A conveyor 60 m long is inclined at an angle of 18° to the horizontal and deposits material 8 m below its top end. Determine the distance and displacement of material carried by this conveyor.
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The distance the material travels is simply the length of the conveyor, which is 60 m.
The displacement, however, is the shortest distance from the starting point to the ending point. In this case, it is the horizontal distance the material travels.
We can use trigonometry to find this distance. The conveyor forms a right triangle with the horizontal and vertical distances the material travels. The cosine of the angle of the conveyor is equal to the adjacent side (the horizontal distance) divided by the hypotenuse (the length of the conveyor).
So, cos(18°) = horizontal distance / 60 m
Solving for the horizontal distance gives us:
Horizontal distance = 60 m * cos(18°) = 56.57 m
However, since the material is also displaced 8 m vertically downwards, we need to take this into account. The total displacement is the square root of the sum of the squares of the horizontal and vertical displacements (by Pythagoras' theorem):
Displacement = sqrt((56.57 m)^2 + (8 m)^2) = 57.07 m
So, the distance the material travels is 60 m and the displacement is 57.07 m.
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