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3 years ago

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:

The answer is: 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.

Explanation:

The Grand Canyon is one of the most impressive erosive features in the world.

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.

The now widely accepted theory 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.

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

Once again, 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.

The answer is: 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.

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You drop an ice cube into an insulated flask full of water and wait for the ice cube to completely melt. The ice cube initially

3241004551 [841]

Answer: $T=12.69^{\circ}C$

Explanation:

Given

Mass of ice $m=60 gm$

mass of water $M=760 gm$

Initial Temperature of water $T_i=20^{\circ}C$

Let T be the Final Temperature of mixture

Latent heat of Fusion $L=334 J/gm$

heat required to melt ice completely is

$Q_1=60\times 334=20.04 kJ$

Heat released by water is taken by ice thus

$Mc_{water}(20-T)=mL+mc_{water}(T-0)$

$0.76\times 4.184\cdot (20-T)=20.04+0.06\times 4.184\cdot (T)$

$T=12.69^{\circ}C$

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2 years ago

A laborer pushes a box to distance of 20m. if the force required to push the box is 25N what will be the quantity of work done

Marrrta [24]

Answer:

500 J

Explanation:

In the case of a rectilinear movement, the work is calculated as the product of Force (N) * movement (m). In your case, unless the angle between the force vector and the displacement vector is different from 0, the work is:

25 N * 20 m = 500 J

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3 years ago

Difference between rest and motion?

Yuki888 [10]

Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

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2 years ago

Please help me with this review question.

jolli1 [7]

Answer:

28.7%

Explanation:

efficiency = work output /work input × 100

8 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago