Question

Tectonic plate moves at rate 0.006 m per year how many meters would it move in 2 times 10^6 years… write in scientific notation

Answer 1

To find out how far a tectonic plate would move in \(2 \times 10^6\) years at a rate of \(0.006\) meters per year, you can use the formula:

\[
\text{Distance} = \text{Rate} \times \text{Time}
\]

Substituting the given values:

\[
\text{Distance} = 0.006 \, \text{m/year} \times (2 \times 10^6 \, \text{years})
\]

Calculating this:

\[
\text{Distance} = 0.006 \times 2 \times 10^6
\]

\[
\text{Distance} = 0.012 \times 10^6
\]

Now, converting \(0.012\) into scientific notation:

\[
0.012 = 1.2 \times 10^{-2}
\]

So, we have:

\[
\text{Distance} = 1.2 \times 10^{-2} \times 10^6
\]

When multiplying powers of ten, you add the exponents:

\[
\text{Distance} = 1.2 \times 10^{6 - 2}
\]

\[
\text{Distance} = 1.2 \times 10^4
\]

Thus, the tectonic plate would move \(1.2 \times 10^4\) meters in \(2 \times 10^6\) years.