Consider the equation =2/3log10()−10.7 relating the

Ответ:

wildfire771003

12.07.2021

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; $M_W = \dfrac{2}{3} \times log_{10} (E) - 10.7$

Where;

M_W = The moment magnitude

E = The energy

If M_W is increased by 6, we have;

Δ M_W = 6

$\Delta M_W = M_{W2} - M_{W1} = 6$

From which we get;

$\Delta M_W = M_{W2} - M_{W1} = 6 = \dfrac{2}{3} \times log_{10} (E_2) - 10.7 - \left(\dfrac{2}{3} \times log_{10} (E_1) - 10.7 \right)$

$6 = \dfrac{2}{3} \times log_{10} (E_1) - 10.7 - \left(\dfrac{2}{3} \times log_{10} (E_2) - 10.7 \right) = \left(\dfrac{2}{3} \right ) \times \left ( log_{10} (E_2) - log_{10} (E_1) \right)$

$6 = \left(\dfrac{2}{3} \right ) \times \left ( log_{10} (E_2) - log_{10} (E_1) \right) = \left(\dfrac{2}{3} \right ) \times log_{10} \left( \dfrac{E_2}{E_1} \right)$

$6 = \left(\dfrac{2}{3} \right ) \times log_{10} \left( \dfrac{E_2}{E_1} \right)$

$log_{10} \left( \dfrac{E_2}{E_1} \right) = 6 \times \left(\dfrac{3}{2} \right ) = 9$

Therefore;

$\dfrac{E_2}{E_1} = 10^9$

E₂ = E₁ × 10⁹ = 1000,000,000 × E₁

Therefore, when the moment magnitude, M_W , increases by 6, the energy increases by a factor of 1000,000,000

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Consider the equation =2/3log10()−10.7 relating the moment magnitude of an earthquake and the energy (in ergs) released by it. If increases by 6, by what factor does the energy increase?

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