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2 years ago

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

Physics

2 answers:

S_A_V [24]2 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.

dsp732 years ago

3 0

Transverse dunes comment if wrong

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Answer:

Explanation:

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$\lambda_{max} T = b$

Where b = 2.898 x 10⁻³ m K

Given ,

$\lambda{max} = 18\times10^{-9}$

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2 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

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Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

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Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

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$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

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