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Elan Coil [88]
3 years ago
6

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

Physics
2 answers:
Svetlanka [38]3 years ago
5 0
What are your answer choices?

horrorfan [7]3 years ago
5 0

The answer to this question is D.

The neritic zone (option a) is a shallow area of ocean (about 200m in depth) which is essentially the area where the ocean interacts with the coast, although in very rare cases this can be further out to sea dependent on the oceanic plate.

The surface zone (option b) is, as the name suggests, where the ocean and the air meet and therefore not exclusive to the "open ocean" as specified in the question.

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Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.

From this question, the plane is still up in the air.

We have wind blowing in [W 60° N ]

To solve the problem we have to make use of the sine rule

\frac{SinA}{a}=\frac{SinB}{b} =\frac{SinC}{c}

We put the values in the equation, we have:

50/Sinθ = 200/sin60°

The next step is to cross multiply

50 x sin60° = 200Sinθ

50 x 0.8660 = 200sinθ

We make Sin θ the subject

Sine θ = 43.30/200

sine θ = 0.2165

we find the value of θ

θ = sine⁻¹(0.2165)

θ = 12.50

So Lindsay has to fly this plane towards this direction

[W 12.5° S]

Here is a similar question brainly.com/question/13338067?referrer=searchResults

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What is newton's first low of motion?​
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Newton's first law of motion is that an object in motion will tend to stay in motion unless an external force acts upon it.
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A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

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a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

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Learn more about coefficient of friction here: brainly.com/question/14121363

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5 0
1 year ago
You lift a 50 N object 2 meters off the ground what work did you do on the object
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Work equals force × displacement (distance between initial point and end point is displacement)
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4 0
3 years ago
A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat
liraira [26]

Answer: E = 1.8 *10 ^{4} N

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E =\frac{Kq}{r^{2} } r

where  k=9* 10^{9}N/m^{2}, q=2 *10 ^{-8} c , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

E=\frac{9 * 10^{9}* 2* 10^{-8}  }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N

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