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 that the tectonic plate would move in 2×10^6 years, we can multiply the average rate of movement (0.006 m/year) by the number of years (2×10^6 years).
(0.006 m/year) * (2×10^6 years) = 0.006 * 2 * 10^6 m
Simplifying the equation, we get:
= 0.012 * 10^6 m
To express this number in scientific notation, we can convert 0.012 to 1.2 and adjust the exponent as necessary. In this case, we move the decimal place 3 places to the right and increase the exponent by 3.
= 1.2 * 10^3 m
Therefore, the tectonic plate would move 1.2 * 10^3 meters in 2×10^6 years.
To find out how many meters a tectonic plate would move in 2x10^6 years, we can multiply the rate of movement per year (0.006 m/year) by the total number of years (2x10^6).
0.006 m/year * 2x10^6 years = 12x10^(-3) x 2x10^6
= 24x10^(3+6)
= 24x10^(9)
= 2.4x10^(10) meters
Therefore, a tectonic plate would move 2.4x10^(10) meters in 2×10^6 years.
To find out how many meters the tectonic plate would move in 2×10^6 years, we can multiply the rate of movement (0.006 m/year) by the number of years (2×10^6).
First, let's convert the years into regular numerical form. "2×10^6" means 2 multiplied by 10 raised to the power of 6, which is equal to 2,000,000.
Now, we can calculate the movement of the tectonic plate by multiplying the rate (0.006 m/year) by the number of years (2,000,000):
0.006 m/year x 2,000,000 years = 12,000 meters
Therefore, the tectonic plate would move 12,000 meters in 2×10^6 years.
In scientific notation, this can be written as 1.2 × 10^4 meters.