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BigorU [14]
3 years ago
13

Where did Earth’s water come from?, select the two competing scientific theories supported by evidence that explain the formatio

n of the hydrosphere.
Physics
1 answer:
liraira [26]3 years ago
7 0

Answer:

Explanation:

There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.

The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.

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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
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a) F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

R-Fsin \theta - mg = 0 (1)

where

R is the normal reaction of the floor (upward)

F sin \theta is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

F cos \theta - \mu_k R = 0 (2)

where

F cos \theta is the horizontal component of the force F (forward)

\mu_k R is the force of friction (backward)

From (1) we get

R=Fsin \theta +mg

And substituting into (2)

F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

b) \mu_s=cot(\theta)

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

R-Fsin \theta - mg = 0 (1)

F cos \theta - \mu_s R = 0 (2)

In this second case, we want to find the critical value of \mu_s such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, F cos \theta.

Therefore, using the same procedure as before,

R=Fsin \theta +mg

F cos \theta - \mu_s (Fsin \theta +mg) = 0

And solving for \mu_s,

F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}

Now we analyze the expression that we found. We notice that if the force applied F is very large, F sin \theta >> mg, therefore we can rewrite the expression as

\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)

So, this is the critical value of the coefficient of static friction.

8 0
3 years ago
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