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 correct answer is 1.2x10 to the power of 4
To find the distance the average tectonic plate would move in 2 x 10^6 years, we can multiply the rate of movement (0.006 m per year) by the number of years (2 x 10^6):
0.006 m/year * 2 x 10^6 years = 12,000,000 m
Expressed in scientific notation, this is 1.2 x 10^7 meters.
To find out how many meters a tectonic plate would move in 2 x 10^6 years, we can multiply the rate of movement (0.006 m/year) by the number of years.
0.006 m/year * 2 x 10^6 years = 1.2 x 10^4 m.
Therefore, the tectonic plate would move 1.2 x 10^4 meters in 2 x 10^6 years.
To calculate the distance that the average tectonic plate moves in 2 x 10^6 years, you simply need to multiply the rate of movement (0.006 m/year) by the number of years (2 x 10^6). Here's how you can do it step by step:
1. Multiply the rate of movement (0.006 m/year) by the number of years (2 x 10^6):
Distance = 0.006 m/year * (2 x 10^6 years)
2. Multiply the numbers:
Distance = 0.006 * 2 * 10^6 m
3. Simplify the expression by multiplying 0.006 and 2:
Distance = 0.012 * 10^6 m
4. Convert 0.012 to scientific notation by moving the decimal place to the right until you have a number between 1 and 10:
Distance = 1.2 * 10^(-2) * 10^6 m
5. Apply the laws of exponents by adding the exponents:
Distance = 1.2 * 10^(6 - 2) m
6. Subtract the exponents:
Distance = 1.2 * 10^4 m
Therefore, the average tectonic plate would move 1.2 * 10^4 meters in 2 x 10^6 years.