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4 years ago

Which of the following natural features is formed by weathering, erosion, and deposition?

Physics

2 answers:

Otrada [13]4 years ago

7 0

The answer is d. sand dunes

Black_prince [1.1K]4 years ago

7 0

Answer:

Sand dunes

Explanation:

Dunes are hills of sand which are formed by the action of winds which is an agent of weathering. Sand dunes usually formed along the beach or in a desert. Sand dunes form when the wind blows, they carry particles of sand before they drop to the ground. As they roll and bounce on the ground, sand particles create small, wave-shaped ripples of sand. These ripples can build up into larger structures.

Dunes can grow very high say up to 4000 feet and be many miles long. Dunes can be various shapes such as crescents, stars, or just repeated lines.

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Gravity is greater when there is

Papessa [141]

-- more mass involved

-- less distance between the two objects

4 0

3 years ago

1. What are three examples of how invasive species spread?

Darya [45]

Answer:

Their primary way of spreading is from human activities, they can quickly travel around the world for example these new "murder" hornets that can kill a large bee hive with one sting, they traveled all the way from Asia. Invasive species can also be through people's luggage, small boats, planes and large shipment like cargo carriers. I hope this helps. : )

Explanation:

8 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler

kirza4 [7]

Answer:

a.) 2.66 seconds for 79.2 km/h and 1.82 seconds for 47.8 km/h

b.)-4.13 m/s2 for 79.2 km/h and -3.63 m/s2 for 47.8 km/h

Explanation:

Now for (a)

time = velocity/distance

for velocity = 79.2 km/h and distance 0.0586 km

t = (0.0586/79.2)*3600

t = 2.66 seconds (Please note that multiplication with 3600 is to convert hours into seconds)

for velocity 47.8 km/h and distance 0.0242 km

t = (0.0242/47.8)*3600

t = 1.82 seconds

Now for (b)

2as = vf2-vi2

so,

2a(58.6) = -(79.2*1000/3600)^2 (Please note that multiplication and division with 1000 and 3600 respectively is to convert speed unit from km/h to m/s)

a = -4.13 m/s2 for 58.6 m

and

2a(24.2) = -(47.8*1000/3600)^2

a = -3.63 m/s2 for 24.2 m.

Please note that "-" sign express the deceleration.

5 0

3 years ago

A 2.3 kg , 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt

Readme [11.4K]

Answer

ω2=82.1 rpm

Explanation:

given required

m1=2.3 kg ω2=?

m2+m3=0.5kg

r1=10 cm

ω1=100 rpm

solution

Using the application of conservation of angular momentum we can solve as follows/

L before collision= L after collision

m1r²ω1=(m1+m2+m3)r²ω2

2.3 kg×0.1² m²×100 rpm=(2.3 kg+0.5 kg)×0.1²m²×ω2

2.3 rpm=0.028×ω2

ω2=82.1 rpm

6 0

4 years ago