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If an earthquake is 121,000 times as intense as the minimum detectable earthquake, what is its Richter number? Round answer to the nearest tenth. Thank you!
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Do you happen to know the equation for richter numbers?
R=log(I/Io) ?
That looks right. \[\log_{10}(\frac{121000}{1}) \]
So basically you are looking for what power of 10 equals 121000. That is also the Richter number.
5.08?
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That's what I got.
Thanks so much!
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