I have tried to look this question up on so many search engines, but nothing I want ever comes up. I want to know what natural/tectonic force acts upon  horizontal strata to tilt it.
211923
Sedimentary rock forms relatively flat. Youngest always on top of the oldest. Just like a stack of newspapers if you were to accumulate them everyday.\n\nFolding from compressional stress is one. Rock is literally pushed together creating anticlines & synclines, upward folding & downward folding arches respectively.\n\nFaults can also do this. This can be accomplished through compressional (pushed together) or tensional (pulled apart) stresses.\n\nI am not too sure about this one but isostatic adjustment could possibly tilt rocks.\n\nHope this helps.\n\n:)
440306
The vertical, horizontal components of all forces should balance.\n       Let the tensions in the ropes be T1 and T2 newtons.\n       T1*cos(90-71) + T2*cos(90-28) = 254.8\n       i.e., T1*sin71 + T2*sin28 = 254.8\n\n       For horizontal componets (which are opp. & equal)\n       T1*cos71 + T2*cos28 = 0\n\nSolve for T1, T2 please.
243191
The original theory was called Continental Drift, proposed by Alfred Wegener in the first edition of his book "The Origin of Continents and Oceans" around 1915. Very few scientists accepted the idea, and it was not until actual measurements (such as from satellites) about 35 or 40 years ago that the idea began to gain popularity. The modern theory which covers continental drift and more is called Plate Tectonics.\n\nSeafloor spreading is thought to be caused by convection currents in the Earths magma. As material is added at the mid-ocean ridges, other material is pulled back under at subduction zones along continents, making the seafloor act similar to a very slow conveyor belt. Other areas of interest might be the Rift Valley in Africa and a failed rift in the Mississippi Valley of the USA related to the New Madrid Earthquake.
473909
In the vertical direction earth is pulling the object downward with a force.\n\nAnd all objects are pulled down by the earth with an acceleration given by g = 9.8 m/ s^2.\n\nFirst we consider the vertical motion alone.\n\nIn the vertical direction \nthe initial velocity u = 0.\nthe acceleration  a = g. \nThe time t = t to be found.\nDisplacement in the vertical direction   s  =   180 m.\n\nWe use the equation s = ut + 05 a tt.\n\nUsing this we find t = 6 second.\n\nNow let us consider the horizontal motion.\n\nThe object is released from the plane which was moving with a speed of 20m/s.\n\nThere fore all objects which was in the plane will also have a speed of 20m/s in the horizontal direction.\n\nHence in the horizontal direction the initial speed is 20m/s.\n\nThe acceleration in this direction is zero since no force is acting in this direction.\n\nThe time t is the same as 6 second.\n\nDistance we have to find.\n\nWe use the equation s = ut + 05 a tt.\n\nSince a = 0 the equation reduces to s = ut. = 20 x 6 = 120m\n\nThe horizontal displacement is 120 m.
541861
When a bolt is tightened, it stretches.\nThink of the thread as just two, long inclined planes in contact with each other. The tension in the bolt results in the two planes being pressed together, the harder you press, the greater the friction. As you undo the bolt, the tension is reduced, the pressure is reduced and so the friction is reduced.
508171
Angle of elevation should be that raised from the ground until it is vertical (90 degrees), correct?  The wall which the ladder is placed against will have a vertical face (90 degrees).  The ladder, therefore, represents the hypotenuse of the triangle so formed.\nCosine is the length of the adjacent sided divided by the hypotenuse.  So cos 70 = X/40 or 40*(cos 70) = X.\nWith me so far?\nOnce you have X, you have the hypotenuse and 1 side of a right angle triangle and finally a use for Pythagoras's theorem the square on the hypotenuse equals the sum of the squares on the other 2 sides.\nCan you do the rest of the arithmetic?  I hope this problem was in Trigonometry, otherwise I have given you the wrong way to solve it. But it is the easier way.
