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andrew-mc [135]
3 years ago
12

What causes a metamorphic rock to change into an igneous rock?

Physics
1 answer:
Natasha_Volkova [10]3 years ago
8 0
It has to be melted or cooled to become or change into an igneous rock.
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Distinguish between refraction,reflection and total internal reflection
Ipatiy [6.2K]

The contrast between total internal reflection and refraction is that total internal reflection is the full underwater build with no defeat of brightness, whereas refraction is the shift in the path of a tide that is parting from one medium to another.

7 0
2 years ago
How can gases have such a large impact on earths atmosphere
kozerog [31]

Answer:

well, the air in our atmosphere stays there and nothing can get past the last layer. Think of is as a car. if there gets to be too much gas in it, it can harm you. I would also say it is one of the causes of global warming. Also air pollution is a big problem.

Explanation:

8 0
3 years ago
The planet Neptune is the farthest planet from the Sun. A satellite is orbiting around Neptune at a distance of 180 km. As the s
stich3 [128]
2,992² + r² = ( r + 180 )²
8,952,064 + r² = r² + 360 r + 32,400
360 r = 8,952,064 - 32,400
360 r = 8,919,664
r = 8,919,664 : 360
r = 24,776.844 km
d = 24,776.844  ·  2 = 49,553688 ≈ 49,554 km
Answer:
The approximate diameter of the planet Neptune is 49,554 km.
4 0
4 years ago
Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a
Tasya [4]

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,\rm L_f = 1.45 m = 1450 \ mm

Initial length,\rm L_i = 7.5 mm

\rm F  = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F  = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

8 0
2 years ago
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
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