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Question Number 10704 by Saham last updated on 23/Feb/17

If the bulk modulus for water is 2.28 × 10^9 Nm^(−2)  , determine the   time it takes for a sound to travel a distant of 110m in water of   density 1000kgm^(−3) .

$$\mathrm{If}\:\mathrm{the}\:\mathrm{bulk}\:\mathrm{modulus}\:\mathrm{for}\:\mathrm{water}\:\mathrm{is}\:\mathrm{2}.\mathrm{28}\:×\:\mathrm{10}^{\mathrm{9}} \mathrm{Nm}^{−\mathrm{2}} \:,\:\mathrm{determine}\:\mathrm{the}\: \\ $$$$\mathrm{time}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{for}\:\mathrm{a}\:\mathrm{sound}\:\mathrm{to}\:\mathrm{travel}\:\mathrm{a}\:\mathrm{distant}\:\mathrm{of}\:\mathrm{110m}\:\mathrm{in}\:\mathrm{water}\:\mathrm{of}\: \\ $$$$\mathrm{density}\:\mathrm{1000kgm}^{−\mathrm{3}} . \\ $$

Answered by ridwan balatif last updated on 23/Feb/17

E=2.28×10^9 Nm^(−2)   x=110m  ρ=1000kgm^(−3)   v=(√(E/ρ))  v=(√((2.28×10^9 )/(1000)))  v=1509.96m/s  t=(x/v)  t=((110)/(1509.96))  t≈0.0728s

$$\mathrm{E}=\mathrm{2}.\mathrm{28}×\mathrm{10}^{\mathrm{9}} \mathrm{Nm}^{−\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{110m} \\ $$$$\rho=\mathrm{1000kgm}^{−\mathrm{3}} \\ $$$${v}=\sqrt{\frac{\mathrm{E}}{\rho}} \\ $$$${v}=\sqrt{\frac{\mathrm{2}.\mathrm{28}×\mathrm{10}^{\mathrm{9}} }{\mathrm{1000}}} \\ $$$${v}=\mathrm{1509}.\mathrm{96m}/\mathrm{s} \\ $$$${t}=\frac{\mathrm{x}}{{v}} \\ $$$${t}=\frac{\mathrm{110}}{\mathrm{1509}.\mathrm{96}} \\ $$$${t}\approx\mathrm{0}.\mathrm{0728s} \\ $$

Commented by Saham last updated on 23/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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