23149
Plate tectonics is the answer to your questions about weathering and erosion, as well as thickness of the continental crust. Plate tectonics moves and lifts continental plates continuously, and faster than erosion can carry the sediments away. Mt. Everest is a good example of this, as pressure from the adjoining crustal plates is lifting the mountain at a rate of a few centimeters a year. \n\nAs this pressure against the continental plates builds they often fold and break, and this causes parts of the contintal plates to slide up over other parts of the plate, making them thicker that the oceanic plates. \n\nAt present the only way to find the top of the mantle is using seismology. Siesmologists measure the way that shock waves (sound waves) move through the earth when earthquakes happen. By studying the way these sound waves are deflected from different rocks they can tell about how deep the boundary is between the crust and the mantle. \n\nHere is one of the best websites that explains plate tectonics. It has animated illustrations that show moving plates and how they have changed over time:\nhttp://www.ucmp.berkeley.edu/geology/tectonics.html
597435
It is caused by the movement of large plates on the earth's crust.  When these plates move into each other it causes a collision, which basicallly push them together - hence the reason for all the Mountains out west - 2 plates pushed together caused the ends to go up...hence mountains...but when these plates move you get the shaking, thus causing an earthquake.
431225
Because we know F=ma and the cart is moving at a constant speed, we know the total horizontal force=ma=0.  If the 91N force is acting down at 45 degrees, then only half of the force is down (vertical) and half is horizontal.  Since 45.5 N is acting horizontal forward, then drag and friction must total to 45.5 N actinig horizontal backward.
353994
The equation necessary to find the horizontal and vertical positions of the stone are\nm horizontal = (Initial horizontal velocity)(time) + (.5)(horizontal acceleration)(time)^2\nm vertical = (Initial vertical velocity)(time) + (.5)(vertical acceleration)(time)^2\nSince your horizontal acceleration is zero, that part goes away.\nYour vertical acceleration is caused by gravity, so it should be about (-9.8m/s^2).\nYour Initial horizontal velocity is (28.0 m/s)(cos (46.0)) = 19.45 m/s\nYour Initial vertical velocity is (28.0 m/s)(sin (46)) = 20.14 m/s\nThus, your two equations are\nm (horizontal) = (19.45 m/s) (time) and\nm (vertical) = (20.14 m/s)(time) + (.5)(-9.8m/s^2)(time)^2\nPlugging in your numbers, you get\na) m(horizontal) = 19.45 m; m(vertical) = 15.24 m\nb) m(horizontal) = 33.1 m; m(vertical) = 25.9 m\nc) m(horizontal) = 79.75 m; m(vertical) = 0 m (back on the ground)
246469
Shear walls resist horizontal forces on a wall.  Imagine that a giant pair of shears is trying to distort the wall by placing opposite horizontal forces at the top and bottom of the wall in the direction of its length.  Obviously any wall section with a large opening in it (windows, doors etc.) will have less resistance to such shear. See the link and use arrows for more detail.  Good luck!
89242
Maybe you have a large ore deposit beneath your location. This could generate a strong field that the polarized molecules would be attracted to. If it's closer to the upper part of the hill and big enough, it could also provide a small amount of gravitational attraction that could be pulling the water uphill.\n\nThe gravitational field of the Earth varies from place to place at the surface. We see this all the time when we take very sensitive readings. It is possible that a gravitational anomalies can be enough to cause odd effects at the surface.
