Tag Archives: Approximation

Solution Guide to College Physics by Openstax Chapter 1 Banner

Chapter 1: Introduction: The Nature of Science and Physics

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Physical Quantities and Units

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Accuracy, Precision, and Significant Figures

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

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Approximation

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

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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.

max firing rate=1 nerve impulse10−3 sec=103 impulses/sec  (Answer)\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

10−210(10−15)=1012 cells  (Answer)\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.

10210(10−15)=1016 cells  (Answer)\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

Ocean’s DepthEarth’s Diameter=104107=11000  (Answer)\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.

104107=11000  (Answer)\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

no. of atoms=dm2dH=10−82(10−10)=50 atoms  (Answer)\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

no. of atoms=mbact10 mH=10−1510(10−27)=1011 atoms  (Answer)\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

ThTn=2×109 sec10−22 sec=2×1031 times  (Answer) \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

1 history=1 history×1011 sec1 history×1 generation13 lifetime×0.5 lifetime109 sec=150 generations  (Answer)\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

1 lifetime=(1 lifetime)(109 sec0.5 lifetime)(1 heartbeat1 sec)=2×109 heartbeats  (Answer)\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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