Tag Archives: College Physics Solutions

Problem 2-6: The average speed and average velocity of a spinning helicopter blade

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PROBLEM:

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.

(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference.

(b) What is its average velocity over one revolution?

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SOLUTION:

Part A

The average speed of the blade tip is equal to the distance traveled divided by the time elapsed.

\begin{align*} \text{speed} & =\frac{ \text{distance traveled}}{ \text{time elapsed}} \\ \\ & =\frac{2\pi \text{r}}{ \text{t}} \\ \\ & =\frac{2\pi \left(5.00\text{m}\right)}{60\:\text{s}}\times 100\:\text{rev} \\ \\ & =52.36\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

After one revolution, the tip of the blade is at the same position as it is originally. This means that the displacement is zero. Thus, the velocity is zero.

\text{v}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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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.

\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.

\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

\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

\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

\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.

\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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