Category Archives: Sciences

Includes Engineering Sciences such as Statics, Dynamics, Strength of Materials, Physics, Fluid Mechanics, Chemistry

Problem 1-31: The lifetime of a human vs the mean life of an extremely unstable atomic nucleus

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PROBLEM:

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of  10-22 s .)


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SOLUTION:

The lifetime of a human is 2×109 s, while the lifetime of an unstable atomic nucleus is 10-22 s.

Therefore, a human lifetime is much longer by

 \frac{T_h}{T_n}=\frac{2\times 10^9\:\text{sec}}{10^{-22}\:\text{sec}}=2\times 10^{31}\:\text{times}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-30: The number of generations passed since the year 0 AD

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PROBLEM:

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?


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SOLUTION:

We are looking for generations passed in history. The general assumptions are:

  • There are 1011 seconds in 1 history
  • In 1 generation, there is 1/3 of a lifetime.
  • In half a lifetime, there are 109 seconds

Therefore, the number of generations passed is

\begin{align*}
1\ \text{history} & =1\:\text{history}\times \frac{10^{11}\:\text{sec}}{1\:\text{history}}\times \frac{1\:\text{generation}}{\frac{1}{3}\:\text{lifetime}}\times \frac{0.5\:\text{lifetime}}{10^9\:\text{sec}} \\
& =150\:\text{generations} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Problem 1-29: The number of heartbeats in a lifetime

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PROBLEM:

How many heartbeats are there in a lifetime?


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SOLUTION:

The general assumptions:

  • There is 1 heartbeat in 1 second
  • In half a lifetime, there are 109 seconds.

Therefore, in a lifetime, the number of heartbeats is

\begin{align*}
1\:\text{lifetime} & =\left(1\:\text{lifetime}\right)\left(\frac{10^9\:\text{sec}}{0.5\:\text{lifetime}}\right)\left(\frac{1\:\text{heartbeat}}{1\:\text{sec}}\right) \\
& =2\times 10^9\:\text{heartbeats} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Problem 1-28: Calculating the volume and its uncertainty of a car piston with dimensional uncertainties

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PROBLEM:

A car engine moves a piston with a circular cross section of  7.500±0.002 cm diameter a distance of 3.250±0.001 cm  to compress the gas in the cylinder.

(a) By what amount is the gas decreased in volume in cubic centimeters?

(b) Find the uncertainty in this volume.


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SOLUTION:

Part A

The average volume is

\begin{align*}
V & =\pi r^2h \\
& =\pi \left(\frac{7.5\:\text{cm}}{2}\right)^2\left(3.25\:\text{cm}\right) \\
& =143.5806\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part B

Solve for the percent uncertainties of each dimension

\begin{align*}
\%\:unc_r & =\frac{0.002\:\text{cm}}{7.500\:\text{cm}}\times 100\%=0.027\% \\
\%\:unc_h & =\frac{0.001\:\text{cm}}{3.25\:\text{cm}}\times 100\%=0.031\% \\
\end{align*}

The percent uncertainty in the volume is the combined effect of the uncertainties of the dimensions

\text{\%\:unc}_{vol}=0.027\%+0.031\%=0.058\%

The uncertainty in the volume is

 \delta _{vol}=\frac{0.058}{100}\times 143.5806=0.083\:\text{cm}^3  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-27: Calculating the area and its uncertainty of a room with a given dimensional uncertainties

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PROBLEM:

The length and width of a rectangular room are measured to be 3.955 ±0.005 m and 3.050 ± 0.005 m . Calculate the area of the room and its uncertainty in square meters.


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SOLUTION:

The average area of the room is

\begin{align*}
A & =l\times w \\
& =3.955\:\text{m}\times 3.050\:\text{m} \\
& =12.06\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Compute for the percent uncertainties of each dimension.

\begin{align*}
\text{\%\:unc}_{width} & =\frac{0.005\:\text{m}}{3.050\:\text{m}}\times 100\%=0.1639\% \\
\text{\%\:unc}_{length} & =\frac{0.005\:\text{m}}{3.955\:\text{m}}\times 100\%=0.1264\:\%
\end{align*}

The percent uncertainty in the area is the combined effect of the uncertainties of the length and width.

