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Chapter 2: Problem 21

The following expression has occurred in a problem solution: $$R=\frac{(0.6700)(264,980)(6)\left(5.386 \times 10^{4}\right)}{(3.14159)\left(0.479 \times 10^{7}\right)}$$ The factor 6 is a pure integer. Estimate the value of \(R\) without using a calculator, following the procedure outlined in Section 2.5b. Then calculate \(R\), expressing your answer in both scientific and decimal notation and making sure it has the correct number of significant figures.

Short Answer

Expert verified
The estimated value of \(R\) without using a calculator is \(1.4 x 10^{5}\), and the more precise calculated value of \(R\) is \(2.0 x 10^{5}\) (in scientific notation) or \(200,000\) (in decimal notation).

Step by step solution

01

Estimating the Value

Begin by rounding all the numerical factors into one significant figure except pure integers. Hence, \n\(0.6700\) becomes \(0.7\), \n\(264,980\) becomes \(300,000\) (only one significant figure), \n\(5.386 脳 10^{4}\) becomes \(5 脳 10^{4}\), and \n\(0.479 脳 10^{7}\) becomes \(5 脳 10^{7}\). \nThe integer \(6\) remains the same. Now, replace the original numbers in \(R\) by these estimated values. The fraction becomes \((0.7)(300,000)(6)(5 x 10^{4}) / ((3)(5 x 10^{7}))\).
02

Simplify the Fraction

Next step is to simplify the fraction. We can cancel out equal terms in the numerator and denominator and rewrite \(300,000\) as \(3 x 10^{5}\) . After this simplification, the fraction becomes \((0.7)(6)(3 x 10^{5}) / (3)\).
03

Perform Calculations

The fraction now becomes \(1.4 x 10^{5}\). After performing these calculations, we find that the estimated value of \(R\) is \(1.4 x 10^{5}\).
04

Calculating Exact Value

Now, calculate the exact value using the original numbers in \(R\) given in the problem. \n\(R=(0.6700)(264,980)(6)(5.386 x 10^{4}) / ((3.14159)(0.479 x 10^{7})\). \nAfter calculating more precisely, \(R\) becomes \(1.997 x 10^{5}\), in scientific notation.
05

Final Output

The final output would be correctly rounded to the number of least significant figures from the given data, which is 2. Therefore, the final value of \(R\) is \(2.0 x 10^{5}\) in scientific notation and \(200,000\) in decimal notation.

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Key Concepts

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

Understanding Scientific Notation
Scientific notation is a way to express very large or very small numbers in a concise and consistent manner. This method helps simplify calculations and reduces the risk of errors in lengthy or complex expressions. The format involves writing numbers as a product of a coefficient and a power of ten, like this: \( a \times 10^n \)where \( a \) is a number between 1 and 10, and \( n \) is an integer.For example:
  • The number 5,386 can be written as \( 5.386 \times 10^3 \), and
  • The number 0.00479 can be expressed as \( 4.79 \times 10^{-3} \).
In the given exercise, we notice scientific notation when numbers like \( 5.386 \times 10^{4} \) are used. This helps efficiently handle the magnitude of numbers, making it easier to manage multiplication or division in mathematical expressions.
When simplifying or estimating expressions using scientific notation, it becomes easy to approximate calculations without a calculator, significantly aiding problem-solving tasks.
Mastering Estimation Techniques
Estimation techniques are essential tools for quickly approximating values and making decisions when exact numbers are not required. These techniques involve rounding numbers to the nearest significant figure to simplify calculations, allowing for quick mental math.In our exercise, we see that each number is rounded to one significant figure for estimation purposes:
  • \( 0.6700 \) is rounded to \( 0.7 \)
  • \( 264,980 \) becomes \( 300,000 \) (or \( 3 \times 10^5 \))
  • And \( 5.386 \times 10^4 \) simplifies to \( 5 \times 10^4 \)
  • \( 0.479 \times 10^7 \) turns into \( 5 \times 10^7 \)
Using estimation techniques, complex expressions like fractions can be substantially simplified, making them much easier to compute intuitively.
Rounding off to the nearest significant figures can sometimes change the result slightly, but it is often adequate for quickly understanding an expression's magnitude.
Simplifying Fractions
Fraction simplification is about reducing a fraction to its simplest form so that the numbers are as small and manageable as possible. This involves canceling out common factors in the numerator and the denominator and rearranging the expression for clarity.To illustrate this, consider the fraction from our exercise:\[\frac{(0.7)(300,000)(6)(5 \times 10^4)}{(3)(5 \times 10^7)}\]Here, \( (5 \times 10^4) \) and \( (5 \times 10^7) \) were part of larger numbers that had factors in common. By simplifying \( 300,000 \) to \( 3 \times 10^5 \) and canceling equal terms, the expression was reduced:
  • Cancel \( (5 \times 10^4) \) with \( (5 \times 10^7) \) by reducing their powers of ten,
  • Similarly, the constant numerical factors allow us to strike out \( 3 \) from the denominator since \( 3 \) is a factor of \( 3 \times 10^5 \) in the numerator.
After simplification, this results in much more manageable numbers, making it easier to compute either mentally or with minimal tools.
Mastering fraction simplification can greatly aid in estimating and solving problems, leading to more efficient and straightforward calculations.

