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3 years ago

Why do you think the rate of weathering tends to increase at hotter temperatures and higher amounts of rainfall?

Physics

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

Explanation:

Both increase in temperature and availability of water drives physical and chemical weathering and increases their rate.

Increase in temperature causes rocks to expand and molecules are more free to move about. Constantly heating and cooling of a rock can lead to mechanical fracturing or cracks developing.

A high amount of rainfall facilitates chemical weathering. Chemical weathering is the decomposition of rocks.

The main agent that is important for natural chemical processes to occur is the presence water.

Water contains some other dissolved substances in it.

This is evident in tropical regions of the world

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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

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(#1). (D).

(#2). (C).

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2 years ago

A lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before

Zina [86]

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<span>PE=mass∗gravity∗height=0.3kg∗9.8m/s2∗1.8m=?

</span>The answer will typically be given in joules:

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2 years ago

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An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation