Tag Archives: College Physics Solutions

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.

Δx=7 m−0 m=+7 m  (Answer)\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.

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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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

V=πr2h=π(7.5 cm2)2(3.25 cm)=143.5806 cm3  (Answer)\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

% uncr=0.002 cm7.500 cm×100%=0.027%% unch=0.001 cm3.25 cm×100%=0.031%\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

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

The uncertainty in the volume is

δvol=0.058100×143.5806=0.083 cm3  (Answer) \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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