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sasho [114]
4 years ago
7

The flow rate over Niagara falls is 84,760 cfs (cubic feet per second), and the drop from top to bottom is 167 feet. If this flo

w were harnessed to generate electricity, at what rate would power be produced? Assume 100% conversion of potential energy into electrical energy. How does this compare with the power generated by the Three Gorges Dam?
Physics
1 answer:
jekas [21]4 years ago
8 0

To develop this problem we will calculate the volume from the given flow (volume per unit of time). This process will be accompanied by transforming the units given in the International system.

Through the density-volume-mass ratio, we will calculate the mass. Finally we will proceed to calculate the total power generated.

1 feet = 0.3048 m

\dot{V} = 84760ft^3/s

\dot{V} = 84760ft^3/s (\frac{(0.3058m)^3}{(1ft)^3})

\dot{V} = 2400m^3/s \text{and the Volume each second is } V= 2400m^3

Now the mass of water falling down each second would be

\rho = \frac{m}{V} \rightarrow m = \rho V

m = (1000kg/m^3)(2400m^3)

m = 2.4*10^6kg

Now Height in meters is

h = 167 ft (\frac{(0.3058m)}{(1ft)})

h= 50.9 m

The power produced is equivalent to the work done by gravity over a certain time. At this case that can be expressed as,

P = \frac{W}{t}\\P = \frac{F*h}{t}\\P = \frac{mgh}{t}\\P = \frac{2.4*10^6*9.8*50.9}{1}\\P = 1.197*10^9 Watts

The power of Niagara Falls then would be 1.197GW while the power generated by the Three Gorges Dam is  22,500 MW. This indicates that the Niagara Falls, if it were a dam that covered 100% of the fluid it carries, would exceed the production of the dam in China by thousands.

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Answer:

A <u>scientific model</u> is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted.

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3 years ago
Read 2 more answers
A +5.00 pC charge is located on a sheet of paper.
emmainna [20.7K]

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

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3 years ago
A 1,70 atm, una muestra de gas ocupa 4,25 litros. Si la presión en el gas aumenta a 2.40 atm, ¿cuál será el nuevo volumen?
statuscvo [17]

At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?

Answer:

3.01L

Explanation:

Given parameters:

Initial pressure, P1  = 1.7atm

Initial volume, V1  = 4.25L

Final pressure, P2  = 2.4atm

Unknown:

Final or new volume, V2  = ?

Solution:

To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

            P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 final pressure

V2 final volume

        1.7 x 4.25  = 2.4 x V2

             V2  = 3.01L

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