If a Tsunami in Deep Water Has a Period of 10

Question

Examlex · Accepted Answer

The Answer is The power per

\mathrm { km } ext { is }

P / \mathrm { km } = ( 1000 \mathrm {~m} / \mathrm { km } ) imes \left( 1.96 imes 10 ^ { 3 } \mathrm {~W} / \left( \mathrm { m } ^ { 3 } \cdot \mathrm { s } ight) H ^ { 2 } T ight.

Solving for

H

gives

H = \left[ \frac { P / \mathrm { km } } { ( 1000 \mathrm {~m} / \mathrm { km } ) imes \left( 1.96 imes 10 ^ { 3 } \mathrm {~W} / \left( \mathrm { m } ^ { 3 } \cdot \mathrm { s } ight) imes T ight. } ight] ^ { 1 / 2 }

For a period of 10 minute

= 600 \mathrm {~s}

and a power per

\mathrm { km }

of

1000 \mathrm { MW }

then

H = \left[ \frac { 10 ^ { 9 } \mathrm {~W} / \mathrm { km } } { ( 1000 \mathrm {~m} / \mathrm { km } ) imes \left( 1.96 imes 10 ^ { 3 } \mathrm {~W} / \left( \mathrm { m } ^ { 3 } \cdot \mathrm { s } ight) imes ( 600 \mathrm {~s} ) ight. } ight] ^ { 1 / 2 } = 0.92 \mathrm {~m}

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