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Anni [7]
3 years ago
11

What is the primary way that metamorphic rocks forms?

Physics
2 answers:
sergiy2304 [10]3 years ago
8 0

Answer:

high heat, high pressure, hot mineral-rich fluids or commonly a combination of these found deep in the Earth or where tectonic plates meet.

Explanation:

I hope this helps :3

Sladkaya [172]3 years ago
5 0

Answer:

where the rocks are exposed to really hot conditions and high pressure....these conditions are found deep where tectonic plates meet or deep inisde the earth past our core

hope this helps !!!!!

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What is the magnitude of the electric field at the dot in the figure? (figure 1)
blondinia [14]

The magnitude of the electric field at the dot is : 10⁴ v/m

<h3>What is electric field?</h3>

Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

Determine the magnitude of the electric field at the dot

change in voltage  = E .d

100 - 0 = E \times ( 1 \times 10^{-2}\  m ) ----- ( 1 )

From equation ( 1 )

The magnitude of electric field will be given as

E = \dfrac{100 }{ ( 1 \times  10^{-2})\ }

E = 10^4\ \ \frac {v}{   m}

Hence we can conclude that The magnitude of the electric field at the dot is :  10⁴ v/m

Learn more about electric field :

brainly.com/question/14372859

#SPJ4

5 0
2 years ago
You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
Paladinen [302]

Answer:

(A)

\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

Solving for t

\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

Now, let's put in numbers:

g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec

\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0

Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

Solving for D, we have two possible solutions:

D=54.71,\ D=26,651.11

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec

\displaystyle t_2=\frac{54.71}{345}=0.16\ sec

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0
3 years ago
An engineer measured the length of a building which was equal to 55.2 m. If there was an error of 0.02 m, what are the probabili
Vinvika [58]

Answer:

Explanation:

https://web.lemoyne.edu/courseinformation/mth%20112/rinaman/instman/SOLUTION.PDF

6 0
2 years ago
What does displacement describe?
Katarina [22]
<span>A vector which implies that an object has been moved or has changed its position is called displacement. Displacement is usually associated with length and direction of an imaginary straight point. It is the shortest distance from the initial point to the final point of final position (P). Displacement can also be described as the length between the final and initial point on the shortest path. It means an overall change in direction  of the object or point of a body.</span>


5 0
3 years ago
A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?
Anuta_ua [19.1K]

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

5 0
2 years ago
Read 2 more answers
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