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2 years ago

Why can rock avalanches move downslope at speeds exceeding 200 km/hr (125 mi/hr)?

1 answer:

5 0

Rock avalanches move downslope at speeds exceeding 200 km/hr (125 mi/hr) because when the soil expands and compresses, it lifts certain particles and drops them downslope a short distance.

Landslides are the common name for soil or rock avalanches. The most frequent type of avalanche, snowslides, can travel downhill more quickly than a skier. When an unstable snow mass detaches from a slope, a snow avalanche starts.

Rock avalanches are caused by the rapid breakup of initially intact, highly fast-moving rock masses during transportation. In mountainous areas, rock avalanches, which are unexpected rock slope failures marked by high velocities, lengthy runouts, and massive volumes, are among the most deadly and costly geological hazards.

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You might be interested in

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

$U = \mu B$

where;

$\mu$ is the dipole moment
B is the magnetic field

$B = \frac{\mu_0 I}{2\pi r}$

$U = \mu\times (\frac{\mu_0 I}{2\pi r} )$

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

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7 0

2 years ago

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

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8 0

2 years ago

A. 24.89<br> B. 25.89<br> C. 17.74<br> D. 19.73

Veronika [31]

Answer: D

Explanation:

Just did it got an 100

5 0

2 years ago

Scientists measure the time between the arrival of an earthquake's _______ and _______ waves to help determine the distance betw

soldier1979 [14.2K]

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake, the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

Learn More:

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7 0

4 years ago

Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro

Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0

3 years ago