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3 years ago

Where do most metamorphic processes take place?

Physics

1 answer:

inysia [295]3 years ago

8 0

Answer:

Most metamorphic processes take place deep underground, inside the earth's crust.

Explanation:

During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)

You might be interested in

A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65Â°C and 435Â°C if

const2013 [10]

Answer:

The value is $P_e = 31275.2 \ W$

Explanation:

From the question we are told that

The efficiency of the carnot engine is $\eta$

The efficiency of a heat engine is $k = \frac{3}{4} * \eta$

The operating temperatures of the carnot engine is $T_1 = 65 ^oC =338 \ K$ to $T_2 = 435 ^oC = 708 \ K$

The rate at which the heat engine absorbs energy is $P = 44.0 kW = 44.0 *10^{3} \ W$

Generally the efficiency of the carnot engine is mathematically represented as

$\eta = [ 1 - \frac{T_1 }{T_2} ]$

=> $\eta = [ \frac{T_2 - T_1}{T_2} ]$

=> $\eta = 0.3856$

Generally the efficiency of the heat engine is

$k = \frac{3}{4} * 0.3856$

=> $k = 0.2892$

Generally the efficiency of the heat engine is also mathematically represented as

$k = \frac{W}{P}$

Here W is the work done which is mathematically represented as

$W = P - P_e$

Here $P_e$ is the heat exhausted

$k = \frac{P - P_e}{P}$

=> $0.2892 = \frac{44*10^{3} - P_e}{44*10^{3}}$

=> $P_e = 31275.2 \ W$

8 0

3 years ago

During spring tide, the sun, earth, and moon are in a straight line. This causes...............

labwork [276]

D causes higher high tide

8 0

3 years ago

How are energy, work ang power related?

jolli1 [7]

Answer:

Work is the change in energy and power is the rate of doing work.

5 0

2 years ago

PLEASE HELP MEE

ExtremeBDS [4]

Answer:

<h2>work = mgh</h2><h2> = 125 × 2</h2><h2> = 250 joules</h2>

7 0

3 years ago

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

Strike441 [17]

The solution would be like this for this specific problem:

Given:

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m 

Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]

∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m

6 0

3 years ago