Category Archives: Engineering Mathematics Blog

Problem 1-18: Significant figures, uncertainty, and accuracy of the numbers 99 and 100

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PROBLEM:

(a) How many significant figures are in the numbers 99 and 100?

(b) If the uncertainty in each number is 1, what is the percent uncertainty in each?

(c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

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SOLUTION:

Part A

99 has 2 significant figures

100 has 3 significant figures

Part B

\begin{align*}
\frac{1}{99}\times 100\% & =1.0\:\%  \\ 
\frac{1}{100}\times 100\% & =1.00\%
\end{align*}

Part C

Based on the results of parts a and b, the percent uncertainties are more meaningful.

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Problem 1-17: Stating the correct significant figures in a given calculation

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PROBLEM:

State how many significant figures are proper in the results of the following calculations:

\begin{align*}
\left(a\right)\:\frac{\left(106.7\right)\left(98.2\right)}{\left(46.210\right)\left(1.01\right)} \ \qquad \ \left(b\right)\:\left(18.7\right)^2 \ \qquad \ \left(c\right)\:\left(1.60\times 10^{-19}\right)\left(3712\right)
\end{align*}
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SOLUTION:

Part A

The answer is limited by 98.2 and 1.01. They are both 3 significant figures. So the result should be 3 significant figures.

Part B

The answer is limited by 18.7. The answer should be 3 significant figures. So the result should be 3 significant figures.

Part C

The answer is limited by 1.60 which is 3 significant figures. So the result should be 3 significant figures.

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Problem 1-16: Solving for the remaining soda after the removal of some volume

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PROBLEM:

A can contains 375 mL of soda. How much is left after 308 mL is removed?

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SOLUTION:

What is left is calculated as

\begin{align*}
\text{Volume left} & =375 \ \text{mL}-308 \ \text{mL} \\
& = 67 \ \text{mL} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, 67 mL of soda is left.

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Problem 1-15: Solving for the number of heartbeats of a person with the correct significant figure

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PROBLEM:

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?

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SOLUTION:

Part a

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

Convert 72 beats/min to beats/year.

\begin{align*}
72 \ \text{beats/min} & = \left(\frac{72.0\:\text{beats}}{1\:\bcancel{\text{min}}}\right)\left(\frac{60.0\:\bcancel{\text{min}}}{1.00\:\bcancel{\text{hr}}}\right)\left(\frac{24.0\:\bcancel{\text{hr}}}{1.00\:\bcancel{\text{day}}}\right)\left(\frac{365.25\:\bcancel{\text{days}}}{1\:\text{year}}\right)\\
& =3.7869\times 10^7\:\text{beats/year}
\end{align*}

In 2 years, the number of beats of an average person is

\begin{align*}
& = 3.7869 \times 10^{7} \ \text{beats/year} \times 2 \ \text{years} \\
& = 7.5738 \times 10^7 \ \text{beats} 
\end{align*}

For part a, the answer is limited to 2.0 which is 2 significant figures. The answer is 7.6 \times 10^{7} \ \text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part b

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part c

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is  7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

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Problem 1-14: Solving for the percent uncertainty of 130±5 beats/min

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PROBLEM:

An infant’s pulse rate is measured to be 130±5 beats/min. What is the percent uncertainty in this measurement?

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SOLUTION:

The percent uncertainty can be calculated as

\begin{align*}
\%\:\text{uncertainty}\:& =\frac{\delta _A}{A}\times 100\% \\
&  =\frac{5\:\text{beats/min}}{130\:\text{beats/min}}\times 100\% \\
& =3.85\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the percent uncertainty is 3.85%.

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Problem 1-13: Computing for the range of possible speeds given 90 km/h and 5% uncertainty

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PROBLEM:

(a) A car speedometer has a  5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h?

(b) Convert this range to miles per hour. (1 km=0.6214 mi)

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SOLUTION:

Part A

The uncertainty in the velocity of the car is computed as

\begin{align*}
\delta _v & =\frac{5.0\:\%}{100\:\%}\times 90.0\:\text{km/hr} \\
& = 4.5\:\text{km/hr}
\end{align*}

Therefore, the range of the possible speeds is 

\begin{align*}
\text{Range} & =90.0\:\pm 4.5\:\text{km/hr} \\
\text{Range} & :85.8\:\text{km/hr}\:-\:94.5\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

So the range of the possible speeds is 85.5 km/hr to 94.5 km/hr.

