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3 years ago

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 charge of an atom with 21 protons and 6 electrons

Svetach [21]

Answer:

Scandium with an ion charge of +3

Explanation:

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2 years ago

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A small hot-air balloon is filled with 1.01×106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.10

lara31 [8.8K]

<span>Volume of air in the balloon 1.01 x 10^6 L
Density of air is 1.20 g/l
Mass = Density X Volume

So mass of the air in the Balloon= ( 1.01 x 10^6) X 1.20 = 1.212 x 10^6 g
As the air is heated, the volume of air in the balloon expands to 1.10x 10^6 L Density= Mass/ voume So the Density of heated air = 1.212 x 10^6/ 1.10x 10^6 = 1.101 g/l The answer is 1.101 g/l.</span>

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3 years ago

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

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2 years ago

An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?

Serjik [45]

The image of the object is 8cm to the left of the lens (D)

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

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2 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

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3 years ago