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2 years ago

What two factors determine the energy production of the Hoover Dam

2 answers:

swat322 years ago

6 0

Mar 28, 2011 · The Hoover Dam generates electricity using hydropower. Hydropower is the power generated by moving water. Dams are often built on a large river, so the water falls down from the top of the dam. At the bottom of the dam, the water rotates a large turbine

Hope this helps :p

scZoUnD [109]2 years ago

4 0

Answer:

Two factors are water speed and amount of water.

Explanation:

The production of hydro electrical power can be done when the water passes from the dam to the rivers which are below dam.

Turbines which are surrounded by magnets produce the electricity when the flowing water is hit on the turbines than turbines rotates and produce electricity and turbines are present in the dam.

The energy production of the Hoover dam is based on the amount or pressure of the water if the more water is passing through the intake pipe then there is more production of energy. An artificial lake is created behind the dam.

And energy production also depends on the speed of water means if the speed of water is large then it rotate the turbine faster and increase the production.

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Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

A bike that is coasting down a steep hill increases its speed from 8.0 m/s to 14 m/s. The length of the hill is 55 meters. How m

masya89 [10]

t=5s

it was correct on my do-now

so I hope it was useful for you

8 0

2 years ago

Read 2 more answers

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0

2 years ago

A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is

vova2212 [387]

solution:

$consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\$ $270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm$

7 0

2 years ago

A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.

AfilCa [17]

The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle $\theta$ between the direction of the force and the direction of the displacement:
$W=Fd \cos \theta$
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so $\cos \theta = \cos 0=1$
Therefore, the work done is
$W=Fd=(5 N)(2 m)=10 J$

6 0

2 years ago

Read 2 more answers