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3 years ago

Where are short distance surface currents located

Physics

2 answers:

mylen [45]3 years ago

7 0

https://quizlet.com/11912008/chapter-12-flash-cards/
this sight might help

marysya [2.9K]3 years ago

7 0

Answer:

As explained below.

Explanation:

The air and the ocean transport heat from the lower surface to the areas of higher surfaces and hence the cool and warm current makes a trip back to the globe i.e there is a redistribution of heat across the latitudes.

Thus the formation of currents takes place where the distribution of water takes place under the effects of the sun's gravity and air and thus moves rapidly in the surroundings. Like the wrong gulf stream and the Kuroshio current that transports water from the north pacific ocean.

As these are produced by the rotation of the earth and the axial tilt of the planet, and the impact that they derive from the Coriolis force. The deep oceanic currents also are affected by the Thermohaline belts.

The shorter duration of surface current is thus a product of the swirling motion of the particles from side to side and effected by the can be aided by the tremors felt by volcanic activity.

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A 2.6 kg mass attached to a light string rotates on a horizontal,

Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

maximum mass the string can support before breaking, m = 17.9 kg
radius of the circle, r = 0.525 m

The maximum speed the mass can have before it breaks is calculated as follows;

$T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s$

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

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3 years ago

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d1i1m1o1n [39]

If "0.3 minute" is correct, then it's 9,543,272 Joules.

If it's supposed to say "0.3 SECOND", then the KE is 2,651 Joules.

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3 years ago

Which of the following is a zone found in the open ocean

Svetlanka [38]

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3 years ago

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A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

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3 years ago

A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

lidiya [134]

The position vector can be transcribed as:

A = 6 i + y j

i points in the x-direction and j points in the y-direction.

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|A|2 = A·A

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3 years ago