Question
Find the absolute and relative uncertainty and express each answer with a reasonable number of significant figures. a. 6.2(+-0.2)-4.1(+-0.1)
b. 9.43(+-0.05)×0.016(+-0.001)
b. 9.43(+-0.05)×0.016(+-0.001)
Answers
Answered by
GPT-4o mini
To find the absolute and relative uncertainties for the given problems, we will follow the rules for combining uncertainties in addition/subtraction and multiplication/division.
### a. \( 6.2 \pm 0.2 - 4.1 \pm 0.1 \)
First, we perform the subtraction:
\[
6.2 - 4.1 = 2.1
\]
Now, we calculate the absolute uncertainty for the result of the subtraction. When subtracting two values, the absolute uncertainties are added:
\[
\text{Absolute uncertainty} = 0.2 + 0.1 = 0.3
\]
So we have:
\[
2.1 \pm 0.3
\]
Now we can express the result in the correct format with appropriate significant figures. The value of \(2.1\) has 2 significant figures, so we round \(2.1\) and \(0.3\) appropriately, which leads to:
**Final result for Part a:**
\[
2.1 \pm 0.3
\]
### b. \( 9.43 \pm 0.05 \times 0.016 \pm 0.001 \)
First, we perform the multiplication:
\[
9.43 \times 0.016 = 0.15088
\]
Next, we will calculate the absolute uncertainty. When multiplying, the relative uncertainties need to be calculated first:
1. Calculate the relative uncertainty for \(9.43\):
\[
\text{Relative uncertainty of } 9.43 = \frac{0.05}{9.43} \approx 0.005303 \quad \text{(or } 0.5303\% \text{)}
\]
2. Calculate the relative uncertainty for \(0.016\):
\[
\text{Relative uncertainty of } 0.016 = \frac{0.001}{0.016} = 0.0625 \quad \text{(or } 6.25\% \text{)}
\]
Now, add the relative uncertainties together for the product:
\[
\text{Total relative uncertainty} \approx 0.005303 + 0.0625 \approx 0.067803
\]
Now, calculate the absolute uncertainty in the result by multiplying the relative uncertainty by the calculated product:
\[
\text{Absolute uncertainty} = 0.15088 \times 0.067803 \approx 0.01021
\]
Thus, we can express the total product with its absolute uncertainty:
\[
0.15088 \pm 0.01021
\]
Now we round to the appropriate significant figures. The value \(0.15088\) has 5 significant figures, but the uncertainty \(0.01021\) has 4 significant figures. Since the final result should be expressed to the least number of significant figures in the uncertainties:
**Final result for Part b:**
\[
0.151 \pm 0.010
\]
### Summary of Results:
a. \( 2.1 \pm 0.3 \)
b. \( 0.151 \pm 0.010 \)
### a. \( 6.2 \pm 0.2 - 4.1 \pm 0.1 \)
First, we perform the subtraction:
\[
6.2 - 4.1 = 2.1
\]
Now, we calculate the absolute uncertainty for the result of the subtraction. When subtracting two values, the absolute uncertainties are added:
\[
\text{Absolute uncertainty} = 0.2 + 0.1 = 0.3
\]
So we have:
\[
2.1 \pm 0.3
\]
Now we can express the result in the correct format with appropriate significant figures. The value of \(2.1\) has 2 significant figures, so we round \(2.1\) and \(0.3\) appropriately, which leads to:
**Final result for Part a:**
\[
2.1 \pm 0.3
\]
### b. \( 9.43 \pm 0.05 \times 0.016 \pm 0.001 \)
First, we perform the multiplication:
\[
9.43 \times 0.016 = 0.15088
\]
Next, we will calculate the absolute uncertainty. When multiplying, the relative uncertainties need to be calculated first:
1. Calculate the relative uncertainty for \(9.43\):
\[
\text{Relative uncertainty of } 9.43 = \frac{0.05}{9.43} \approx 0.005303 \quad \text{(or } 0.5303\% \text{)}
\]
2. Calculate the relative uncertainty for \(0.016\):
\[
\text{Relative uncertainty of } 0.016 = \frac{0.001}{0.016} = 0.0625 \quad \text{(or } 6.25\% \text{)}
\]
Now, add the relative uncertainties together for the product:
\[
\text{Total relative uncertainty} \approx 0.005303 + 0.0625 \approx 0.067803
\]
Now, calculate the absolute uncertainty in the result by multiplying the relative uncertainty by the calculated product:
\[
\text{Absolute uncertainty} = 0.15088 \times 0.067803 \approx 0.01021
\]
Thus, we can express the total product with its absolute uncertainty:
\[
0.15088 \pm 0.01021
\]
Now we round to the appropriate significant figures. The value \(0.15088\) has 5 significant figures, but the uncertainty \(0.01021\) has 4 significant figures. Since the final result should be expressed to the least number of significant figures in the uncertainties:
**Final result for Part b:**
\[
0.151 \pm 0.010
\]
### Summary of Results:
a. \( 2.1 \pm 0.3 \)
b. \( 0.151 \pm 0.010 \)
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