First this sounds like a sin curve.
Amplitude: 10
Period: 15 = 2pi/(k)
Solve for k to find frequency:
Multiply both sides by k.
15k = 2pi
Divide both sides by 15.
k = 2pi/15
y = 10sin((2pi/15)x)
Normally, I'd want to use 9 as a vertical shift, but the lowest point of the curve would then be one foot below the ocean floor.
So I guess you could use the model: y = 10sin((2pi/15)x) + 9. But like I said that nine makes me nervous
Ci. Let x = 2 minutes
y = 10sin((2pi/15)2) + 9
y = 10(.7431) + 9
y = 7.431 + 9
y = 16.431
Cii. Let x = 4 minutes
y = 10sin((2pi/15)4) + 9
y = 10(.9945) + 9
y = 9.945 + 9
y = 18.945
Ciii. Let x = 12 minutes
y = 10sin((2pi/15)12) + 9
y = 10(-.9511) + 9
y = -9.511 + 9
y = -.511
According to the model the minumum depth would be negative one foot.
2007-01-29 08:46:17
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answer #1
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answered by mirramai 3
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A and B. This would be a sine wave, in the form y = n + A*sin(bx). A is your amplitude, while the period would be 2*pi/b. You've been given A = 10 and 2*pi/b = 15 (minutes, I assume), which gives you b = 2*pi/15. n is the normal water level, which you were given as 9 m. The sine curve mus be negative, though, because a sine curve normally goes up first, but you need to go down. That means the depth of the water is y = 9 - 10*sin(2*pi*x / 15). You can graph this using the normal shape of a sine function, but with f(0) = 9 instead of 0, and additional zeroes at 7.5 and 15. Don't forget to go down first, then up.
C. Using the given values in place of x in my equation, I got
i. 1.57 m
ii. -0.95 m
iii. 18.51 m
D. The minimum depth would be 9 - 10 = -1 m. This probably means that not only does the depth of the water decrease at the shore, but the shoreline actually recedes. This really is what happens when a tsunami strikes.
2007-01-29 08:44:16
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answer #2
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answered by DavidK93 7
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The equation is a sine wave with amplitude 10 and wavelength =15. So the equation is y = 10sin(2pi x/15).
In an x-y coordinate system, this wave would start at x= 0 and curveupwards reaching a maximum of 10 when x = 3.775, then would curv back dowwn crossing the x-axis at x = 7.5, continue on downward reaching a minimum of -10 at x = 11.25 and then start back up hitting the x-axis at x = 15,thus completing one complete cycle. The cycle would repeat over and over as it moves toward the pier. This should give you an idea of what the wave looks like and you should be able to sketch it. Look up sine wave on the internet for help.
When the wave arrives at the pier, water level is normal at 9m; however, the water will soon rise dramatically.
You have not indicated how fast the tsunami is moving so I will guess it is moving, so I will guess that it completes on cycle in
15 minutes. If this is the case then when
x= 2, we have y = 7.43 meters.
when x=4, y=9.94 meters ,
when t = 6 , y = -9.51 meters
So depending on exactly where the tsunami is in its cycle when it
arrives at the pier, the water will first lower from its normal 9 meters to -1 meter, then rise to normal level of 9 meters, then two minutes later be at 16.43 meters, and 2 minutes later at 18.94 meters and crest at 19 meters.
I hope this helps you.
2007-01-29 09:52:44
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answer #3
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answered by ironduke8159 7
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I think it's h(t) = 9 + 10(sin (t - 15)). It might be sin (t/15).
h is height
t is time
it looks like a sine wave, moved up 9 places, but instead of going from 1 to -1 it goes to (9+10) to (9-10).
C. The period is 15, but in what units? seconds? minutes? Whatever it is, just plug & play.
D. I think the minimum water depth would be negative. Ths means that all the water on would be pulled back, exposing the ocean floor.
2007-01-29 08:48:49
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answer #4
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answered by bequalming 5
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hello i will say 4 mins for it to pick up a good speed to hit land, let me know what you think
2007-01-29 08:45:47
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answer #5
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answered by BROKEN AND BEAUTFUL 2
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