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3 years ago

11

What function do the wind-driven currents play in the transport of heat energy in the oceans?

1 answer:

aleksklad [387]3 years ago

7 0

Wind driven currents create in the equatorial and warm zones of the earth’s oceans. The currents then transmit the warm water north, where the heat in the water is misplaced to the colder atmosphere. In the instance of the Gulf Stream, this transmission of heat to the air warms up England and Europe.
As the water cools, its' density upsurges because of chilling and increasing salinity. This cold water drops to the bottom of the ocean, where it is then stimulated by another current back south, similar to a train creating a big circuit.

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When you rub your hands together, you cannot push harder on one hand than the other. can push harder on one hand than the other.

Sholpan [36]

I think it means that your putting in an equal amount of force on each hand because it's the same brain that controls what you do. I think that's what it means, not entirely sure but hope this helps

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3 years ago

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ipn [44]

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3 years ago

How can I connect projectile motion into inclines. plsss give me ideas

Alenkasestr [34]

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3 years ago

An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:

kozerog [31]

Answer: $p_{SO_2}=0.017atm$

Explanation:

We are given:

$K_c=1.7\times 10^8$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $600K$

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6$

The chemical reaction follows the equation:

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}$

We are given:

$p_{SO_3}=300atm$ $p_{O_2}=100atm$

Putting values in above equation, we get:

$3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}$

$p_{SO_2}=0.017atm$

Hence, the partial pressure of the $SO_2$ at equilibrium is 0.017 atm.

3 0

3 years ago

An MP3 player draws a current of 0.120 ampere

WARRIOR [948]

Answer: (2) 108 C

Explanation:

Electric charge of 1 coulomb has said to have passed when a current of 1 Ampere is passed for 1 second.

$Q=I\times t$

Q= charge in coloumb = ?

I = current in amperes = 0.120 A

t= time in seconds = 900 sec

$Q=0.120A\times 900sec=108C$

3 0

3 years ago

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