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goldenfox [79]
3 years ago
7

Which term describes an advantage of wind energy?

Physics
2 answers:
atroni [7]3 years ago
7 0

The answer is "D". This is because wind is very abundant on earth...

Radda [10]3 years ago
7 0

Answer: D. Abundant

Explanation:

Wind is a mass of gases of different origin and form. It can be define as a flow of gases in a particular direction. Wind is a renewable resource which means that it is abundantly available in nature and can be replenished again by the wind cycle. The abundance of wind makes it suitable for the generation of wind energy by the wind mills.

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the refractive index is the blank of the speed of light in a medium to the speed of light in a vacuum
andrew11 [14]

Answer:ratio

Explanation:

8 0
3 years ago
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what is your weight in newtons if your potential energy is equal to 1000 joules and your mass is 60 kg
Darina [25.2K]

Your potential energy and mass don't tell what your weight is.

If I walk up from the first floor to the second floor, my weight hasn't
changed even though my potential energy has increased.

6 0
3 years ago
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Please help will mark brainliest
andreyandreev [35.5K]

Answer:

b

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it melts yes but it's still water [same element] so there fore it's a physical change

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3 years ago
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Whqt do you mean by geo-stationery satellite? Explain.​
Ratling [72]

Answer:

Description: When a geosynchronous satellite is placed directly above the Equator with a circular orbit and angular velocity identical to that of the Earth, the satellite is known as a geostationary satellite

Explanation:

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8 0
3 years ago
The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm
DIA [1.3K]

Answer:

T_{max} = 4.735\,kN\cdot m

Explanation:

The shear stress due to torque can be calculed by using the following model:

\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}

The maximum torque on the section is:

T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}

The Torsion Constant for the circular tube is:

J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})

J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]

J_{tube} = 4.560\times 10^{-6}\,m^{4}

Now, the require output is computed:

T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}

T_{max} = 4.735\,kN\cdot m

7 0
3 years ago
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