The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×106

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:

\[
\boxed{1.2 \times 10^4} \, \text{m}
\]