What represents the third dimension in a topographic map and shows elevation?

Physics

2 answers:

Elis [28]3 years ago

8 0

Answer:

The contour lines are the third dimension, showing the elevation of an area

Explanation:

Topographic maps are those that shows the topography of a particular area, type of relief and the location of hills and plains. They are provided with the north direction and the index that helps in a better understanding of the maps. They are commonly used by geologists and geographers.

The third dimension of the map i.e the height of the surface features is shown in terms of contour lines. Contour lines are the lines of equal elevations. They are shown by a number that depicts the height at that point from the mean sea level. The high mountains, valleys, and basins are easily identified from these maps.

IgorC [24]3 years ago

3 0

An object or any location can be represented on a two-dimensional surface like a paper or computer screen. This representation is known as map. Most maps do not take into account the elevation of the object or the location they representing. On the other hand, Topographic maps use contour lines to represent the third dimension and to show elevation change on or below the surface of the earth.

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Which statement describes how nuclear power generation systems work?

galben [10]

Answer:

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Explanation:

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2 years ago

A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$