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vekshin1
3 years ago
13

Why are my rift zones common places for igneous rocks to form

Physics
1 answer:
andrew-mc [135]3 years ago
6 0
The thin crust of a rift zone causes melting in the upper mantle, resulting in volcanic activity.
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Help please !
Leviafan [203]

Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.

3 0
3 years ago
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PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST!!!!!
umka2103 [35]

Answer:

i think its The movement of large pieces of ice from one place to another.

Explanation:

3 0
2 years ago
Karen has a mass of 57.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis
liq [111]

Answer:22537.92 J

Explanation:

Given

mass of karen \left ( m\right )=57.9 kg

Distance covered by Karen =69.6 m

acceleration due to gravity=9.8

inclination=34.8^{\circ}

vertically karen move

h=69.6\times \sin \left ( 34.8\right )

h=39.72 m

Work done by escalator on karen is equal to change in potential energy of  karen

\Delta P.E.=m\times g\times \Delta h

\Delta P.E.=57.9\times 9.8\times 39.72=22537.92 J

8 0
3 years ago
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Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont
koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn   ⇒   µ = f/n ≈ 0.216

7 0
2 years ago
KINEMATICS: MOTION ALONG STRAIGHT LINE
Vinil7 [7]

Explanation:

hope this helps, cheers!

6 0
3 years ago
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