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Aleksandr [31]
3 years ago
11

Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex

t high tide will occur 12 hours later (at noon), find the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight.
Physics
1 answer:
aev [14]3 years ago
5 0

We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

t=10.0hours

From the Question we are told that

Maximum height h_{max}=3m

Minimum height  H_{min}=0.5m

Time for  next high tide will occurT=12 hours =>720 min

Generally Average Height

h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75

Therefore determine Amplitude to be

A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25

Generally, the equation for Time is mathematically given by

At t=0

h(x)=Acos(Bx)+h_{avg}

Where

B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}

Therefore

h(t)=Acos8.73*10^{-3}(t)+h_{avg}

Hence the Time at T=1.125 is

1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75

-0.1249t=1.75

t=10.0hours

For more information on this visit

brainly.com/question/22361343

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The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

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From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

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