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3 years ago

4application of energy from water

1 answer:

5 0

Fuel Extraction and Production – Water is a critical resource for the drilling and mining of natural gas, coal, oil, and uranium. In many cases, fuel extraction also produces wastewater, as with natural gas and oil wells and coal slurry ponds.

Fuel Refining and Processing – Oil, uranium, and natural gas all require refining before they can be used as fuels – a process that uses substantial amounts of water.

Fuel Transportation – Water is used to transport coal through slurries — pipelines of finely ground coal mixed with water — and to test energy pipelines for leaks.[1]Emissions Control – Many thermoelectric power plants emit sulfur, mercury, particulates, carbon dioxide, and other pollutants, and require pollution control technologies. These technologies also require significant amounts of water to operate.

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Electromotive force in a circuit;

tester [92]

Answer:

A causes free electrons to flow

Explanation:

The amount of force that causes electrons to flow in a conductor is called electromotive force.

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3 years ago

Which of the following is NOT proof of a chemical change

Blababa [14]

Answer:

Explanation:

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3 years ago

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The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV

Ratling [72]

The calculated magnitude is 6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

= 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

=6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

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1 year ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

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3 years ago

I NEED a LOT of help PLEASE!

Minchanka [31]

Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.

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3 years ago

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