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White raven [17]
3 years ago
14

Which of the following lists contains only examples of renewable sources of energy?

Physics
2 answers:
anyanavicka [17]3 years ago
5 0
<h3><u>Answer;</u></h3>

A. wind, tidal, geothermal, and hydroelectric

<h3><u>Explanation</u>;</h3>
  • Renewable energy sources are the energy sources that are constantly being replenished, such as sunlight, wind, and water. This means that we can use them as much as we want, and we do not have to worry about them running out.
  • Additionally, renewable energy sources are usually much more environmentally friendly than fossil fuels. Examples of renewable sources of energy includes; solar energy, wind energy, hydroelectric power, tidal energy, geothermal energy, etc.
Allisa [31]3 years ago
3 0
A wind, tidal, geothermal, and hydroelectric
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Average velocity is different than average speed because calculating average velocity involves a)Distance b)Time c)Motion d)Disp
likoan [24]

Answer:

The answer is D.

Explanation:

Average speed involve just distance and time but average velocity includes displacement and time.

(Correct me if I am wrong)

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A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
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Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
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