PROBLEM:

Assuming one nerve impulse must end before another can begin, what is the maximum firing rate of a nerve in impulses per second?

Solution:

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:

(a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium.

(b) Making the same assumption, how many cells are there in a human?

SOLUTION:

Part A

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)

Part B

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:

(a) What fraction of Earth’s diameter is the greatest ocean depth?

(b) The greatest mountain height?

SOLUTION:

Part A

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)

Part B

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:

Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10^-27 kg; and the mass of a bacterium is on the order of 10^-15 kg)

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:

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 .)

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)