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Need help ASAP please!!! The formula determines the magnitude of an earthquake, where I is the intensity of the earthquake and S is the intensity of a “standard earthquake.” How many times stronger is an earthquake with a magnitude of 8 than an earthquake with a magnitude of 6? Show your work.

Respuesta :

Answer:

100

Step-by-step explanation:

M = log(I/S)

The intensity of the first earthquake:

8 = log(I₁/S)

10⁸ = I₁/S

I₁ = 10⁸ S

The intensity of the second earthquake:

6 = log(I₂/S)

10⁶ = I₂/S

I₂ = 10⁶ S

The ratio is:

I₁/I₂ = (10⁸ S) / (10⁶ S)

I₁/I₂ = 100

It is 100 times stronger.

Lanuel

An earthquake with a magnitude of 8 is 100 times stronger than an earthquake with a magnitude of 6.

Given the following data:

  • Magnitude A = 8
  • Magnitude B = 6

To determine how many times stronger is an earthquake with a magnitude of 8 than an earthquake with a magnitude of 6:

Mathematically, the magnitude of an earthquake is given by the formula:

[tex]Magnitude = Log(\frac{I}{S} )[/tex]

Where:

  • I is the intensity of the earthquake.
  • S is the intensity of a standard earthquake.

For earthquake A:

[tex]8=Log (\frac{I_A}{S} )\\\\10^8=\frac{I_A}{S}\\\\I_A=S10^8[/tex]

For earthquake B:

[tex]6=Log (\frac{I_B}{S} )\\\\10^6=\frac{I_B}{S}\\\\I_B=S10^6[/tex]

For strength:

[tex]Strength = \frac{I_A}{I_B} \\\\Strength = \frac{S10^8}{S10^6} \\\\Strength = 10^2[/tex]

Strength = 100

Read more: https://brainly.com/question/21620194

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