All categories

4 years ago

What types of geologic structures would you expect to find in an area that had undergone compressional tectonic stress?

Physics

1 answer:

romanna [79]4 years ago

3 0

Answer:

Folds.

Explanation:

A fold occurs when an originally flat or a plane surface of sedimentary rocks undergoes a bent or a curve and that may lead to permanent deformation.

And are formed due to the slumping of the sedimentary materials. They can form under the conditions of the compressional stress, hydrostatic pressure, and temperature gradient.

A set of folds is distributed at a region scale makes the fold belt and is commonly found in the orogenic zones or the mountain building areas. These folds are of various types as Anticlines, Synclines, and Antiforms.

You might be interested in

What information do you need to describe an object's location

stira [4]

longitude and latitude<span />

4 0

3 years ago

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

oksano4ka [1.4K]

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

6 0

3 years ago

Read 2 more answers

Question 4 Unsaved If element X has 93 protons, how many electrons does it have? Your Answer:

9966 [12]

It have 93 electrons

8 0

4 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

6 0

3 years ago

As Clinton walks he pushes his shoe against the track.

Valentin [98]

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

4 0

3 years ago