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Oksanka [162]
2 years ago
15

A trough is 5 meters long, 2 meters wide, and 3 meters deep. The vertical cross-section of the trough parallel to an end is shap

ed like an isoceles triangle (with height 3 meters, and base, on top, of length 2 meters). The trough is full of water (density 1000kgm3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top.
Mathematics
1 answer:
SSSSS [86.1K]2 years ago
6 0

We need a work of 294210 watts to pump the water over the top. \blacksquare

<h3>Work needed to pump all the water over the top</h3>

Since the <em>cross section</em> area of the trough (A), in square meters, varies with the height of the water (h), in meters, and considering that <em>pumping</em> system extracts water at <em>constant</em> rate, then the work needed to pump all the water (W), in joules, is:

W = \int\limits^{V_{max}}_0 {p} \, dV (1)

Where:

  • p - Pressure of the infinitesimal volume, in pascals.
  • V - Volume, in cubic meters.
  • V_{max} - Maximum volume allowed by the trough, in cubic meters.

The <em>infinitesimal</em> volume is equivalent to the following expression:

dV = A\, dh (2)

Since the area is directly proportional to the height of the water, we have the following expression:

A = \frac{A_{max}}{H_{max}}\cdot h (3)

Where:

  • A_{max} - Area of the base of the trough, in square meters.
  • H_{max} - Maximum height of the water, in meters.

In addition, we know that pressure of the water is entirely hydrostatic:

p = \rho \cdot g \cdot h (4)

Where:

  • \rho - Density of water, in kilograms per cubic meters.
  • g - Gravitational acceleration, in meters per square second.

By (2), (3) and (4) in (1):

W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}}{H_{max}} \int\limits^{H_{max}}_{0} {h^{2}} \, dh   (5)

Where:

  • W_{max} - Width of the base of the triangle, in meters.
  • L_{max} - Length of the base of the triangle, in meters.
  • H_{max} - Maximum height of the triangle, in meters.

The resulting expression is:

W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}\cdot H_{max}^{2}}{3}   (5b)

If we know that \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}}, W_{max} = 2\,m, L_{max} = 5\,m and H_{max} = 3\,m, then the work needed to pump the water is:

W = \frac{(1000)\cdot (9.807)\cdot (2)\cdot (5)\cdot (3)^{2}}{3}

W = 294210\,W

We need a work of 294210 watts to pump the water over the top. \blacksquare

To learn more on work, we kindly invite to check this verified question: brainly.com/question/17290830

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Step-by-step explanation:

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2 years ago
Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a
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Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

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