Part B

Convert the range to mi/h

For 85.5 km/hr

\begin{align*}
85.5\:\text{km/hr}=\left(85\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=53.13\:\text{mi/hr}
\end{align*}

For 94.5 km/hr

94.5\:\text{km/hr}=\left(94.5\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=58.72\:\text{mi/hr}

Therefore, the range can be represented as 53.13 mi/hr to 58.72 mi/hr.

\text{Range} : 53.13 \ \text{mi/hr} - 58.72 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-12: Solving for the percent uncertainty of a 20-m tape that is off by 0.50 cm

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PROBLEM:

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

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SOLUTION:

The percent uncertainty can be calculated by the uncertainty divided by the total measure, but the units should be the same. That is,

\begin{align*}
\%\:\text{uncertainty} & =\frac{\delta _l}{l}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{20\:\text{m}\left(\frac{100\:\text{cm}}{1\:\text{m}}\right)}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{2000\:\text{cm}}\times 100\% \\
\\
&=0.025\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the percent uncertainty is 0.025%.

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Problem 1-11: Solving for the uncertainty given mass and percent uncertainty

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PROBLEM:

Suppose that your bathroom scale reads your mass as 65 kg with a 3%  uncertainty. What is the uncertainty in your mass (in kilograms)?

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SOLUTION:

The uncertainty is computed as 

\delta _m=\frac{3\:\%}{100\:\%}\times 65\:\text{kg}=1.95\:\text{kg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the uncertainty in the given mass is 1.95 kg.

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Problem 1-10: The average speed of the earth in its orbit in kilometers per second and in meters per second

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PROBLEM:

a) The average distance between the Earth and the Sun is  108 km. Calculate the average speed of the Earth in its orbit in kilometers per second. 

b) What is this is meters per second?

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SOLUTION:

Part A

We assume that the earth revolves around the sun in a circular manner. Therefore, the distance between the earth and the sun will be the radius of its orbit.

The total distance traveled the earth in full revolution is

\begin{align*}
d & =2\pi r \\
& = 2\pi \left(10\right)^8\:\text{km} \\
\end{align*}

The total time of travel is

\begin{align*}
t & =365.25\:\text{days}\left(\frac{24\:\text{hr}}{1\:\text{day}}\right)\left(\frac{3600\:\text{s}}{1\:\text{hr}}\right) \\
& = 3.15576\times 10^7\:\text{sec} \\
\end{align*}

Therefore, the average speed is

\begin{align*}
\text{speed} & =\frac{\text{distance}}{\text{time}}=\frac{d}{t} \\
 & = \frac{2\pi \left(10\right)^8\:\text{km}}{3.15576\times 10^7\:\text{s}} \\
& = 19.91\:\text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the average speed of the earth is 19.91 km/s.

Part B

Convert 19.91 km/s to m/s.

\begin{align*}
s & =\left(19.91\:\frac{\text{km}}{\text{s}}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right) \\
& =19\,910\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the velocity is 19 910 m/s. 

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Problem 1-9: The distance moved in 1 sec and the speed in kilometers per million years of tectonic plates

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PROBLEM:

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year.
(a) What distance does it move in 1.0 s at this speed?
(b) What is its speed in kilometers per million years?

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SOLUTION:

Part A

We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by d=r\times t.

The rate in cm/s is 

\begin{align*}
r & =\left(4\:\frac{\text{cm}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{year}}}{365\:\frac{1}{4}\:\bcancel{\text{days}}}\times \frac{1\:\bcancel{\text{day}}}{24\:\bcancel{\text{hours}}}\times \frac{1\:\bcancel{\text{hour}}}{60\:\bcancel{\text{mins}}}\times \frac{1\:\bcancel{\text{min}}}{60\:\text{s}}\right) \\
& = 1.2675 \times 10^{-7} \  \text{cm/s}
\end{align*}

The distance traveled is

\begin{align*}
d & = r \times t \\
d & = \left( 1.2675\times 10^{-7} \ \text{cm/s} \right)\left( 1.0  \ \text{s} \right) \\
d & = 1.3\times 10^{-7} \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We need to convert 4.0 cm/year to km/My.

\begin{align*}
4.0 \ \frac{\text{cm}}{\text{year}} & = 4.0\:\frac{\bcancel{\text{cm}}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{m}}}{100\:\bcancel{\text{cm}}}\times \frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\times \frac{1\:000\:000\:\bcancel{\text{years}}}{1\:\text{My}} \\ \\
& =40\:\text{km/My} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the speed of the tectonic plates is 40 km/My. 

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