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Juliette [100K]
1 year ago
10

Q/C An undersea earthquake or a landslide can produce an ocean wave of short duration carrying great energy, called a tsunami. W

hen its wavelength is large compared to the ocean depth d, the speed of a water wave is given approximately by v = √(gd). Assume an earthquake occurs all along a tectonic plate boundary running north to south and produces a straight tsunami wave crest moving everywhere to the west. (c) If the wave has amplitude 1.80 m when its speed is 200 m/s, what will be its amplitude where the water is 9.00 m deep?
Physics
1 answer:
iragen [17]1 year ago
7 0

The amplitude A_{2} is 8.307 m

When describing the peak value of a quantity, such as the level of sound waves or the power and voltage in electrical and electronic systems, the term amplitude is employed. A louder sound has a bigger amplitude, and a softer sound has a smaller amplitude.

As we previously discussed in the concept session, the square root of the water's speed is inversely related to the amplitude of its waves.

A\alpha\frac{1}{\sqrt{v} }

Solve for first case:

A_{1}=\frac{1}{\sqrt{v_{1} } }

Solve for second case:

A_{2}=\frac{1}{\sqrt{v_{2} } }

So, the two equations is equal

A^{2} _{1}  v_{1} = A^{2} _{2} v_{2}

        =A^{2} _{2} \sqrt{g d_{2} }

Rearrange and solve for the amplitude A_{2} :

A_{2} =\sqrt{\frac{A^{2}_{1} v_{1} }{v_{2} } }

    =\sqrt{\frac{A^{2}_{1} v_{1}  }{\sqrt{g d_{2} } } }

    =\sqrt{\frac{(1.8m)^{2} X200 m/s }{\sqrt{9.8m/s^{2}X9 m } } }

    = 8.307 m

So, The amplitude A_{2} is 8.307 m.

Learn more about amplitude here:

brainly.com/question/3613222

#SPJ4

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Answer:

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As we know that the momentum of the car and truck is same

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What will most likely happen if a light wave moves through a solid?
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Answer:

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(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

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when the distance between the charges changes to 13 cm (0.13 m)

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Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

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F = \frac{KQ^2}{r^2}

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F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

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{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

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