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1 year ago

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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irina [24]

Answer:

42 m

Explanation:

$v_{o}$ = initial velocity of the car = 24 m/s

$v_{f}$ = final velocity of the car = 0 m/s

μ = coefficient of kinetic friction = 0.7

g = acceleration due to gravity = 9.8 m/s²

a = acceleration due to kinetic frictional force = - μg = - (0.7)(9.8) = - 6.86 m/s²

d = distance through which the car skids

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d$

Inserting the values

$0^{2} = 24^{2} + 2 (- 6.86) d$

d = 42 m

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3 years ago

A 25 N force is used to push a box across a floor against a frictional force of 14 N. The box

Svetlanka [38]

Answer:

$m = 44~Kg$

Explanation:

<u>Net Force</u>

Newton's second law explains the dynamics principles when a number of forces are applied to an object.

The net force vector is the sum of the individual vector forces applied. The magnitude of the net force is related to the magnitude of the acceleration of the body as follows:

$Fn=m.a$

Furthermore, the acceleration can be calculated if we know the kinematic behavior of the body:

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf, vo, and t are the final speed, initial speed, and time, respectively.

The box is pushed across the floor with a force of 25 N against a frictional force of 14 N.

The net force applied to the box is:

$Fn = 25 N - 14 N = 11 N$

We also know the box is accelerated from rest (vo=0) to vf=4 m/s in t=16 seconds, thus:

$\displaystyle a=\frac{4-0}{16}$

$a = 0.25~m/s^2$

From the equation:

$Fn=m.a$

We solve for m:

$\displaystyle m=\frac{Fn}{a}=\frac{11~N}{0.25~m/s^2}$

$\mathbf{m = 44~Kg}$

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kirza4 [7]

Answer:

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Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87 m

Work done by the engine is given as

W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

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KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

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W' = W + W''

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