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Q9PE

Expert-verifiedFound in: Page 11

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 \(40.0{\rm{ }}{{km} \mathord{\left/{\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}\).

**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 \({\rm{4}}{\rm{.0 }}{{{\rm{cm}}} \mathord{\left/ {\vphantom {{{\rm{cm}}} {{\rm{year}}}}} \right. \\} {{\rm{year}}}}\).

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,

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

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

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,

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

- 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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