Physical Quantities and Units
Accuracy, Precision, and Significant Figures
Approximation
PROBLEM:
One nerve impulse lasts for 10-3 s.
\text{max firing rate}=\frac{1\:\text{nerve impulse}}{10^{-3}\:\text{sec}}=10^3\:\text{impulses/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)PROBLEM:
SOLUTION:
The mass of a hummingbird is 10-2 kg, while the mass of a cell is 10-15 kg. The number of cells in the hummingbird is
\frac{10^{-2}}{10\left(10^{-15}\right)}=10^{12}\:\text{cells} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)The mass of a person is 102 kg.
\frac{10^2}{10\left(10^{-15}\right)}=10^{16}\:\text{cells}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)PROBLEM:
SOLUTION:
The greatest ocean depth is 104 m, while the earth’s diameter is 107 m
\frac{\text{Ocean's Depth}}{\text{Earth's Diameter}}= \frac{10^4}{10^7}=\frac{1}{1000}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)The highest mountain is also roughly 104 m.
\displaystyle \frac{10^4}{10^7}=\frac{1}{1000}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)PROBLEM:
SOLUTION:
The cell membrane is 10-8 m while the hydrogen atom is 10-10 m. The number of atoms in the cell membrane is
\begin{align*}
\text{no. of atoms} & =\frac{\text{d}_{\text{m}}}{2\text{d}_{\text{H}}}\\ \\
& =\frac{10^{-8}}{2\left(10^{-10}\right)} \\ \\
& =50\:\text{atoms} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}PROBLEM:
SOLUTION:
The number of atoms is
\begin{align*}
\text{no. of atoms} & =\frac{m_{bact}}{10\:m_H} \\ \\
& = \frac{10^{-15}}{10\left(10^{-27}\right)}\\\\
&=10^{11}\:\text{atoms} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}PROBLEM:
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)PROBLEM:
SOLUTION:
We are looking for generations passed in history. The general assumptions are:
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*}PROBLEM:
SOLUTION:
The general assumptions:
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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