Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of\({\bf{4}}.{\bf{0}}{\rm{ }}{{{\rm{cm}}} \mathord{\left/ {\vphantom {{{\rm{cm}}} {{\rm{year}}}}} \right. \\} {{\rm{year}}}}\). (a) What distance does it move in \({\rm{1 s}}\) at this speed? (b) What is its speed in kilometers per million years?

The distance does it move in \({\rm{1 s}}\) at given speed is \(d = 1.268{\rm{ }}nm\)
What is its speed in kilometers per million years is \(40.0{\rm{ }}{{km} \mathord{\left/{\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: A concept of velocity and distance.

Speed is a measure of the distance traveled by objects over a period of time. Here is a word count that shows the relationship between distance, speed and time: Speed is equal to the distance traveled divided by the time it takes to get there.

Given data

Consider the given data as below.

The average speed of Tectonic plates is \({\rm{4}}{\rm{.0 }}{{{\rm{cm}}} \mathord{\left/ {\vphantom {{{\rm{cm}}} {{\rm{year}}}}} \right. \\} {{\rm{year}}}}\).

Step 2: (a) Distance in \({\rm{1 sec}}\):

The distance defines by using the following formula.

\(d = vt\) ….. (1)

Here, \(d\) is the distance, \(v\) is the velocity, and \(t\) is the time.

The conversion if year to second is,

\(\begin{array}{c}1{\rm{ }}year = 1{\rm{ }}year \times \frac{{365{\rm{ }}days}}{{1{\rm{ }}year}} \times \frac{{24{\rm{ }}hour}}{{1{\rm{ }}day}} \times \frac{{60{\rm{ }}\min }}{{1{\rm{ }}hour}} \times \frac{{60{\rm{ }}\sec }}{{1{\rm{ }}\min }}\\ = 31536000{\rm{ }}\sec \end{array}\)

Substitute \({\rm{4}}{\rm{.0 }}{{{\rm{cm}}} \mathord{\left/

{\vphantom {{{\rm{cm}}} {{\rm{year}}}}} \right.

\kern-\nulldelimiterspace} {{\rm{year}}}}\) for \(v\) and \(\frac{1}{{31536000}}{\rm{ }}\sec \) for \(t\) in the above equation.

\(\begin{array}{c}d = 4.0 \times \frac{1}{{31536000}}cm\\ = 1.268 \times {10^{ - 7}}{\rm{ }}cm\\ = 1.268 \times {10^{ - 7}}{\rm{ }}cm\left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1.0{\rm{ }}cm}}} \right)\\ = 1.268 \times {10^{ - 9}}{\rm{ }}m\end{array}\)

The relation between nanometer and meter is as follows.

\(1{\rm{ }}nm = {10^{ - 9}}{\rm{ }}m\)

Therefore, the distance will be,

\(d = 1.268{\rm{ }}nm\)

Step 3: (b) Velocity in

\({{{\rm{km}}} \mathord{\left/ {\vphantom {{{\rm{km}}} {{\rm{million yr}}}}} \right. \\} {{\rm{million yr}}}}\) .

The relation between year and million years is as follow.

\(1{\rm{ }}year = 1{\rm{ }} \times {10^{ - 6}}million{\rm{ }}yr\)

Therefore, the speed of the tectonic plates is,

\(\begin{aligned}{c}v = 4.0{\rm{ }}{{cm} \mathord{\left/{\vphantom {{cm} {year}}} \right.\\} {year}}\\ = \left( {4.0{\rm{ }}\frac{{cm}}{{year}} \times \frac{{1 \times {{10}^{ - 5}}{\rm{ }}km}}{{1{\rm{ }}cm}}} \right) \times \left( {\frac{{1{\rm{ }}year}}{{1 \times {{10}^{ - 6}}{\rm{ }}millon{\rm{ }}yr}}} \right)\\ = 4.0 \times {10^1}{\rm{ }}{{km} \mathord{\left/{\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}\\ = 40.0{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}\end{aligned}\)

Step 4: Conclusion:

The distance does it move in \({\rm{1 s}}\) at given speed is \(d = 1.268{\rm{ }}nm\).
What is its speed in kilometers per million years is \(40.0{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}\).

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Short Answer

Step by Step Solution

Step 1: A concept of velocity and distance.

Step 2: (a) Distance in \({\rm{1 sec}}\):

Step 3: (b) Velocity in

Step 4: Conclusion:

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: A concept of velocity and distance.

Step 2: (a) Distance in \({\rm{1 sec}}\):

Step 3: (b) Velocity in

Step 4: Conclusion:

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