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vivado [14]
3 years ago
13

Development of the Grand Canyon through the uplift of the plateau and erosion by the Colorado River illustrates that:

Physics
2 answers:
MAXImum [283]3 years ago
7 0
Landforms result from a combination constructive and destructive forces.
Hope this helps
Fittoniya [83]3 years ago
4 0

Answer:

<em><u>The answer is</u></em>: <u>The Grand Canyon is a testament to the erosive power of water. This modified model seems to be the predominant theory among today's evolutionary geologists.</u>

<u />

Explanation:

<u>The Grand Canyon is one of the most impressive erosive features in the world</u>.

Carved through sedimentary layers of sandstone, limestone and shale and in the basement formations of mostly metamorphic shales and igneous granites, the Grand Canyon is a testament to the erosive power of water.

<u>The now widely accepted theory</u> has the capture of the current that takes place in one of the northwest drains that is believed to have existed before the elevation of the plateau.

<u>It was believed that the ancestral Colorado River had turned north,</u> draining into the Great Salt Lake region.

<u>Once again</u>, once the capture was carried out, the plateau rose, which caused the drains that tend to the northwest to flow into the Colorado River.

This modified model seems to be the predominant theory among today's evolutionary geologists.

<em><u>The answer is</u></em>: <u>The Grand Canyon is a testament to the erosive power of water. This modified model seems to be the predominant theory among today's evolutionary geologists.</u>

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Study the four transverse waves shown. Compare the properties of waves B, C, &amp; D to that of wave A.
GalinKa [24]
Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.
7 0
3 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
Can someone please help me?
IRINA_888 [86]
It might be to late but the answer is C
7 0
3 years ago
Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m
ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

x^2=y^2+145^2

On differentiating

2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}

\dfrac{dx}{dt}=14.89\ miles/hr

Hence, The rate of change of the distance is 14.89.

8 0
3 years ago
A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele
ludmilkaskok [199]

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for y, and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

8 0
3 years ago
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