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TiliK225 [7]
2 years ago
5

Arrange the steps describing how a wind turbine functions to provide electricity to homes and businesses.

SAT
1 answer:
aksik [14]2 years ago
3 0

Wind turbines generate electricity by following a simple principle. Moving wind transfers energy to the blades of the windmill, which results in spinning of the blades. These blades are connected to an internal shaft which also starts spinning. This spinning of the shaft generates electricity, which is further distributed to electrical substations to provide electricity to homes and businesses.

<h3>What is a wind turbine?</h3>

Wind turbines work by creating kinetic energy when the power of the wind makes their blades spin. This kinetic energy is converted into electrical energy.

Thus, the steps describing how a wind turbine functions are,

Step 1: Wind moves the blades of the turbine.

Step 2: Internal shaft spins

Step 3: Generator produces electricity

Step 4: Distribution lines carry electricity to substation

Learn more about wind turbine here,

brainly.com/question/928271

#SPJ1

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Based on the calculations, the pressure of this gas sample at 27°C is equal to 0.17 atm.

<h3>How to determine the final pressure.</h3>

In order to calculate the final pressure of an ideal gas at constant volume, we would apply Gay Lussac's law.

Mathematically, Gay Lussac's law is given by this formula;

P\;\alpha \;T\\\\P=kT

<u>Where:</u>

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Substituting the given parameters into the formula, we have;

\frac{P_1}{T_1} =\frac{P_2}{T_2} \\\\\frac{0.80}{127} =\frac{P_2}{27} \\\\0.0063 =\frac{P_2}{27}\\\\P_2 = 0.0063 \times 27

Final pressure = 0.17 atm.

Read more on temperature here: brainly.com/question/888898

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suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a
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Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.

Explanation:

Let a Force pushes the hockey puck of mass m.

Then acceleration, a= \frac{F}{m}a=mF

From the equation of motion,

\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)

In the second case, when mass is 2m, then acceleration,

a'=\frac{F}{2m}a′=2mF

and t' is the time taken.

The final speed is v,

\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)

Hence, it would take two times the previous amount of time to push the pluck of double mass.

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