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viva [34]
4 years ago
13

During the rock cycle, sedimentary rocks are returned to the Earth's interior and become metamorphic rocks through metamorphism.

Metamorphic rocks can then move even deeper into the Earth's interior and melt into magma.
What phenomenon is responsible for moving sedimentary and metamorphic rock into the Earth's interior?
A. plate tectonics
B. infiltration
C. condensation
D. the carbon cycle

Physics
2 answers:
atroni [7]4 years ago
6 0
A plate tectonics I think
Tom [10]4 years ago
4 0
A. plate tectonics is the correct answer.
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we can reduce the friction between its moving parts.

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3 years ago
Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus
miss Akunina [59]

Answer:

(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2

Explanation:

Energy density in magnetic field is given as;

U_B = \frac{1}{2 \mu_o} B^2

where;

B is the magnetic field strength

Energy density of electric field

U_E = \frac{1}{2}\epsilon E^2

where;

E is electric field strength

Take the ratio of the two fields energy density

\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o}  *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2

Now replace E x L with IR

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2

Also, B = μ₀I / 2πr, substitute this value in the above equation

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2

cancel out the current "I" and factor out μ₀

\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2

Finally, the equation becomes;

\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

3 0
3 years ago
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