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sergeinik [125]

3 years ago

15

Which energy output objects work with the turbine

1 answer:

viktelen [127]3 years ago

6 0

Answer:

The energy output object that works with the turbine is the alternator (generator)

Explanation:

In energy generation the turbine receives input energy from high pressured steam, high energy water etc. Which impinges and turn the blades of the turbine, this turbine is connected by means of a shaft to the alternator that converts the rotational motion of the shaft to electrical energy through Electro magnetic induction principles and also outputs the energy for consumption.

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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

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Give an example of when a conscientious objector might exercise their right:

Mariana [72]

Am sorry what can you be more specific

7 0

3 years ago

You push on a box with a force of 300 N directly north. Another person pushes the box with a

tangare [24]

Answer:

$resultant \\ \: F = \sqrt{ {300}^{2} + {600}^{2} } \\ = \sqrt{450000} \\ = 670.82 \: newtons$

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Angelina_Jolie [31]

Answer:

I think (d) is right answer

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How would an unexpected result to an experiment affect the original hypothesis?

Kipish [7]

Answer:

it would by them changing their hypothisis and retrying the test again

Explanation:

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3 years ago