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Perform the following calculations and retain the appropriate number of significant figures in each result. (a) \(\left(38.4 \times 10^{-3}\right) \times\left(6.36 \times 10^{5}\right)=\) (b) \(\frac{\left(1.45 \times 10^{2}\right) \times\left(8.76 \times 10^{-4}\right)}{\left(9.2 \times 10^{-3}\right)^{2}}=\) (c) \(24.6+18.35-2.98=\) (d) \(\left(1.646 \times 10^{3}\right)-\left(2.18 \times 10^{2}\right)+\left[\left(1.36 \times 10^{4}\right)\right.\) [Hint: The significant figure rule for the extraction of a root is the same as for multiplication.] \(\left.\times\left(5.17 \times 10^{-2}\right)\right]=\) (e) \(\frac{-7.29 \times 10^{-4}+\sqrt{\left(7.29 \times 10^{-4}\right)^{2}+4(1.00)\left(2.7 \times 10^{-5}\right)}}{2 \times(1.00)}\)

Short Answer

Expert verified
The results of the calculations are: (a) 24.38, (b) 152.183, (c) 40.0, (d) 715.94, and (e) 0.00.

Step by step solution

01

Calculating Part (a)

Firstly, calculate the result for part (a) by multiplying the given numbers. This operation is as follows: \(38.4 \times 10^{-3} \times 6.36 \times 10^{5} = 24.38\). The result has to be rounded to the second decimal place as it contains a multiplication of two numbers and one of the numbers (38.4) contains two decimal positions.
02

Calculating Part (b)

The next operation involves the division. The calculation is \(\frac{1.45 \times 10^{2} \times 8.76 \times 10^{-4}}{9.2 \times 10^{-3}^{2}} = 152.183\). The result should be rounded up to three decimal positions because the number with the least number of decimal places in the divisor (9.2) has one significant spot behind the decimal point.
03

Calculating Part (c)

Part (c) involves an addition and subtraction process. Hence, perform the calculation: \(24.6 + 18.35 - 2.98 = 39.97\). The number of decimal positions equals the least number of decimal places in the given numbers, i.e., one (in 24.6). Hence, the result should be presented as 40.0.
04

Calculating Part (d)

In part (d), subtract and multiply the given numbers to get the result: \(1.646 \times 10^{3} - 2.18 \times 10^{2} + 1.36 \times 10^{4} \times 5.17 \times 10^{-2} = 715.94\). The result should be rounded to two decimal places as indicated by the least decimal positions number \((2.18 \times 10^{2})\).
05

Calculating Part (e)

In this final operation, take the root and carry out the additions, subtractions and division as shown: \(\frac{-7.29 \times 10^{-4} + \sqrt{\left(7.29 \times 10^{-4}\right)^{2} + 4 \times 1.00 \times 2.7 \times 10^{-5}}}{2 \times 1.00} = 0.000183\). The result should retain the same decimal positions as the number with least decimal positions in the equation. Here, after performing the square root operation, the number of decimal positions becomes 2, as in \(1.00\). Hence, round the result \((0.000183)\) up to two decimal places.

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Key Concepts

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

Multiplication and Division Rules
When you are multiplying or dividing numbers, the key rule to remember is that your result should have the same number of significant figures as the number with the fewest significant figures in the problem. This rule stems from the principle that your precision cannot exceed the least precise measurement involved in the calculation.

Let's break it down with part (a) from our exercise:
  • The expression is \( (38.4 \times 10^{-3}) \times (6.36 \times 10^{5}) \).
  • Here, \( 38.4 \times 10^{-3} \) has 3 significant figures and \( 6.36 \times 10^{5} \) also has 3 significant figures.
  • Thus, the final answer should be rounded to 3 significant figures.
Now consider part (b):
  • The expression is \( \frac{(1.45 \times 10^{2}) \times (8.76 \times 10^{-4})}{(9.2 \times 10^{-3})^{2}} \).
  • The significant figures of the terms are as follows: \( 1.45 \times 10^{2} \) has 3 significant figures, \( 8.76 \times 10^{-4} \) also has 3 significant figures, and \( 9.2 \times 10^{-3} \) has 2 significant figures.
  • The calculated result should, therefore, be reported with 2 significant figures, as 9.2 limits the precision.
Addition and Subtraction Precision
For addition and subtraction, the rule is all about decimal places, not significant figures. The result should be rounded off to the least number of decimal places found in the numbers being added or subtracted.

Consider part (c) of the exercise:
  • The calculation is \( 24.6 + 18.35 - 2.98 \).
  • Here, the decimal places are: 1 for 24.6, 2 for 18.35, and 2 for 2.98.
  • The number with the least decimal places is 24.6, which has 1 decimal place.
  • Therefore, the result should be rounded to 1 decimal place, giving us 40.0 after calculations.
This method ensures that your answer doesn’t suggest a level of precision that wasn't present in your measurements.
Rounding Off
Rounding is a vital skill to master for maintaining the correct number of significant figures or decimal places in your calculated results. This ensures that results neither overstate nor understate the precision of the original measurements.

Here's how you round a number:
  • Identify which digit will be the last one retained based on your significant figures or decimal place requirements.
  • If the digit immediately following this is 5 or higher, round the last retained digit up by one.
  • Otherwise, leave the last retained digit as it is.
For example, consider part (e) of the exercise:
  • Our result is \(0.000183\) and needs to be rounded to match the least number of decimal places in the operation components, which here is 2.
  • This means the final value should be \(0.00\), reflecting that only zeros appear in the first two decimal places.
Correct rounding aligns your answers with the precision of the data used, maintaining scientific integrity.

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Most popular questions from this chapter

As mentioned on page \(13,\) the MCO was lost because of a mix-up in the units used to calculate the force needed to correct its trajectory. Ground-based computers generated the force correction file. On September \(29,1999,\) it was discovered that the forces reported by the ground-based computer for use in MCO navigation software were low by a factor of \(4.45 .\) The erroneous trajectory brought the MCO \(56 \mathrm{km}\) above the surface of Mars; the correct trajectory would have brought the MCO approximately \(250 \mathrm{km}\) above the surface. At \(250 \mathrm{km},\) the MCO would have successfully entered the desired elliptic orbit. The data contained in the force correction file were delivered in lb-sec instead of the required SI units of newton-sec for the MCO navigation software. The newton is the SI unit of force and is described in Appendix B. The British Engineering (gravitational) system uses a pound (lb) as a unit of force and \(\mathrm{ft} / \mathrm{s}^{2}\) as a unit of acceleration. In turn, the pound is defined as the pull of Earth on a unit of mass at a location where the acceleration due to gravity is \(32.174 \mathrm{ft} / \mathrm{s}^{2} .\) The unit of mass in this case is the slug, which is \(14.59 \mathrm{kg}\). Thus, BE unit of force \(=1\) pound \(=(\text { slug })\left(\mathrm{ft} / \mathrm{s}^{2}\right)\) Use this information to confirm that BE unit of force \(=4.45 \times\) SI unit of force 1 pound \(=4.45\) newton

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