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larisa86 [58]
3 years ago
6

What process, driven by heat generated from within the Earth's core, occurs in the mantle and is a primary cause of plate tecton

ics?
A.faulting
B.mantle convection
C.subduction
D.trench formation
Physics
2 answers:
timofeeve [1]3 years ago
5 0
The answer is B. Mantle convection.
Lilit [14]3 years ago
3 0
<h2>Answer:</h2>

<u>The correct option is</u><u> (B)mantle convection</u>

<h2>Explanation:</h2>

Tectonic plates at the earth's surface move because of the intense heat in the Earth's core. this heat causes molten rock in the mantle layer to move. It moves in a pattern called a convection cell and producing a movement called mantle convection which is the slow creeping motion of Earth's solid silicate.

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A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point
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Answer:

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

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What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel
Verizon [17]

Answer:

b. mg ( μ · cos θ + sin θ)

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg ·  cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ      (N = wy = mg ·  cos θ)

F = μ · mg ·  cos θ  + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

8 0
3 years ago
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