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Luden [163]
4 years ago
5

Describe the difference between the ridge push hypothesis and slab pull process.

Advanced Placement (AP)
2 answers:
AVprozaik [17]4 years ago
8 0

Slab pull is the cool dense crust sinking into the earth in subduction zones, this literally pulls the entire plate down with it. Ridge push is the lifting and pushing apart of the plates at spreading zones.  

The Pacific ocean floor moves much quicker than the Atlantic because it has both slab-pull and ridge push acting on it, while the Atlantic only had ridge push (probably the weaker of the two forces)

Murrr4er [49]4 years ago
3 0

The correct answer to this open question is the following.

The difference between the ridge push hypothesis and slab pull process is the following.

During earthquakes, we are so scared to think about the reason why the enormous mass of the land can move. These geological terms of plaque tectonics can be understood like this. The excess height of the mid-ocean ridge push exerts so much pressure. This is the ridge push. The force that the weight of the subducted slab exerts on the plate that is attached is called the slab pull. These forces drive the tectonic plates of planet Earth.

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3 years ago
I can’t figure this problem out
Natalka [10]

Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

\int\limits^4_1 {(3f(x)+2)} \, dx

So lets break this up into two different integrals that represent the same area.

\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

\int\limits^4_1 {f(x)} \, dx

But since we have these values we can say that:

-3 - 2 = -5

Which means that \int\limits^4_1 {f(x)} \, dx = -5

So now we can evaluate \int\limits^4_1 {(3f(x)+2)} \, dx

Lets first break up our integrand into two integrals

\int\limits^4_1 {(3f(x)+2)} \, dx = 3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx

Now we can evaluate this:

We know that \int\limits^4_1 {f(x)} \, dx = -5

So:

3(-5)+2[x] where x is evaluated at 4 to 1 so

-15 + 2(3)

So we are left with -15 + 6 = -9

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3 years ago
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