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Most popular questions from this chapter

State what you would plot to get a straight line if experimental ( \(x, y\) ) data are to be correlated by the following relations, and what the slopes and intercepts would be in terms of the relation parameters. If you could equally well use two different kinds of plots (e.g., rectangular or semilog), state what you would plot in each case. [The solution to Part (a) is given as an example.] (a) \(y^{2}=a e^{-b / x}\). (b) \(y^{2}=m x^{3}-n\) (c) \(1 / \ln (y-3)=(1+a \sqrt{x}) / b\) (d) \((y+1)^{2}=\left[a(x-3)^{3}\right]^{-1}\) (e) \(y=\exp (a \sqrt{x}+b)\) (f) \(x y=10^{\left[a\left(x^{2}+y^{2}\right)+b\right]}\) (g) \(y=[a x+b / x]^{-1}\)

The temperature in a process unit is controlled by passing cooling water at a measured rate through a jacket that encloses the unit. The exact relationship between the unit temperature \(T\left(^{\circ} \mathrm{C}\right)\) and the water flow rate \(\phi(\mathrm{L} / \mathrm{s})\) is extremely complex, and it is desired to derive a simple empirical formula to approximate this relationship over a limited range of flow rates and temperatures. Data are taken for \(T\) versus \(\phi\). Plots of \(T\) versus \(\phi\) on rectangular and semilog coordinates are distinctly curved (ruling out \(T=a \phi+b\) and \(T=a e^{b \phi}\) as possible empirical functions), but a log plot appears as follows: A line drawn through the data goes through the points \(\left(\phi_{1}=25 \mathrm{L} / \mathrm{s}, T_{1}=210^{\circ} \mathrm{C}\right)\) and \(\left(\phi_{2}=40 \mathrm{L} / \mathrm{s},\right.\) \(\left.T_{2}=120^{\circ} \mathrm{C}\right)\). (a) What is the empirical relationship between \(\phi\) and \(T ?\) (b) Using your derived equation, estimate the cooling water flow rates needed to maintain the process unit temperature at \(85^{\circ} \mathrm{C}, 175^{\circ} \mathrm{C},\) and \(290^{\circ} \mathrm{C}\). (c) In which of the three estimates in Part (b) would you have the most confidence and in which would you have the least confidence? Explain your reasoning.

A horizontal drum, a cross-section of which is shown below, is being filled with benzene (density \(\left.=0.879 \mathrm{g} / \mathrm{cm}^{3}\right)\) at a constant rate \(\dot{m}(\mathrm{kg} / \mathrm{min}) .\) The drum has a length \(L\) and radius \(r,\) and the level of benzene in the drum is \(h\). The expression for the volume of benzene in the drum is \(V=L\left[r^{2} \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{r^{2}-(r-h)^{2}}\right]\) (a) Show that the equation gives reasonable results for \(h=0, h=r,\) and \(h=2 r\). (b) Estimate the mass of benzene \((\mathrm{kg})\) in the tank if \(L=10 \mathrm{ft}, r=2 \mathrm{ft},\) and \(h=4\) in. (c) Suppose there is a sight glass on the side of the tank that allows observation of the height of liquid in the tank. Use a spreadsheet to prepare a graph that can be posted next to the sight glass so that an operator can estimate the mass that is in the tank without going through calculations like that in Part (b).

Fit (a) a line and (b) a line through the origin to the following data using either the method of least squares (Appendix A.1) or graphics software: $$\begin{array}{|l|l|l|l|}\hline x & 0.3 & 1.9 & 3.2 \\\\\hline y & 0.4 & 2.1 & 3.1 \\\\\hline\end{array}$$ On a single plot, show both fitted lines and the data points.

A waste treatment pond is \(50 \mathrm{m}\) long and \(25 \mathrm{m}\) wide, and has an average depth of \(2 \mathrm{m}\). The density of the waste is \(75.3 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\). Calculate the weight of the pond contents in \(\mathrm{lb}_{\mathrm{f}},\) using a single dimensional equation for your calculation.
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