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Murljashka [212]
3 years ago
12

Sarah wants to install a wind turbine on her farm, which would convert wind energy to electricity to power her home. She can onl

y afford one wind turbine, so she wants to determine the best location around her farm to harness the wind’s energy most effectively throughout the year. Sarah decides to conduct a science investigation to answer her question A.She could build a miniature model of her farm, and blow air through a straw in different locations to
simulate wind.
B. She could create a survey for her friends and neighbors, asking them where they think the wind
turbine would look the coolest.
C. She could conduct an experiment, using a tool to measure wind speed at multiple locations around
her farm, and collect data over a period of one year.
D. She could go on a nature walk every weekend for one year, recording her observations to make an
inference about the best location for the wind turbine.
Physics
2 answers:
Marina86 [1]3 years ago
6 0

Answer:

The best choice would be c

Explanation: Sarah wants this wind turbine to efficient since she can only get one. C has the most reasonable option data collected will help her know the best wind speed over her farm.

aleksklad [387]3 years ago
4 0
C it’s the best answer like the other one did
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Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

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Komok [63]

Answer:

graph A

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the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)

but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

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kondor19780726 [428]

Answer:(a)9.685 mm

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(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

Y_1=4.84mm

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