\text{\%\:unc}_{area}=0.1639\%+0.1264\%=0.2903\%

The uncertainty in the area is

\delta _{area}=\frac{0.2903\:\%}{100\:\%}\times 12.06\:\text{m}^2=0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the area is

A=12.06\pm 0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-26: Uncertainty and percent uncertainty of a pound-mass (lbm) unit

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PROBLEM:

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm=0.4539 kg.

(a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty?

(b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?


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SOLUTION:

Part A

The percent uncertainty of the lbm is

\text{\%\:uncertainty}_{\text{lbm}}=\frac{0.0001\:\text{kg}}{0.4539\:\text{kg}}\times 100\%=0.022\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For uncertainty of 1 kg, the corresponding lbm is

\begin{align*}
\text{lbm} & =\frac{\delta_{\text{lbm}}}{\text{\%\:uncertainty}_{\text{lbm}}}\times 100\% \\
\\
 & =\frac{1\:\text{kg}}{0.02\:\%}\times \frac{1\:\text{lbm}}{0.04539\:\text{kg}}\times 100\% \\
\\
& =11015.64\:\text{lbm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) 
\end{align*}

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Problem 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

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PROBLEM:

The sides of a small rectangular box are measured to be 1.80±0.01 cm, 2.05±0.02 cm, and 3.1±0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.


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SOLUTION:

The average volume of the box is

\begin{align*}
\text{Volume} & =l\times w\times h \\
& =\left(1.80\:\text{cm}\right)\left(2.05\:\text{cm}\right)\left(3.1\:\text{cm}\right) \\
&= 11.4\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

The percent uncertainty for each of the dimensions:

\begin{align*}
1.80\pm 0.01\:\text{cm}\:\rightarrow & \:\frac{0.01\:\text{cm}}{1.80\:\text{cm}}\times 100\%=0.556\% \\
\\
2.05\pm 0.02\:\text{cm}\:\rightarrow & \:\frac{0.02\:\text{cm}}{2.05\:\text{cm}}\times 100\%=0.976\% \\
\\
3.1\pm 0.1\:\text{cm}\:\rightarrow &\:\frac{0.1\:\text{cm}}{3.1\:\text{cm}}\times 100\%=3.226\% \\
\end{align*}

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions.

\begin{align*}
\%\:\text{uncertainty}_{\text{volume}} & =0.556\%+0.976\%+3.226\% \\
& =4.758\%
\end{align*}

The uncertainty of the volume is

\begin{align*}
\delta _{\text{volume}} & =0.04758\times 11.4\:\text{cm}^3 \\
& =0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the volume is

\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-24: Calculating uncertainties in distance, time and speed of a marathon runner

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PROBLEM:

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?


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SOLUTION:

Part A

The percent uncertainty in the distance is

\begin{align*}
\text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\
 & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part B

The uncertainty in time is

\begin{align*}
\text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\
& =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part C

The average speed is

\begin{align*}
\text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\
& = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

\begin{align*}
\text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\
& =0.0593\%+0.0111\% \\
&  =0.0704\% \\
\end{align*}

Therefore, the uncertainty in the speed is

\begin{align*}
\delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\
& = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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Problem 1-23: Calculating for the time it takes a marathon runner to finish 26.22 miles given a speed of 9.5 mi/h

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PROBLEM:

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?


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SOLUTION:

The relationship of the velocity (or speed), time, and distance is

v=\frac{d}{t}

So, by rearranging the equation, we can equate time to

t=\frac{d}{v}

By direct substitution

t=\frac{26.22\:\text{mi}}{9.5\:\text{mi/hr}}=2.76\:\text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-22: Solving for the area of a circle with a given diameter

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PROBLEM:

What is the area of a circle 3.102 cm in diameter?


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SOLUTION:

The area of a circle can be computed using the formula below when the radius is given.

A=\pi r^2

We also know that the radius is half the diameter, so the area can be calculated using the formula,

A=\pi \left(\frac{d}{2}\right)^2

So, by direct substitution

A=\pi \left(\frac{3.102\:\text{cm}}{2}\right)^2=7.557\:\text{cm}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The area of the circle is 7.557 square centimeters.


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