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Express the result of each of the following calculations in exponential form and with the appropriate number of significant figures. (a) \(\left(4.65 \times 10^{4}\right) \times\left(2.95 \times 10^{-2}\right) \times\left(6.663 \times 10^{-3}\right) \times 8.2=\) (b) \(\frac{1912 \times\left(0.0077 \times 10^{4}\right) \times\left(3.12 \times 10^{-3}\right)}{\left(4.18 \times 10^{-4}\right)^{3}}=\) {c} \(\left(3.46 \times 10^{3}\right) \times 0.087 \times 15.26 \times 1.0023=\) (d) \(\frac{\left(4.505 \times 10^{-2}\right)^{2} \times 1.080 \times 1545.9}{0.03203 \times 10^{3}}=\) (e) \(\frac{\left(-3.61 \times 10^{-4}\right)+\sqrt{\left(3.61 \times 10^{-4}\right)^{2}+4(1.00)\left(1.9 \times 10^{-5}\right)}}{2 \times(1.00)}\) [Hint: The significant figure rule for the extraction of a root is the same as for multiplication.]

Short Answer

Expert verified
The answers to the calculations are: (a) \(8.9 * 10^{3}\) (b) \(6.32 * 10^{6}\) (c) \(4.4 * 10^{3}\) (d) \(1.057 * 10^{-2}\) (e) \(3.00 * 10^{-4}\)

Step by step solution

01

Calculation (a)

This one involves multiplication of four numbers. We start off the calculation by multiplying the numbers as follows: \(4.65 * 10^{4} * 2.95 * 10^{-2} * 6.663 * 10^{-3} * 8.2\). The result of this operation is 8939.71111. We then present it in exponential form; this would be \(8.93971111 * 10^{3}\). We now need to apply the rule of significant figures. In multiplication, the rule is to use the same number of significant figures in the result as in the number with the smallest number of significant figures. Here, the number with the least significant figures is 8.2 with two significant figures, so our answer should also have two significant figures. Therefore, our final answer is \(8.9 * 10^{3}\).
02

Calculation (b)

This one involves both multiplication and division. The numerator is calculated as: \(1912 * 0.0077 * 10^{4} * 3.12 * 10^{-3}\), giving 0.46211616. The denominator involves a power of \(4.18 * 10^{-4}\) cubed, which gives 0.0000000730728. Dividing the numerator by the denominator gives us \(0.46211616 / 0.0000000730728 = 6321720.03556\). We now express this in exponential form, \(6.32172003556 * 10^{6}\). The least significant figures in our numbers is three (from 1912), so our final answer should also have three significant figures. Therefore, our final answer is \(6.32 * 10^{6}\).
03

Calculation (c)

In the third calculation, we simply multiply the four participating numbers. Thus, \(3.46 * 10^{3} * 0.087 * 15.26 * 1.0023 = 4433.38214\). We change this into exponential form to give \(4.43338214 * 10^{3}\). Considering significant figures, our answer should have two significant figures (from 0.087) so our final answer is \(4.4 * 10^{3}\).
04

Calculation (d)

This problem is somewhat more complex, as we need to take the square of \(4.505 * 10^{-2}\). This gives us 0.000203025, and multiplied by 1.080 and 1545.9 gives us 0.338582174. We divide this by \(0.03203 * 10^{3}\) which gives us \(0.338582174 / 32.03 = 0.01057\), after rounding. We change this into exponential form to give \(1.057 * 10^{-2}\). Taking into account significant figures, the least significant figures are four (from 1.080), so our final answer becomes \(1.057 * 10^{-2}\).
05

Calculation (e)

This last exercise involves taking a square root. First we compute the square root of \((3.61 * 10^{-4})^2 + 4(1.00)(1.9 * 10^{-5})\) which equates to \(6.000025 * 10^{-4}\). We divide this number by \(2 * 1.00\) to have \(3.0000125 * 10^{-4}\). Our answer needs to have three significant figures because the number with the smallest number of significant figures, \(1.9 * 10^{-5}\), has two. Therefore, our final answer is \(3.00 * 10^{-4}\)

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Key Concepts

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

Exponential Form
When dealing with large or small numbers, the exponential form provides a convenient way to express quantities concisely. This form utilizes powers of ten to indicate the scale of the number. For example, the number 3000 can be written as \(3 \times 10^{3}\). This notation is especially helpful in calculations involving orders of magnitude, since it simplifies multiplication and division.
To express a number in exponential form, identify the most significant digit (non-zero) and then count how many places you move the decimal point to reach that digit. The number of places moved determines the exponent of ten. Here are a few steps to convert a number into exponential form:
  • Identify the significant figures.
  • Place the decimal after the first significant digit.
  • Count the number of places you moved the decimal point and use that as the exponent.
This method keeps the number manageable and easy to read, especially when dealing with extremely large or small values.
Scientific Notation
Scientific notation is a standardized way of expressing numbers as a product of a number between 1 and 10 and a power of 10. It's essentially a form of exponential notation specifically used in scientific calculations. For instance, instead of writing 0.000032, you can use \(3.2 \times 10^{-5}\) in scientific notation. This makes it easier to compare sizes and manage units.
To convert numbers into scientific notation, follow these straightforward steps:
  • Move the decimal point so the number becomes a value between 1 and 10.
  • Count how many places the decimal has been moved; this becomes the exponent of 10. Moving the decimal to the left results in a positive exponent, while moving it to the right gives a negative exponent.
  • Write the number as a product of the number from the first step and \(10^{\text{(exponent)}}\).
Scientific notation is especially beneficial in making calculations less cumbersome and reducing errors when dealing with very large or very small quantities.
Multiplication and Division
When performing multiplication and division in scientific calculations, there are specific rules regarding the treatment of significant figures and exponential values. These processes often involve numbers expressed in exponential form, and understanding how to handle these ensures your results are accurate and consistent.
**Significant Figures:**In multiplication and division, the number of significant figures in the result should be the same as in the factor with the smallest number of significant figures. For example, if multiplying \(2.3 \times 3.56\), the result should be rounded to two significant figures--the same as in 2.3, which has the fewest.
**Exponential Values:**When multiplying numbers in exponential form, multiply the base numbers and add the exponents. For example: \((4 \times 10^{2}) \times (3 \times 10^{4}) = 12 \times 10^{6}\). For division, divide the base numbers and subtract the exponents: \((4 \times 10^{6}) / (2 \times 10^{2}) = 2 \times 10^{4}\).
Understanding these principles will help you correctly manipulate numbers in exponential form, ensuring your calculations in scientific contexts are both efficient and accurate.

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Most popular questions from this chapter

The following equation can be used to relate the density of liquid water to Celsius temperature in the range from \(0^{\circ} \mathrm{C}\) to about \(20^{\circ} \mathrm{C}:\) $$d\left(\mathrm{g} / \mathrm{cm}^{3}\right)=\frac{0.99984+\left(1.6945 \times 10^{-2} t\right)-\left(7.987 \times 10^{-6} t^{2}\right)}{1+\left(1.6880 \times 10^{-2} t\right)}$$ (a) To four significant figures, determine the density of water at \(10^{\circ} \mathrm{C}\). (b) At what temperature does water have a density of \(0.99860 \mathrm{g} / \mathrm{cm}^{3} ?\) (c) In the following ways, show that the density passes through a maximum somewhere in the temperature range to which the equation applies. (i) by estimation (ii) by a graphical method (iii) by a method based on differential calculus

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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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