Question

the bulk modulus of the earth is 40gpa and shear modulus of the earth is 25gpa.the P wave is arrived 5 min before S wave.calculate the distance of earthquake.here, density of earth is 5.51 g/cm3

Answer 1

Answer:

Distance of the earth quack is given as

$d = 3.05 \times 10^6 m/s$

Explanation:

As we know that the speed of P wave and S wave is given as

$v_p = \sqrt{\frac{B}{\rho}}$

$v_p = \sqrt{\frac{40 \times 10^9}{5510}}$

$v_p = 2694 m/s$

now similarly we have

$v_s = \sqrt{\frac{B}{\rho}}$

$v_s = \sqrt{\frac{25 \times 10^9}{5510}}$

$v_s = 2130 m/s$

now we know that

$\frac{d}{v_s} - \frac{d}{v_p} = 5 min$

$d(\frac{1}{2130} - \frac{1}{2694}) = 5 \times 60$

$d = \frac{300}{9.83 \times 10^{-5}}$

$d = 3.05 \times 10^6 m/s$

#Learn

Topic : Kinematics

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