Question

Find the absolute uncertainty and express each answer with a reasonable number of significant figures. [12.41(+-.09)/4.16(+-.01)]×7.0682(+-.0004)

Answer 1

To find the absolute uncertainty of the expression \((12.41 \pm 0.09) / (4.16 \pm 0.01) \times (7.0682 \pm 0.0004)\), we can use the following rules for uncertainties.

1. **Division**: When dividing quantities, we calculate the relative uncertainty (percentage uncertainty) of each quantity, sum those for the final result, and convert back to absolute uncertainty.

2. **Multiplication**: When multiplying quantities, we again calculate the relative uncertainties, sum those, then convert back to absolute uncertainty.

**Step 1: Finding the quotient**

Let's first calculate \( \frac{12.41 \pm 0.09}{4.16 \pm 0.01} \).

- For \( 12.41 \), the absolute uncertainty is \(0.09\).

Relative uncertainty of \(12.41\) is given by:

\[
\frac{0.09}{12.41} \approx 0.00724 \quad (0.724\%)
\]

- For \( 4.16 \), the absolute uncertainty is \(0.01\).

Relative uncertainty of \(4.16\) is given by:

\[
\frac{0.01}{4.16} \approx 0.002403 \quad (0.2403\%)
\]

Now, we sum the relative uncertainties:

\[
\text{Total relative uncertainty} = 0.00724 + 0.002403 \approx 0.009643
\]

Now we calculate \( \frac{12.41}{4.16} \):

\[
\frac{12.41}{4.16} \approx 2.985576
\]

To keep an appropriate number of significant figures, we round this to 2.986 (3 significant figures).

Now, calculate the absolute uncertainty using the total relative uncertainty:

\[
\text{Absolute uncertainty} = 2.986 \times 0.009643 \approx 0.0288
\]

So we have:

\[
\frac{12.41 \pm 0.09}{4.16 \pm 0.01} \approx 2.986 \pm 0.029
\]

**Step 2: Multiplying with \(7.0682 \pm 0.0004\)**

Now we compute:

\[
(2.986 \pm 0.029) \times (7.0682 \pm 0.0004)
\]

- For \(7.0682\), the absolute uncertainty is \(0.0004\).

Relative uncertainty of \(7.0682\):

\[
\frac{0.0004}{7.0682} \approx 0.0000566 \quad (0.00566\%)
\]

Now we compute the relative uncertainty for the result from the previous step:

Let's find the product first:

\[
2.986 \times 7.0682 \approx 21.132334
\]

This rounds to \(21.132\) (5 significant figures).

Now, we combine the relative uncertainties:

\[
\text{Total relative uncertainty} = 0.009643 + 0.0000566 \approx 0.0096996
\]

Now find the absolute uncertainty for the final result:

\[
\text{Absolute uncertainty} = 21.132 \times 0.0096996 \approx 0.205
\]

So, rounding to three significant figures for consistency (as the 21.132 means we should have three significant figures in the final result):

\[
\text{Final result} = 21.132 \pm 0.205
\]

Thus, expressing our final result:

\[
21.1 \pm 0.2
\]

Putting it all together, the complete calculation yields:

\[
\frac{12.41 \pm 0.09}{4.16 \pm 0.01} \times (7.0682 \pm 0.0004) \approx 21.1 \pm 0.2
\]