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Pue to continental drift. the North American and Furnnean continents are drifting apart at an average speed of about \(3 \mathrm{cm}\) per year. At this speed, how long (in years) will it take for them to drift apart by another \(1500 \mathrm{m}\) (a little less than a mile)?

Short Answer

Expert verified
50,000 years.

Step by step solution

01

Convert meters to centimeters

Since the speed is given in centimeters per year, convert the distance they drift apart to centimeters. We know that 1 meter equals 100 centimeters, so we multiply 1500 meters by 100 to get the distance in centimeters.
02

Calculate centimeters

Convert the distance of 1500 meters into centimeters using the conversion from Step 1. The calculation becomes:\[1500 \text{ meters} \times 100 = 150,000 \text{ centimeters}\]Therefore, the distance is 150,000 centimeters.
03

Calculate time in years

To find out how long it will take for them to drift apart by 150,000 centimeters at a rate of 3 centimeters per year, divide the total distance by the speed of drift.The calculation becomes:\[\frac{150,000 \text{ cm}}{3 \text{ cm/year}} = 50,000 \text{ years}\]
04

Final Solution: Conclusion

The calculation in the previous step shows that it will take 50,000 years for North America and Europe to drift apart by another 1500 meters at the given rate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Drift
The concept of "Speed of Drift" is key in understanding continental drift. It's essentially how fast two landmasses, like the North American and European continents, are moving away from each other. In our example, the speed of drift is given as approximately 3 centimeters per year. This measurement tells us how much distance is added between the two continents yearly.

Continental drift is a slow process, and the speed is often given in very small units, such as centimeters, because of how gradual these shifts are. When we talk about 3 centimeters per year, it may seem minuscule — like a snail's pace in the grand scheme of things. But over thousands or even millions of years, these tiny increments add up to significant changes in the Earth's geography.

It’s important to distinguish that the speed of drift refers to average estimates over long periods, as tectonic movement can vary due to numerous geophysical factors.
Distance Conversion
When working with the speed of continental drift or any other similar measurements, converting the units of distance can be crucial, particularly when the speed and distance are given in different units. In our exercise, while the drift speed is given in centimeters per year, the distance that needs to be measured is originally in meters.

To address this, we convert the meters to centimeters, since the conversion is straightforward:
  • 1 meter equals 100 centimeters.
  • If we have a distance of 1500 meters, multiplying this by 100 gives us 150,000 centimeters.
By converting all measurements to the same unit (centimeters in this case), calculations are simplified, allowing us to directly compare or utilize these values in further mathematical processes.
Time Calculation
After converting distances into compatible units, the next step involves calculating how long the continental drift will take, known as "Time Calculation." When you calculate time from speed and distance, the formula is straightforward:
  • Divide the total distance by the speed.
In our example, the total distance is 150,000 centimeters, and the speed is 3 centimeters per year. Clearly dividing these figures, we get: \[\frac{150,000 \text{ centimeters}}{3 \text{ centimeters per year}} = 50,000 \text{ years}\]Thus, based on this speed, it will take an astonishing 50,000 years for the continents to drift apart by another 1500 meters. This lengthy timespan highlights the incredibly slow nature of continental drift, despite the seemingly simple calculations involved.

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Most popular questions from this chapter

A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1} .\) For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2} .\) The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

A skydiver is falling straight down, along the negative \(y\) direction. (a) During the initial part of the fall, her speed increases from 16 to \(28 \mathrm{m} / \mathrm{s}\) in \(1.5 \mathrm{s},\) as in part \(a\) of the figure. (b) Later, her parachute opens, and her speed decreases from 48 to \(26 \mathrm{m} / \mathrm{s}\) in \(11 \mathrm{s},\) as in part \(b\) of the drawing. Concepts: (i) Is her average acceleration positive or negative when her speed is increasing in part \(a\) of the figure? (ii) Is her average acceleration positive or negative when her speed is decreasing in part \(b\) of the figure? Calculations: In both instances (parts \(a\) and \(b\) of the figure) determine the magnitude and direction of her average acceleration.

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0\) s is 2.0 s. $$ \begin{array}{lcc} & \text { Initial velocity } v_{0} & \text { Acceleration } a \\\\\hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\ \hline\end{array}$$

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board ), assuming that the downward direction is chosen as the negative direction. \(J\) (b) What is the highest point he reaches above the water?

A golfer rides in a golf cart at an average speed of \(3.10 \mathrm{m} / \mathrm{s}\) for 28.0 s. She then gets out of the cart and starts walking at an average speed of \(1.30 \mathrm{m} / \mathrm{s} .\) For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is \(1.80 \mathrm{m} / \mathrm{s} ?\)

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