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An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?

Respuesta :

Answer:

The S wave arrives 6 sec after the P wave.

Explanation:

Given that,

Distance of P = 45 km

Speed of p = 5000 m/s

Speed of S = 3000 m/s

We need to calculate the time by the P wave

Using formula of time

[tex]t = \dfrac{D}{v}[/tex]

Where, D = distance

v = speed

t = time

Put the value in to the formula

[tex]t_{p} =\dfrac{45\times1000}{5000}[/tex]

[tex]t_{p} = 9\ sec[/tex]

Now, time for s wave

[tex]t_{s}=\dfrac{45000}{3000}[/tex]

[tex]t =15\ sec[/tex]

The required time is

[tex]\Delta t=t_{s}-t_{p}[/tex]

[tex]\Delta t=15-9[/tex]

[tex]\Delta t =6\ sec[/tex]

Hence, The S wave arrives 6 sec after the P wave.

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