Question 52207This question is from textbook
: a quake sends out two waves at 8.9km/s and 5.1km/s. a siesmograph records the arrival of the 8.9km/s wave 73 seconds before the 5.1km/s wave. what is the distance from the siesmograph to the center of the quake?
This question is from textbook
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Use the distance formula: d = rt where: d = distance, r = rate (speed), and t = time.
For the fast wave (8.9 km/s):
d = (8.9 km/s)t secs.
For the slower wave (5.1 km/s):
d = (5.1 km/s)(t+73)secs. (It takes 73 seconds longer for this wave to reach the seismograph)
Since the distances, d, are the same, you can write:
(8.9 km/s)t = (5.1 km/s)(t+73)secs. Simplify and solve for t.
8.9t = 5.1t + 372.3 Subtract 5.1t from both sides of the equation.
3.8t = 372.3 Divide both sides by 3.8
t = 98 Rounded to the nearest second. (t = 97.973684...) Now substitute this value of t into either of the two initial equations and solve for d, the distance. Let's do both just to check!
d = (8.9 km/s)(98 s)
d = 872.2 km
d = (5.1 km/s)(98+73) s.
d = (5.1)(171)
d = 872.1 km
The epicenter of the 'quake is 872 km (rounded to the nearest km.) from the seismograph.
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