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3 years ago

What is the most important factor affecting the movement of water in an ocean?

Physics

2 answers:

insens350 [35]3 years ago

8 0

Answer:

It would be currents. Currents bring cold, rich nutrients to the surface of the water while the warmer and less dense air is getting pushed downward creating a movement of the water. I hope this helps.

arlik [135]3 years ago

6 0

Probably ocean currents since these use heat to move large amounts of water throughout the ocean, and can you make this the brainliest answer

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What is the acceleration of an object with a mass of

aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

Δ Being Δ

F = Force = 183 N

m = Mass = 367 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=183 \ N / 367\ kg}$

⇒ Resolving

$\boxed{ a = 0.49 \ m/s^{2}}$

Result:

The aceleration is 0,49 meters per second squared (m/s²)

Good Luck!!

3 0

3 years ago

A pendulum is set in motion and time. The time measured for 20 complete swings is 308.

laiz [17]

Answer:

solution

308=20 swings

?::: =1

( 308×1)÷20

308÷20

154÷10

=15.4

=15.4

6 0

3 years ago

How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats

3 0

3 years ago

Read 2 more answers

In a pilot study to help train a group of data collectors who will evaluate childrens pain using a FACES rating scale, the resea

muminat

Answer:

Option c. Inter-rater Reliability

Explanation:

Here, the rating is done by a group of data collectors under training for evaluation children's pain on Faces scale which is a scale ranging from 0 to 10 with different expressions or faces with a happy face at 0 to a crying face at 10.

Also in Inter-rater Reliability, the relative consistency of a study or test is assessed and the extent to which different group members rated the same behavior, the consistency of which is evaluated.

Thus it can also be helpful in interviews, etc.

6 0

3 years ago

An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.

vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0

3 years ago