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4 years ago

Describe the process by which wind causes erosion

1 answer:

4 0

Wind starts to take tiny little bits of gravel, sand, or dirt and erodes it to a spot until a certain height because at that height wind can't erode that high.

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King arthurs knights fire a cannon from the top of the castle wall. the cannonball is fired at a speed of 50m/s and an angle of

erma4kov [3.2K]

Answer:

Explanation:

initial height = 0.5*gt^2 = 0.5*9.8*1.5 = 7.35 m

now,

solving this and taking only the positive value of t

t = (sqrt(76906)+250)/98 or <em>5.3808 secs</em>

s = 50*cos(30 degree)*(sqrt(76906)+250)/98

<h2><u>s = 232.9959 m or 234 approx.</u></h2>

5 0

3 years ago

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

m_a_m_a [10]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to = 4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω² .................1

and kinetic energy is directly proportional to square of the amplitude.

$\frac{KE2}{KE1} = \frac{A2^2}{A1^2}$ .............2

$\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}$

$\frac{KE2}{KE1}$ = 0.52284

so factor that bug maximum KE change is 0.52284

5 0

4 years ago

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

notka56 [123]

You must assume that the mass of the rocket and engine remains constant - even though the engine is burning.

You know the engine produces 13.8N for a distance of 14.6m

The total energy expended (work done) by the engine is FxD so you can calculate that

Now - some of that is given to the rocket as kinetic and potential energy, and some is expended against the drag force.

At the peak of its flight ALL the energy given to the rocket is potential energy (its velocity is zero) and that is calculated as mgh

So
Energy given to rocket = mgh
Energy expended by engine = F x D (D= height where engine stops)
Energy 'lost' to drag is the difference between the two values.

7 0

3 years ago

Read 2 more answers

A world war ii bomber flies horizontally over level terrain, with a speed of 273 m/s relative to the ground and at an altitude o

Alex_Xolod [135]

Given:
u = 273 m/s, horizontal launch speed
h = 3.27 km = 3270 m, altitude.
v = 0, vertical launch speed

g = 9.8 m/s².
Wind resistance is ignored.

The time, t, for the bomb to reach the ground is given by
0t + (1/2)gt² = h
4.9t² = 3270
t² = 667.3469
t = 25.833 s

The horizontal distance traveled is
273*25.833 = 7052.4 m = 7.0524 km

Answer: 7.05 km

6 0

4 years ago

5. Based on the data collected by the students, what is the tangential velocity of

malfutka [58]

The tangential velocity of the ball is 0.63 m/s.

<h3>What is tangential speed?</h3>

Tangential speed is the linear component of the speed of any object moving along a circular path.

v = ωr

where;

v is the tangential speed
ω is the angular speed (rad/s)
r is the radius of the circle

At a given radius of the circle = 0.3 m

Let the number of revolution per min = 20 rev/min

$\omega = 20 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 2.1 \ rad/s$

v = ωr

v = 2.1 x 0.3

v = 0.63 m/s

Thus, the tangential velocity of the ball is 0.63 m/s.

Learn more about tangential velocity here: brainly.com/question/25780931

8 0

2 years ago