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Geologists collect data on friction along the side of faults so that they can predict how much pressure is applied on the faults so they can predict how strong the earthquake is.
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Why do scientists collect data on friction along the sides of faults?
i dont know really
How do geologists collect data on friction along the sides of faults?
Geologists study the types of movement that occurs along faults. How rocks move along a fault depends on how much friction there is between the sides of the fault. Friction is the force that opposes the motion of one surface as it moves across another surface. Friction exists because surfaces are not perfectly smooth.Along parts of some faults, the rocks on both sides of the fault slide by each other without much sticking. Therefore stress does not build up, and big earthquakes are unlikely. Along many faults, the rocks lock together. In this case, stress increases until it is large enough to overcome the friction force. Stress builds up until an earthquake occurs.
Is a basketball shape a sphere?
technically, no because it has thousands of ridges on it for friction, along with the black part that kind of dips down. if you are looking for a nonspecific answer that is simple, then yes, it is.
Explain why do bus tires can greatly increase when traveling over along distance at high speed?
Vehicles ove forward because of friction between the tyres and the road. Friction causes heat so the tyres and the air inside them heat up. A confined gas (air) tries to expand when it is heated. Tyres will expand, when the air inside applies sufficient pressure.
What is the coefficient of friction between the box and the floor A 10kg box is pulled at a uniform velocity with a rope that forms a angle of 60 degrees A tension of 80N is constantly maintained?
Before tackling this one, I must clean it up a bit.-- I'll assume that the floor is horizontal under the box.-- "Angle" means the difference between two directions, but the question specifiesonly one of them ... the direction of the rope. I'll assume that the 60° is the anglebetween the rope and the horizontal travel of the box, and that the rope and thetension in it are both directed above the horizontal, i.e., sloped toward the ceiling,not toward the floor.Now we have something we can work with.-- The horizontal component of the tension in the rope is 80 cos(60) = 40 N.-- The box is sliding along at constant speed, so the horizontal forces on it are balanced.That means that the friction force is also 40 N but backwards.-- The weight of the box is (m g) = (10 x 9.8) = 98 N.-- The coefficient of friction is friction force/weight = 40/98 = 40.8 %=======================Why this solution is bogus, at least in part:The other component of the tension in the rope ... the vertical one ... is 80 sin(60) = about 69.3 N.That force is applied to the box at the point where the rope connects, and pullsstraight up at that point. Its effect must be to reduce the box's apparent weightat that end, and by some complicated amount everywhere along the length of thebox. So the force of friction is also distributed along the length of the box in somenon-uniform and complicated way, and the aggregate apparent coefficient of frictionis some ugly integral of the contributions due to an element of weight at everyelement of length/area from one end of the box to the other.Am I over-thinking this ? ? Perhaps it would be best if I take a nap.
