on april 11, 2012, two earthquakes were measured off the northwest coast of sumatra. the first had a…

on april 11, 2012, two earthquakes were measured off the northwest coast of sumatra. the first had a magnitude of 8.6. the second had a magnitude of 8.2. by what approximate factor was the intensity of the first earthquake greater than the intensity of the second earthquake?\n\nm = logleft(\frac{i}{i_0}\right)\n\n$m =$ the magnitude of an earthquake\n$i =$ the intensity of an earthquake\n$i_0=$ the smallest seismic activity that can be measured, which is 1\n\no 0.40\no 1.02\no 1.05\no 2.51

on april 11, 2012, two earthquakes were measured off the northwest coast of sumatra. the first had a magnitude of 8.6. the second had a magnitude of 8.2. by what approximate factor was the intensity of the first earthquake greater than the intensity of the second earthquake?\n\nm = logleft(\frac{i}{i_0}\right)\n\n$m =$ the magnitude of an earthquake\n$i =$ the intensity of an earthquake\n$i_0=$ the smallest seismic activity that can be measured, which is 1\n\no 0.40\no 1.02\no 1.05\no 2.51

Answer

Explanation:

Step1: Write equations for each earthquake

For the first earthquake with magnitude $M_1 = 8.6$, we have $8.6=\log\left(\frac{I_1}{I_0}\right)$. Since $I_0 = 1$, then $I_1 = 10^{8.6}$. For the second earthquake with magnitude $M_2=8.2$, we have $8.2=\log\left(\frac{I_2}{I_0}\right)$. Since $I_0 = 1$, then $I_2 = 10^{8.2}$.

Step2: Find the ratio of intensities

We want to find the ratio $\frac{I_1}{I_2}$. Substitute $I_1 = 10^{8.6}$ and $I_2 = 10^{8.2}$ into the ratio: $\frac{I_1}{I_2}=\frac{10^{8.6}}{10^{8.2}}$. Using the rule of exponents $\frac{a^m}{a^n}=a^{m - n}$, we get $\frac{10^{8.6}}{10^{8.2}}=10^{8.6 - 8.2}=10^{0.4}$.

Step3: Calculate the value of $10^{0.4}$

$10^{0.4}\approx2.51$.

Answer:

D. 2.51