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Karolina [17]
2 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]2 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]2 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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Construct a graph of position versus time for the motion of a dog, using the data in the table below. Explain how the graph indi
Lynna [10]

Answer:

The dog is moving at a constant speed

Explanation:

Given that,

Position : 5, 10, 15, 20, 25

Time = 5. 10, 15, 20, 25

We need to draw a position time graph

Using given data

A graph of position and time shows the speed.

According to graph,

The graph indicates that the dog is moving at a constant speed because the graph is straight line.

Hence, The dog is moving at a constant speed

6 0
3 years ago
To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo
Setler [38]

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\  P = -2mv_{i}

For Beanbag

value of impulse will large if velocity is zero.

v_{f}=0\\ P = -mv_{i}

Explanation:

8 0
2 years ago
If your heart is beating at 76 beats per minute what is the frequency of your heart's oscillations
PSYCHO15rus [73]
That would be a frequency of 1.2666... beats per second. This can be phrased as your heart beats at 1.27 Hz.
4 0
3 years ago
A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0
3 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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