101300
Interesting puzzle.  Let's talk aboout a rigid beam projecting from a wall first.  If there is a weight downward anywhere on the beam, it will produce a shear force and if the weight were sufficient the beam would be cut off at the wall (shear) as though a knife passed downward through it flush with the wall because this is where the downward beam force is resisted by the upward force of the wall.\n\nIf the weight is at the end of the rigid beam it tends to rotate the beam downward at the base as though there were a hinge there but the wall produces a counter-rotary force resisting motion.  The "moment" equals the weight times the distance from the wall and decreases to zero at the wall.\n\nTherefore your beam must consist of fittings that will resist shear at all times but release when the moment is reduced to zero at the wall.  Assuming the fittings can be greased to reduce friction to nearly zero, the beam fitting should catch the wall at the top like placing your hand up into a drawer handle to open it.  The lower fitting should be an inclined outward and downward plane such when there was insufficient moment to rotate the beam against the wall, the base of the beam will start to slide down the ramp.  There could be several configurations of hardware to achieve the same effect.
520682
Well, yes, the gavity is pulling you down, and you are not moving. So you are subjected to an acceleration and are not moving.\nMind you, one can argue that the *resulting* force is zero, as the ground is reacting with an equal and opposite force to your weight; if you take it that way, to have an unbalanced force should result in motion.
414747
It happens in the brittle part of the crust, which is not usually deeper than 20 km.\n\nHere's the jist:\n\nIt all comes down to stress and strain. It is not 'the plates rubbing together' as many people say; or at least that's a very simplistic view.\n\nPlate tectonics causes stress on the continents and oceans, all over the surface of the earth. In some places, the stress is very small (usually within the plates). Elsewhere, the stress is high, usually where the plates meet each other. Since each plate moves, when two come in contact, the stress each other. They can push on each other and cause compressional stress, they can pull with extensional stress, and they can slide past or shear each other with tensional stress.\n\nFaulting, causing earthquakes, comes from the fact that this stress is building up all the time, but rocks and continents are strong materials. Just like hitting a rock with a small hammer, you do put stress on it, but a small amount. It would take a sledgehammer to put enough strain (effects and accumulation of stress) to build up and cause breakage. In the earth, the area around an active fault builds up strain from the stress of plate tectonics. Most faults become locked, because of this strength, and thus can not release their strain. Away from the fault, the stress produces very small and slow movement of the rock masses as a whole. Eventually, the strain is too much and the rest of the plate has moved too far and the fault releases the strain build-up all at once in a big stress release called an earthquake. This is called the elastic rebound theory, and it explains most (but not all) movements.\n\nThis is why an earthquake's size is relative to the fault size. The bigger the fault, the bigger the strain build up, and the bigger the release in an earthquake. Subduction zones and collision zones, where large portions of plates actually can rub together as a whole, have the really big earthquakes, like the 2004 Sumatra earthquake. In other plate boundaries, the entire plate boundary does not act together, so the faults become spread out and many faults take up the strain from the tectonic stress, like with the San Andreas Fault in California (it only takes up ~3/4 of the stress between the plates).\n\nEarthquakes are really tricky things; there is still so much we need to learn. As of now, there is no way to predict them, but we can say where the danger is highest and about how long between events.
519680
Trenches define one of the most important natural boundaries on the Earth’s solid surface, that between two lithospheric plates. There are three types of lithospheric plate boundaries: divergent (where lithosphere and oceanic crust is created at mid-ocean ridges), convergent (where one lithospheric plate sinks beneath another and returns to the mantle), and transform (where two lithospheric plates slide past each other). The deepest ocean depth to be sounded is in the Challenger Deep of the Mariana Trench at a depth of 10,911 m (35,798 ft) below sea level\n\nFormation:\n\nThe distribution of earthquakes reveals the location on the subducted Pacific Plate beneath the Mariana Islands. From the deep Mariana trench to a depth of about 100 km the plate is subducted at a gentle angle. From about 100 to 680 km the plate is subducted at a near vertical angle. Some geologists believe that the downward gravitational force of the steeply dipping plate is enough to pull the Pacific Plate away from the mid-ocean ridge, essentially providing a driving force for plate tectonics. Other geologists believe that the plates are pushed away from the mid-ocean ridge by forces generated at the ridge. Most geologists would agree that both forces are important.
418456
1. Continental drift or plate tectonics.\n\n2. Horizontal fault line.