+\frac{\Delta B}{B}$. $\frac{\Delta A}{A}=\frac{0.59}{39.83}$ and $\frac{\Delta B}{B}=\frac{0.59}{13.39}$. $\frac{\Delta A}{A}+\frac{\Delta B}{B}\approx0.0148 + 0.0441=0.0589$. The percent error is $0.0589\times100\% = 5.89\%$.

Step8: Calculate the absolute error in $\frac{M_1}{M_2}$

The absolute error in $\frac{M_1}{M_2}$ is $\text{Percent Error}\times\frac{M_1}{M_2}=0.0589\times2.9746\approx0.175$

Answer:

Measured ValuePercent ErrorAbsolute ErrorRange
$M_2$$13.39$ cmN/A$0.59$ cm$[12.8,13.98]$ cm
$M_1 - M_2$$26.44$ cm$3.133\%$ (given)$1.18$ cm$[25.26,27.62]$ cm
$\frac{M_1}{M_2}$$2.9746$$5.89\%$$0.175$N/A