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1 March, 11:05

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A seismograph records the arrival of the transverse waves 68 s before the arrival of the longitudinal waves. How far away is the earthquake?

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  1. 1 March, 14:11
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    Answer: The distance is 723.4km

    Explanation:

    The velocity of the transverse waves is 8.9km/s

    The velocity of the longitudinal wave is 5.1 km/s

    The transverse one reaches 68 seconds before the longitudinal.

    if the distance is X, we know that:

    X / (9.8km/s) = T1

    X / (5.1km/s) = T2

    T2 = T1 + 68s

    Where T1 and T2 are the time that each wave needs to reach the sesmograph.

    We replace the third equation into the second and get:

    X / (9.8km/s) = T1

    X / (5.1km/s) = T1 + 68s

    Now, we can replace T1 from the first equation into the second one:

    X / (5.1km/s) = X / (9.8km/s) + 68s

    Now we can solve it for X and find the distance.

    X / (5.1km/s) - X / (9.8km/s) = 68s

    X (1 / (5.1km/s) - 1 / (9.8km/s)) = X*0.094s/km = 68s

    X = 68s/0.094s/km = 723.4 km
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