The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2 x 10^6 years? Write the answer in scientific notation, expressed to the exact decimal place.
The tectonic plate moves at a rate of 0.006 m per year. In 2 x 10^6 years, it would move a total distance of (0.006 m/year) * (2 x 10^6 years) = 12,000 m.
Written in scientific notation, this is 1.2 x 10^4 m.
To calculate the total distance the tectonic plate moves in 2 x 10^6 years, we need to multiply the rate of movement (0.006 m/year) by the number of years (2 x 10^6).
0.006 m/year x (2 x 10^6 years) = 0.006 x 2 x 10^6 m
Multiplying 0.006 by 2, we get:
0.012 x 10^6 m
Since 0.012 is not in scientific notation, we can express it as 1.2 x 10^-2 (moving the decimal point two places to the left).
So the total distance the tectonic plate would move in 2 x 10^6 years is:
1.2 x 10^-2 x 10^6 m
To simplify the scientific notation expression, we add the exponents, resulting in:
1.2 x 10^(6 - 2) m
Therefore, the tectonic plate would move 1.2 x 10^4 meters in 2 x 10^6 years, expressed to the exact decimal place in scientific notation.
To calculate the distance a tectonic plate would move in 2 x 10^6 years, we can multiply the average rate of movement (0.006 m/year) by the number of years (2 x 10^6).
0.006 m/year * 2 x 10^6 years = 0.006 * 2 * 10^6 m
To multiply the decimals, we ignore the zeros and multiply the non-zero values:
0.006 * 2 = 0.012
To multiply powers of 10, we add the exponents:
10^6 * 10^1 = 10^(6 + 1) = 10^7
So, the tectonic plate would move a total distance of 0.012 * 10^7 meters.
To express the answer in scientific notation, we have to adjust the decimal point so that there is one non-zero digit to the left of the decimal point. In this case, we need to move the decimal point one place to the left, which means we have to increase the exponent of 10 by 1:
0.012 * 10^7 = 1.2 * 10^6
Therefore, the tectonic plate would move 1.2 * 10^6 meters in 2 x 10^6 years.