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The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×10^6 years? Write the answ...
The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×106
years? Write the answer in scientific notation, expressed to the exact decimal place.(1 point)
a
m
years? Write the answer in scientific notation, expressed to the exact decimal place.(1 point)
a
m
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Answered by
GPT-4o mini
To calculate how many meters a tectonic plate would move in \( 2 \times 10^6 \) years at a rate of \( 0.006 \) m per year, you can use the following formula:
\[
\text{Distance} = \text{Rate} \times \text{Time}
\]
Given that the rate is \( 0.006 \) m/year and the time is \( 2 \times 10^6 \) years, we can plug in the values:
\[
\text{Distance} = 0.006 \, \text{m/year} \times 2 \times 10^6 \, \text{years}
\]
Calculating that gives:
\[
\text{Distance} = 0.006 \times 2 \times 10^6
\]
\[
\text{Distance} = 0.012 \times 10^6
\]
\[
\text{Distance} = 1.2 \times 10^4 \, \text{m}
\]
So, the tectonic plate would move \( 1.2 \times 10^4 \) meters in \( 2 \times 10^6 \) years.
Therefore, the answer is:
\[
\boxed{1.2 \times 10^4} \, \text{m}
\]
\[
\text{Distance} = \text{Rate} \times \text{Time}
\]
Given that the rate is \( 0.006 \) m/year and the time is \( 2 \times 10^6 \) years, we can plug in the values:
\[
\text{Distance} = 0.006 \, \text{m/year} \times 2 \times 10^6 \, \text{years}
\]
Calculating that gives:
\[
\text{Distance} = 0.006 \times 2 \times 10^6
\]
\[
\text{Distance} = 0.012 \times 10^6
\]
\[
\text{Distance} = 1.2 \times 10^4 \, \text{m}
\]
So, the tectonic plate would move \( 1.2 \times 10^4 \) meters in \( 2 \times 10^6 \) years.
Therefore, the answer is:
\[
\boxed{1.2 \times 10^4} \, \text{m}
\]
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