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2 years ago

What are the five natural agents of erosion? What is the driving force behind all of these agents of erosion?

Physics

1 answer:

Ne4ueva [31]2 years ago

7 0

Moving water, wind, gravity, and ice are the five natural agents of erosion

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From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 m. The acceleration of gravi

Aleks [24]

Answer:

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8× $t^{2}$

==> $t^{2}$ = $\frac{37.6}{4.9}$

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s

8 0

3 years ago

A magnet can attract or repel another magnet from a distance true or false?

Anuta_ua [19.1K]

False the strength off the magnet lessens the farther you get from it

6 0

3 years ago

Read 2 more answers

One of the largest planes ever to fly, and the largest to fly frequently, is the Ukrainian-built Antonov An-124 Ruslan. The grav

Lerok [7]

Answer:

m = 236212 [kg]

Explanation:

The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

$P=m*g*h\\$

where:

P = potential energy = 3360000000 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1450 [m]

Now, we can clear the mass from the equation above:

$3360000000=m*9.81*1450\\m = 236212 [kg]$

4 0

2 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .