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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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(1) find the density of a substance if the mass of the substance is 150kg and dimension 20mby10mby5m

Vikentia [17]

Answer:

0.15kg/m³

Explanation:

Density = mass/ volume

Given that

Mass = 150kg

Note that volume = length x breadth x height

Volume = 20 x 10 x 5

Volume = 1000m³

Density = mass ➗ volume

Density = 150kg ➗ 1000m³

Density = 0.15kg/m³

I hope this was helpful, Please mark as brainliest

3 0

3 years ago

In the simulation above, as the projectile travels upward, how does the vertical velocity change? Question 9 options:

enot [183]

Answer:

Vertical velocity decreases.

Explanation:

The motion of the ball is a projectile ball, which consists of two independent motions:

- a horizontal motion, with constant velocity

- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground

In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation

$v(t)=v_0 -gt$

And it decreases until the ball reaches its maximum height, then it starts increasing again.

4 0

3 years ago

A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image

lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor. T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0

3 years ago

1. What is the new kinetic energy of the 1900kg ship on the right moving at 4 m/s?

neonofarm [45]

Explanation:

That`s is the answer, just check

6 0

3 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago