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Ainat [17]
1 year ago
14

the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the

hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Physics
1 answer:
Nitella [24]1 year ago
5 0

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

W_{net} = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

W_{net} = m (gh)

m = W_{net} / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

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A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
Nikitich [7]

Answer:

r ≥ R, E = Q / (4πR²ε₀)

r ≤ R, E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

Maximum at r = ⅔ R

Maximum field of E = Q / (3πε₀R²)

Explanation:

Gauss's law states:

∮E·dA = Q/ε₀

What that means is, if you have electric field vectors E passing through areas dA, the sum of those E vector components perpendicular to the dA areas is equal to the total charge Q divided by the permittivity of space, ε₀.

a) r ≥ R

Here, we're looking at the charge contained by the entire sphere.  The surface area of the sphere is 4πR², and the charge it contains is Q.  Therefore:

E(4πR²) = Q/ε₀

E = Q / (4πR²ε₀)

b) r ≤ R

This time, we're looking at the charge contained by part of the sphere.

Imagine the sphere is actually an infinite number of shells, like Russian nesting dolls.  For any shell of radius r, the charge it contains is:

dq = ρ dV

dq = ρ (4πr²) dr

The total charge contained by the shells from 0 to r is:

q = ∫ dq

q = ∫₀ʳ ρ (4πr²) dr

q = ∫₀ʳ ρ₀ (1 − r/R) (4πr²) dr

q = 4πρ₀ ∫₀ʳ (1 − r/R) (r²) dr

q = 4πρ₀ ∫₀ʳ (r² − r³/R) dr

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R) |₀ʳ

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R)

Since ρ₀ = 3Q/(πR³):

q = 4π (3Q/(πR³)) (⅓ r³ − ¼ r⁴/R)

q = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴)

Therefore:

E(4πr²) = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / ε₀

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

When E is a maximum, dE/dr is 0.

First, simplify E:

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

E = Q (4 (r³/R³) − 3 (r⁴/R⁴)) / (4πr²ε₀)

E = Q (4 (r/R³) − 3 (r²/R⁴)) / (4πε₀)

Take derivative and set to 0:

dE/dr = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = 4/R³ − 6r/R⁴

0 = 4R − 6r

r = ⅔R

Evaluating E at r = ⅔R:

E = Q (4 (⅔R / R³) − 3 (⁴/₉R² / R⁴)) / (4πε₀)

E = Q (8 / (3R²) − 4 / (3R²)) / (4πε₀)

E = Q (4 / (3R²)) / (4πε₀)

E = Q (1 / (3R²)) / (πε₀)

E = Q / (3πε₀R²)

3 0
1 year ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

4 0
1 year ago
(Double points) A machine receives electricity that enables it to deliver a total of 8,542 N of force for the completion of its
storchak [24]

Answer: machine's efficiency = 82.2%

Explanation:

Efficiency of a machine is the capability of a machine to convert input to output without waste.

It can be expressed as

Efficiency = output/ input × 100%

Output = 7,023N

Imput = 8,542N

Efficiency = 7,023N/8,542N × 100%

Efficiency = 82.2%

4 0
1 year ago
Nisha was making a list of things that are obtained from plants and Amol was making a list of things that can dissolve in water.
Allisa [31]

Answer:

Sugar

Explanation:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

Nisha's list is made up of things that can be derived from plants. From the list given, oil, sugar and cotton can also be obtained from plants.

Most plant materials are organic matter.

Amol's list is made up of things that can dissolve in water. Sugar and table salt can also be dissolved in water.

Water is able to dissolve these materials because they are polar compounds. One rule of solubility is that like dissolves like.

New list:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

  Sugar                                       Sugar

                                   

Only sugar from the list can be added to both lists. It can be obtained from plant and can also dissolve in water.

                                             

6 0
2 years ago
Drag each tile to the correct location.
In-s [12.5K]

Answer:

Look at the image please

Explanation:

5 0
1 year ago
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