To find out how many meters a tectonic plate would move in \(2 \times 10^6\) years at a rate of \(0.006\) meters per year, we can use the following calculation:
\[ \text{Distance} = \text{Rate} \times \text{Time} \]
\[ \text{Distance} = 0.006 , \text{m/year} \times 2 \times 10^6 , \text{years} \]
Now, performing the multiplication:
\[ \text{Distance} = 0.006 \times 2 = 0.012 \]
Next, we multiply this by \(10^6\):
\[ \text{Distance} = 0.012 \times 10^6 \]
We can express \(0.012\) in scientific notation:
\[ 0.012 = 1.2 \times 10^{-2} \]
Thus,
\[ \text{Distance} = 1.2 \times 10^{-2} \times 10^6 = 1.2 \times 10^{4} \]
So, the distance a tectonic plate would move in \(2 \times 10^6\) years is:
\[ \boxed{1.2 \times 10^4} , \text{m} \]