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Naddik [55]
1 year ago
7

A polar jet is a belt of winds about 600 miles wide, flowing as fast as _________

Physics
1 answer:
maks197457 [2]1 year ago
5 0

A polar jet is a belt of winds about 600 miles wide, flowing as fast as 400km/hr in its central "core".

The wind speeds are greatest where temperature differences between air masses are greatest, and often exceed 92 km/h . Speeds of 400 km/h  have been measured.One factor that contributes to the creation of a concentrated polar jet is the undercutting of sub-tropical air masses by the more dense polar air masses at the polar front. This causes a sharp north-south pressure (south-north potential vorticity) gradient in the horizontal plane, an effect which is most significant during double Rossby wave breaking events.

At high altitudes, lack of friction allows air to respond freely to the steep pressure gradient with low pressure at high altitude over the pole. This results in the formation of planetary wind circulations that experience a strong Coriolis deflection and thus can be considered 'quasi-geostrophic'. The polar front jet stream is closely linked to the frontogenesis process in midlatitudes, as the acceleration/deceleration of the air flow induces areas of low/high pressure respectively, which link to the formation of cyclones and anticyclones along the polar front in a relatively narrow region.

Learn more about jet streams here:

brainly.com/question/248501

#SPJ4

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A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
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Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

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we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

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Answer:

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P = 1.5² × 1.26

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