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Karolina [17]
3 years ago
7

Which location focuses its use on a nonrenewable energy source? a wind mill farm a natural gas power plant a fruit orchard a com

munity swimming pool.
Physics
2 answers:
DedPeter [7]3 years ago
6 0

A natural gas power plant focuses on the use of non renewable energy resources.

<h3>What are non renewable energy?</h3>

Non-renewable energy are energy that comes from sources that cannot be exhausted.

  • This energy can not be replaced because they don't get exhausted.

  • Examples include energy form coal, natural gas, oil.

Therefore, natural gas power plant is an example of non renewable energy source.

For more information on non renewable energy source kindly check

https://brainly.in/question/32015541

Elena L [17]3 years ago
6 0

Answer:

The answer is B a natural gas power plant

Explanation:

because i answered it on edge

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How does the gravitational force between two objects change if the distance
Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

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what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this
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Answer:

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3 years ago
Read 2 more answers
The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.
pogonyaev

Answer:

W=7.56\times 10^{-19}\ J

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

1\ eV=1.6\times 10^{-19}\ J

For 4.73 eV,

4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J

So, the work function for silver is 7.56\times 10^{-19}\ J.

6 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

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