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likoan [24]
3 years ago
11

A trout jumps, producing waves on the surface of a 0.8-mdeep mountain stream. If it is observed that the waves do not travel ups

tream, what is the minimum velocity of the current?
Physics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

The value is v  =  2.8 \  m/s

Explanation:

From the question we are told that  

     The depth of the mountain is  d =  0.8 \  m

Generally the  velocity of the surface  wave  is mathematically represented as

        v =   \sqrt{ g * d }      

=>      v =   \sqrt{ 9.8 * 0.8  }    

=>      v =   2.8 \  m/s

Generally using the Froude number is mathematically represented as

          Fr = \frac{ V }{ v }

Here V is the velocity of the current

 Given that  the waves do not travel upstream, then the flow of the current is  supercritical which means that

            \frac{ V }{ v } > 1

=>         V  > v

=>         V  > 2.8 \  m/s

Hence the minimum velocity of the current  is  

            v  =  2.8 \  m/s

This because the velocity of the  current is greater velocity of the surface  wave , so minimum will be like the lowest possible value of  V  which is v

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Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

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Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

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     x = A cos (w \sqrt{2y_{o}/g})

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a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

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         w = \sqrt{k/2}

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for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

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b) the function is minimun if

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        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

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