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Anni [7]
2 years ago
9

Water flows with a velocity of 3 m/s in a rectangular channel 3 m wide at a depth of 3 m. What is the change in depth and in wat

er surface elevation produced when a gradual contraction in the channel to a width of 2.6 m takes place? Determine the greatest contraction allowable without altering the specified upstream conditions.

Engineering
1 answer:
strojnjashka [21]2 years ago
8 0

Answer: new depth will be 3.462m and the water elevation will be 0.462m.

The maximum contraction will be achieved in width 0<w<3

Explanation:detailed calculation and explanation is shown in the image below

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Java is called portable because you can compile a java code which will spew out a byte-code, and then you run that code with Java Virtual Machine. Java Virtual Machine is like an interpreter, which reads the compiled byte-code and runs it. So first of all, you need to install the JVM on the system you want.

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How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.
Snowcat [4.5K]

Answer:

N_3=\dfrac{T_1}{T_3}N_1

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

Speed\ of\ gear 1=N_1

Number\ of\ teeth\ of\ gear 1=T_1

Speed\ of\ gear 3=N_3

Number\ of\ teeth\ of\ gear 3=T_3

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}

N_3=\dfrac{T_1}{T_3}N_1

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3 years ago
Plant scientists would not do which of the following?
zavuch27 [327]

Explanation:

i think option 4 is correct answer because itsrelated to animal not plants.

6 0
3 years ago
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4 0
2 years ago
A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

7 0
3 years ago
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