Respuesta :
Answer:
The energy, 'E', increases by a factor of 1000,000,000
Step-by-step explanation:
The given equation relating moment magnitude of an earthquake and the energy (in ergs) released by it obtained from a similar question online is presented as follows; [tex]M_W = \dfrac{2}{3} \times log_{10} (E) - 10.7[/tex]
Where;
[tex]M_W[/tex] = The moment magnitude
E = The energy
If [tex]M_W[/tex] is increased by 6, we have;
Δ[tex]M_W[/tex] = 6
[tex]\Delta M_W = M_{W2} - M_{W1} = 6[/tex]
From which we get;
[tex]\Delta M_W = M_{W2} - M_{W1} = 6 = \dfrac{2}{3} \times log_{10} (E_2) - 10.7 - \left(\dfrac{2}{3} \times log_{10} (E_1) - 10.7 \right)[/tex]
[tex]6 = \dfrac{2}{3} \times log_{10} (E_1) - 10.7 - \left(\dfrac{2}{3} \times log_{10} (E_2) - 10.7 \right) = \left(\dfrac{2}{3} \right ) \times \left ( log_{10} (E_2) - log_{10} (E_1) \right)[/tex]
[tex]6 = \left(\dfrac{2}{3} \right ) \times \left ( log_{10} (E_2) - log_{10} (E_1) \right) = \left(\dfrac{2}{3} \right ) \times log_{10} \left( \dfrac{E_2}{E_1} \right)[/tex]
[tex]6 = \left(\dfrac{2}{3} \right ) \times log_{10} \left( \dfrac{E_2}{E_1} \right)[/tex]
[tex]log_{10} \left( \dfrac{E_2}{E_1} \right) = 6 \times \left(\dfrac{3}{2} \right ) = 9[/tex]
Therefore;
[tex]\dfrac{E_2}{E_1} = 10^9[/tex]
E₂ = E₁ × 10⁹ = 1000,000,000 × E₁
Therefore, when the moment magnitude, [tex]M_W[/tex], increases by 6, the energy increases by a factor of 1000,000,000
