Category Archives: Physics

Problem 2-5: The Earth’s average speed and velocity

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PROBLEM:

(a) Calculate Earth’s average speed relative to the Sun.

(b) What is its average velocity over a period of one year?

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SOLUTION:

Part A

The average speed of the earth is equal to the distance traveled divided by time.

Average speed of the Earth=Distance TraveledTotal time of Travel=2πrt=2π(1.50×1011 m)365.25 days×1 day24 hours×1 hour3600 sec=2.99×104 m/s  (Answer)\begin{align*} \text{Average speed of the Earth} & =\frac{\text{Distance Traveled}}{\text{Total time of Travel}} \\ \\ & =\frac{2\pi \text{r}}{\text{t}} \\ \\ & =\frac{2\pi \left(1.50\times 10^{11}\text{ m}\right)}{365.25\:\text{days}}\times \frac{1\:\text{day}}{24\:\text{hours}}\times \frac{1\:\text{hour}}{3600\:\text{sec}} \\ \\ & =2.99\times 10^4\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

After a period of 1 year, the planet Earth has already returned to its original position with respect to the Sun. This means that we do not have any displacement. Therefore, the average velocity is zero.

Average velocity=0 m/s  (Answer)\text{Average velocity}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 2-4: Distance and Displacement of a given Path D

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PROBLEM:

Find the following for path D 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 position of D started from 9 m, then went to 3 m, and then went back to 5 m. The distance traveled by D is the sum of all the paths

distance=6 m+2 m=8 m  (Answer)\text{distance}=6\:\text{m}+2\:\text{m}=8\:\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 sign

∣Δx∣=∣9 m−5 m∣=4 m  (Answer)\left|\Delta x\right|=\left|9\:\text{m}-5\:\text{m}\right|=4\:\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

Δx=5 m−9 m=−4 m  (Answer)\Delta x=5\:\text{m}-9\:\text{m}=-4\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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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.

distance=8 m+2 m+2 m=12 m  (Answer)\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.

∣Δx∣=∣10 m−2 m∣=8 m  (Answer)\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

Δx=10 m−2 m=+8 m  (Answer)\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

Δx=7 m−12 m=−5 meters  (Answer)\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.

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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