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ArbitrLikvidat [17]
3 years ago
14

Long sand ridges oriented at right angles to the wind are called _____ dunes

Physics
2 answers:
S_A_V [24]3 years ago
3 0

The answer is:

Transverse dune

The explanation:

Transverse dune : is abundant barchan dunes It  may merge into barchanoid ridges, which then grade into linear .

The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.

It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.

Transverse dunes have a gently sloping windward side and a steeply sloping leeward side.

They general form in areas of sparse vegetation and abundant sand are transverse dunes.

dsp733 years ago
3 0
Transverse dunes comment if wrong
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A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
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Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  \sqrt{\frac{GM}{( R+h)} }

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x \sqrt{\frac{( R+h)}{GM } }

\frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } = R +h

h = \frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } - R.

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3 years ago
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A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?
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Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

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Which of the following is the best example of a good hypothesis?
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Answer:

Which of the following is the best example of a good hypothesis?

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a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra
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The dimensions of the rectangle are:

l = 50 m

b = 100/\pi m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is =\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l

Therefore,  \pi b+2l=200

b=(200-2l)/\pi

Now,

Area, A = l(200-2l)/\pi=(200l-2l^{2} )/\pi

Now, differentiate A w.r.t l:

Again differentiating w.r.t 'l', we get:

d^{2} A/dl^{2} =-4l/\pi< 0

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Therefore,

(200-4l)/\pi=0

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Now, from

\pi b+2l=200

\pi b=200-2*50

b=100/\pi

r=b/2=50/\pi

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At a distance r from a charge e on a particle of mass m the electric field value is
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<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

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