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Expert-verified Found in: Page 11 ### College Physics (Urone)

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?

1. The distance does it move in $${\rm{1 s}}$$ at given speed is $$d = 1.268{\rm{ }}nm$$
2. What is its speed in kilometers per million years is $$40.0{\rm{ }}{{km} \mathord{\left/{\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}$$.
See the step by step solution

## 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:

1. The distance does it move in $${\rm{1 s}}$$ at given speed is $$d = 1.268{\rm{ }}nm$$.
2. What is its speed in kilometers per million years is $$40.0{\rm{ }}{{km} \mathord{\left/ {\vphantom {{km} {millon{\rm{ }}years}}} \right. \\} {millon{\rm{ }}years}}$$. ### Want to see more solutions like these? 