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Carry out the following calculations, making sure that your answer has the correct number of significant figures: (a) \(\frac{2.795 \mathrm{~m} \times 3.10 \mathrm{~m}}{6.48 \mathrm{~m}}\) (b) \(V=\frac{4}{3} \pi r^{3},\) where \(r=17.282 \mathrm{~mm}\) (c) \(1.110 \mathrm{~cm}+17.3 \mathrm{~cm}+108.2 \mathrm{~cm}+316 \mathrm{~cm}\)

Short Answer

Expert verified
(a) 1.34 m, (b) 21620 mm^3, (c) 443 cm

Step by step solution

01

- Solve part (a)

Calculate \(\frac{2.795 \, \mathrm{m} \times 3.10 \, \mathrm{m}}{6.48 \, \mathrm{m}}\) first. The multiplication yields \(2.795 \times 3.10 = 8.6645 \, \mathrm{m}^2\). Then divide by 6.48 m: \(\frac{8.6645 \, \mathrm{m}^2}{6.48 \, \mathrm{m}} = 1.33788981 \, \mathrm{m}\). The least number of significant figures in the calculation is 3 (from 6.48), so round to 3 significant figures: \(1.34 \, \mathrm{m}\).
02

- Solve part (b)

Calculate the volume using \(V= \frac{4}{3} \pi r^{3}\) where \(r = 17.282 \, \mathrm{mm}\). Substitute \(r\) into the equation: \(V = \frac{4}{3} \pi (17.282)^{3} \, \mathrm{mm}^3 = \frac{4}{3} \pi (5155.310894 \, \mathrm{mm^3}) = 21622.48432 \, \mathrm{mm^3}\). The least number of significant figures in the calculation is 5 (from 17.282), so round to 5 significant figures: \(21620 \, \mathrm{mm^3}\).
03

- Solve part (c)

Add the lengths \(1.110 \, \mathrm{cm} + 17.3 \, \mathrm{cm} + 108.2 \, \mathrm{cm} + 316 \, \mathrm{cm}\). Sum them up: \(1.110 + 17.3 + 108.2 + 316 = 442.61 \, \mathrm{cm}\). The least number of decimal places in the calculation is 0 (from 316), so round to the nearest whole number: \(443 \, \mathrm{cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multiplication and Division with Significant Figures
When multiplying or dividing numbers, you need to pay attention to significant figures. Significant figures reflect the precision of your measurements. Let’s break it down:
First, count the number of significant figures in each number you'll use in your calculation.
For example, in the given calculation for part (a) \(\frac{2.795 \ \mathrm{m} \times 3.10 \ \mathrm{m}}{6.48 \ \mathrm{m}}\), the significant figures are:
  • 2.795 has 4 significant figures
  • 3.10 has 3 significant figures
  • 6.48 has 3 significant figures
Begin by multiplying 2.795 and 3.10, giving you 8.6645. Then divide this result by 6.48, resulting in 1.33788981.
To ensure your final answer is as precise as your least precise measurement, look for the smallest number of significant figures you initially had. Here, it's 3. Thus, round 1.33788981 to 1.34.
This rule ensures your final answer accurately reflects your data's precision.
Volume Calculation
Calculating volume often involves using formulas. With spheres, we use the formula \[ V = \frac{4}{3} \pi r^3 \].
However, it's crucial to manage significant figures correctly.
In part (b) of the original exercise, we calculate the volume of a sphere where the radius \(r\) is 17.282 mm.
Following the formula, plug in the radius:
\[ V = \frac{4}{3} \pi (17.282)^3 \]
First, cube the radius, yielding approximately 5155.310894 \(\mathrm{mm}^3\).
Then, multiply by \ \pi \ and \ \frac{4}{3} \, yielding a volume approximately equal to 21622.48432 \ \mathrm{mm^3} \.
Notice that our initial radius had 5 significant figures. Hence, we round the final volume to 5 significant figures: 21620 \ \mathrm{mm^3} \.
This rounding ensures consistency in the precision throughout your calculation.
Addition with Significant Figures
When adding numbers, it's key to consider decimal places instead of just significant figures.
In part (c) of the exercise, let's add the following lengths: 1.110 cm, 17.3 cm, 108.2 cm, and 316 cm.
Summing them gives you 442.61 cm.
However, the level of precision in your answer must align with the number with the fewest decimal places. Here, 316 cm has 0 decimal places.
Therefore, round 442.61 cm to 443 cm.
By doing this, you maintain the clarity and accuracy of the addition based on the most limiting measurement in your data.

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Most popular questions from this chapter

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