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dangina [55]
2 years ago
7

An earthquake registers a 5.6 on the Richter scale. If the reference intensity of this quake was 7.3 × 109, what was the intensi

ty of the earthquake?
Mathematics
2 answers:
kifflom [539]2 years ago
8 0
Log(x/7300000000) = 5.6 so 

<span>x/7300000000 = 10^5.8 = 398107.17 </span>

<span>x = 2.9061823 x 10^15 </span>
valentinak56 [21]2 years ago
6 0

Answer:

2.91\times 10^{15}

Step-by-step explanation:

Since, an earthquake has a Richter scale magnitude,

M=log(\frac{I}{I_0})

Where, I is the intensity of the earthquake,

While, I_0 is the reference intensity,

Given, M = 5.6,

And, I_0=7.3 \times 10^9

\implies 5.6 = log(\frac{I}{7.3 \times 10^9})

5.6=log I - log(7.3\times 10^9)    ( Because, log (a/b) = log a - log b ),

5.6 = log I - 9.86332286012

log I = 15.4633228601

\implies I=10^{15.4633228601}=2.9061823449\times 10^{15}\approx 2.91\times 10^{15}

Hence, the intensity of the earthquake is 2.91\times 10^{15}.

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I think your question missed key information, allow me to add in and hope it will fit the orginal one. Please have a look at the attached photo,

<em>A solid oblique pyramid has an equilateral triangle as a base with an edge length of 4StartRoot 3 EndRoot cm and an area of 12StartRoot 3 EndRoot cm2. </em>

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