Question

The magnitude, M, of an earthquake is defined to be M=log I/S, where I is the intensity of the earthquake (measured by the amplitude of the seismograph wave) and S is the intensity of a “standard” earthquake, which is barely detectable. Which equation represents the magnitude of an earthquake that is 10 times more intense than a standard earthquake?
A.M=logI/10S
B.M=log(10S)
(It was C i just turned it in)
C.M=log10S/S
D.M=log10/S

Answer 1

Answer:

The required equation is $M=\log \frac{10S}{S}$ i.e. M = 1.

Step-by-step explanation:

We are given that,

Magnitude of an earthquake is defined as $M=\log \frac{I}{S}$,

where I is the intensity of the earthquake measured on seismograph and S is the intensity of the standard earthquake.

Since, it is given that,

The intensity of an earthquake measured is 10 times more than that of the standard earthquake.

i.e. I = 10S

So, we get,

Magnitude of an earthquake is $M=\log \frac{10S}{S}$,

i.e. Magnitude of an earthquake is $M=\log 10$,

i.e. Magnitude of an earthquake is M = 1.

Hence, the required equation is $M=\log \frac{10S}{S}$ i.e. M = 1.

Answer 2

The magnitude of an earthquake which is 10 times more intense than a standard earthquake is $\rm M = log(\frac{10S}{S})$ and this can be determined by using the given data.

Given :

• The magnitude, M, of an earthquake is defined to be $\rm M = log(\frac{I}{S})$.
• 'I' is the intensity of the earthquake (measured by the amplitude of the seismograph wave)
• 'S' is the intensity of a “standard” earthquake, which is barely detectable.

The magnitude of an earthquake is given by:

$\rm M = log(\frac{I}{S})$

Now, if an earthquake is 10 times more intense than a standard earthquake that is:

I = 10S

Now, put the value of 'I' in the equation (1).

$\rm M = log \left(\dfrac{10S}{S}\right)$

M = log (10)

M = 1

Therefore, the correct option is C).

For more information, refer to the link given below:

https://brainly.com/question/22122594