The average tectonic plate moves at the rate of 0.006 m per year. How many meters would it move in 2×106 years? Write the answer in scientific notation, expressed to the exact decimal place
To find the distance the tectonic plate would move in 2x10^6 years, we multiply its average speed by the number of years:
0.006 m/year * 2x10^6 years = 12,000 m
The answer in scientific notation, expressed to the exact decimal place, is 1.2x10^4 m.
To find how many meters a tectonic plate would move in 2×106 years, we can multiply the rate of movement (0.006 m/year) by the number of years.
First, multiply the rate by the number of years:
0.006 m/year × 2×106 years
To simplify multiplication involving scientific notation, we multiply the numerical parts and add the exponents. In this case, the number of meters moved is:
0.006 × 2 × 106 = 0.012 × 106
Now, let's express the answer in scientific notation:
0.012 × 106 = 1.2 × 10-2 × 106
To simplify this further, we can add the exponents:
1.2 × 10-2 × 106 = 1.2 × 104
Therefore, a tectonic plate would move 1.2 × 104 meters in 2×106 years.
To find the distance a tectonic plate would move in 2x10^6 years, we can multiply the average rate of movement per year (0.006 m) by the number of years (2x10^6):
0.006 m/year * 2x10^6 years
Multiplying the numbers:
0.006 * 2 * 10^6 = 12 * 10^(-3) * 2 * 10^6
Combining the factors of 10:
12 * 2 * 10^(-3) * 10^6 = 24 * 10^3 * 10^6
Simplifying the powers of 10:
24 * 10^(3+6) = 24 * 10^9
Expressing the answer in scientific notation:
2.4 x 10^10 meters