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Question Number 8389 by tawakalitu last updated on 09/Oct/16

Answered by ridwan balatif last updated on 10/Oct/16

velocity of alarm clock before strikes the ground is  v=(√(2gh))    =(√(2×10×15))    =(√(300))m/s    =10(√3)m/s  f=((340)/(340+10(√3)))×800    f=((27200)/(34+(√3)))  f=761.22 Hz

$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{alarm}\:\mathrm{clock}\:\mathrm{before}\:\mathrm{strikes}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\mathrm{v}=\sqrt{\mathrm{2gh}} \\ $$$$\:\:=\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{15}} \\ $$$$\:\:=\sqrt{\mathrm{300}}\mathrm{m}/\mathrm{s} \\ $$$$\:\:=\mathrm{10}\sqrt{\mathrm{3}}\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{f}=\frac{\mathrm{340}}{\mathrm{340}+\mathrm{10}\sqrt{\mathrm{3}}}×\mathrm{800}\:\: \\ $$$$\mathrm{f}=\frac{\mathrm{27200}}{\mathrm{34}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{f}=\mathrm{761}.\mathrm{22}\:\mathrm{Hz} \\ $$

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