The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×10^6 years? Write the answer in scientific notation, expressed to the exact decimal place.
To find the total distance the tectonic plate would move in 2×10^6 years, we can multiply the rate of movement by the number of years:
0.006 m/year * 2×10^6 years = 0.006 * 2 * 10^6 = 1.2 * 10^4 m
So, the tectonic plate would move 1.2 * 10^4 meters in 2×10^6 years.
To find out how many meters the average tectonic plate would move in 2×10^6 years, we can multiply the rate of movement by the number of years.
The rate of movement is given as 0.006 m per year, so we can calculate the total distance moved over 2×10^6 years as:
Total distance = rate × time
= 0.006 m/year × 2×10^6 years
To multiply the numbers, we multiply the values and add together the exponents of 10:
0.006 × 2×10^6
= 0.006 × 2 × 10^6
= 0.012 × 10^6
Now we can simplify further by writing 0.012 in scientific notation:
0.012 = 1.2 × 10^(-2)
Thus, the total distance moved by the average tectonic plate in 2×10^6 years is:
1.2 × 10^(-2) × 10^6
= 1.2 × 10^(4)
Therefore, the tectonic plate would move 1.2 × 10^(4) meters in 2×10^6 years.
To determine the distance that a tectonic plate would move in 2x10^6 years, we will multiply the average rate of movement by the given time.
Rate of movement = 0.006 m/year
Time = 2x10^6 years
Distance = Rate of movement × Time
Distance = 0.006 m/year × 2x10^6 years
To multiply these numbers with exponents, we can simply add the exponents: 6 + 6 = 12.
Distance = 0.006 × 10^12 m
So, in scientific notation, a tectonic plate would move 6x10^9 meters in 2x10^6 years.