Tag Archives: College Physics by Openstax Solution Manual

Problem 2-3: Distance and Displacement of a given Path C

PROBLEM:

Find the following for path C in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part a

The distance is the sum of all the paths of C.

\text{distance}=8\:\text{m}+2\:\text{m}+2\:\text{m}=12\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part b

The magnitude of the displacement is the difference between the final position and the initial position without regard to the sign.

\left|\Delta x\right|=\left|10\:\text{m}-2\:\text{m}\right|=8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part c

The displacement is the difference between the final position and the initial position, taking into account the sign

\Delta x=10\:\text{m}-2\:\text{m}=+8\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 2-2: Distance and displacement of a given Path B

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PROBLEM:

Find the following for path B in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part a

Based on the figure, B travels from 12 to 7. The distance traveled is 5 meters.

Part b

The magnitude of the displacement is 5 meters.

Part c

The displacement is calculated keeping in mind the sign. The motion started at 12 and ended at 7. Therefore, the displacement is

\Delta x=7\:\text{m}-12\:\text{m}=-5\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 2-1: Distance and displacement for a given path A

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PROBLEM:

Find the following for path A in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part A

A travels from 0 to 7. The distance traveled is 7 meters.

Part B

The magnitude of the displacement is 7 meters.

Part C

The displacement is the difference between the final and initial positions.

\begin{align*} \Delta x & =7\:\text{m}-0\:\text{m}=+7\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

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Problem 1-36: The maximum firing rate of a nerve

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PROBLEM:

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

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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)

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Problem 1-35: The approximate number of cells in a hummingbird and in a human

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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?

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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)

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Problem 1-34: Earth’s diameter vs greatest ocean depth and vs greatest mountain height

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PROBLEM:

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

(b) The greatest mountain height?

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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)

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Problem 1-33: The number of atoms thick of a cell membrane

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PROBLEM:

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

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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*}

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Problem 1-32: The approximate number of atoms in a bacterium

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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)

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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*}

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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 10¹¹seconds 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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Find the following for path C in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

Part a

Part b

Part c

Find the following for path B in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

Find the following for path A in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

Part A

Part B

Part C

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

Solution:

(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?

Part A

Part B

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

(b) The greatest mountain height?

Part A

Part B

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom?

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)

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 .)

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

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)

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 .)