Comparación de hipotecas Los Chos

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White raven [17]

Answer:

can't understand the writting

4 0

3 years ago

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A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a)

Bess [88]

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

$v=\sqrt{2gh}$

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

$v=\sqrt{2\times 9.81\times 90}=42.02m/s$

now we know the relation between discharge and velocity as dictated by contuinity equation is

$Q=V\times Area$

Applying values in the above equation and solving for area we get

$Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}$

The circular area is related to diameter as

$Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m$

Thus the diameter of the nozzle is 0.246 meters

7 0

3 years ago

The space shuttle fleet was designed with two booster stages. If the first stage provides a thrust of 686.68 Mega-newtons(MN) an

aksik [14]

Answer:

6.30 miles/hour

Explanation:

Newton's second law applies here. In simple terms:

$F = ma$

where F = Force (Thrust) in N

a = acceleration (m/s²)

The acceleration can be given by rearranging the formula to give:

$a = \frac{F}{N}$

= $\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr$

7 0

3 years ago

Cellular network towers are configured in such a way so that they avoid what type of problem?

stiv31 [10]

Avoiding OVERLAP issues

6 0

3 years ago

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An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidi

padilas [110]

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

5 0

3 years ago