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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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In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
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Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

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rotational motion.

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decreases it has less kinetic energy and slower moving particles.

As kinetic energy is 1/2 mV².

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