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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

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

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6 0
2 years ago
A car is driven off a cliff at 39 m/s. It lands 141 m from the base. How high
Mumz [18]

Answer: A

Explanation: STEP BY STEP

4 0
3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
PLEASE HELP!!!!
defon

Answer:

v = 2.45 m/s

Explanation:

first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of cliff = 15 m

Vi = Initial Vertical Velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²

t² = (15 m)/(4.9 m/s²)

t = √3.06 s²

t = 1.75 s

Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,

s = vt

where,

s = horizontal distance covered = 4.3 m

v = original horizontal velocity = ?

Therefore,

4.3 m = v(1.75 s)

v = 4.3 m/1.75 s

<u>v = 2.45 m/s</u>

8 0
3 years ago
A tiger leaps horizontally from a high rock with a speed of 4.5 m/s. She lands 50 m away
VladimirAG [237]
A tigers lost one and yyyyyyyy was the first day in a while and they didn’t have a back up 4.5 50m
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3 years ago
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