The average tectonic plate moves at the rate of 0.006 m per year how many meters would it move in 2 times 10^6 years
To find the total distance the tectonic plate moves in 2 times 10^6 years, we can multiply the average rate of movement by the number of years.
Distance = Rate * Time
Distance = 0.006 m/year * 2 * 10^6 years
Distance = 0.006 * 2 * 10^6 m
Distance = 12 * 10^4 m
Distance = 1.2 * 10^5 m
So, the tectonic plate would move 1.2 * 10^5 meters in 2 times 10^6 years.
To find out how many meters a tectonic plate would move in 2 times 10^6 years at a rate of 0.006 m per year, you need to multiply the rate by the number of years.
Rate of plate movement: 0.006 m/year
Number of years: 2 * 10^6
To calculate the total distance, multiply the rate (0.006 m/year) by the number of years (2 * 10^6):
Distance = Rate * Number of years
Distance = 0.006 m/year * (2 * 10^6) years
Converting 2 * 10^6 to standard notation:
Distance = 0.006 m/year * 2,000,000 years
Now, multiply the numbers together:
Distance = 0.006 * 2,000,000
Calculating the result:
Distance = 12,000 meters
Therefore, a tectonic plate would move approximately 12,000 meters in 2 times 10^6 years at a rate of 0.006 m per year.
To solve this problem, we can use the formula:
Distance = Rate x Time
Given that the average tectonic plate moves at a rate of 0.006 m per year, and we want to find the distance it would move in 2 times 10^6 years, we can substitute these values into the formula:
Distance = 0.006 m/year x 2 x 10^6 years
Now, let's calculate this:
Distance = 0.006 m/year x 2 x 10^6 years
= 0.006 m/year x 2,000,000 years
= 12,000 meters
Therefore, the tectonic plate would move 12,000 meters in 2 times 10^6 years.