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saul85 [17]
3 years ago
6

Wind turbines are becoming an increasingly popular method of generating a community’s electricity. In wind turbines, wind is use

d to power electric generators. How will an increase in the use of wind turbines affect the availability of wind in the future?
Biology
2 answers:
Angelina_Jolie [31]3 years ago
8 0
It will affect the wind in the future because most wind currents will converge into multiple distinct locations like windfarms. This will make wind less prominent in other locations without wind farms
horrorfan [7]3 years ago
5 0
One of the results of the inventions of the microscope and telescope is that it expanded the world as people knew it.<span> Distant stars and microscopic organisms previously unseen and unknown provided a wealth of knowledge to scholars.</span><span> Which of these statements is true regarding these two inventions?</span>
.
<span><span>
</span></span>
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A(n) __________ is a one-celled organism formed by the union of a sperm and an egg.
zhannawk [14.2K]

A zygote.
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5 0
4 years ago
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In congestive heart failure, what does ?congestive? refer to?
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The answer is <span>c. narrowing of the coronary arteries due to plaque build-up</span>
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4 years ago
How does the diameter of each ripple change as time passes
VLD [36.1K]

The diameter of the ripple changes by the following equation = \frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}<u></u>

<u>Explanation:</u>

When a pebble or a rock is dropped in the water then the ripples or waves are formed which starts moving radially outward. As the time passes, the radius of the circle increases.

Let D be the diameter of the circle.

\frac{dD}{dt} will be the rate of change of diameter of the circle with respect to time.

Let A be the area of the circle.

\frac{dA}{dt} will be the rate of change of area of the circle with respect to time.

Area of the circle = \frac{\pi }{4} D^{2}

So,

\frac{dA}{dt} = \frac{d [\frac{\pi }{4}  D^{2}] }{dt}

\frac{dA}{dt} = \frac{\pi }{4} \frac{d[ D(t)^{2}] }{dt}

<u>Using chain rule,</u>

\frac{dA}{dt} = \frac{\pi }{4}  X  \frac{d[D(t)^{2} ]}{d[D(t)]}   X  \frac{d[D(t)]}{dt}

\frac{dA}{dt} = \frac{\pi }{4}  . 2D(t) . \frac{dD}{dt}

\frac{dA}{dt} = \frac{\pi }{2} X D(t) X \frac{d(D)}{dt}

\frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}

Thus, the following equation determines the change in area in terms of diameter with respect to time :  \frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}

7 0
3 years ago
I really need help ASAP please I will mark branlist promise
zubka84 [21]
Organelles maybe??????
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Which diagram demonstrates how crossing-over contributes to genetic variety during meiosis?
Serjik [45]
C. is the answer

A determined the gender
B and D is part of mitosis not meiosis
5 0
4 years ago
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