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andrezito [222]
3 years ago
10

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

ltitude of 2 feet. water is running into the trough at a rate of 1 cubic foot per minute. how fast is the water level rising when the water is 1 foot deep? slader
Physics
1 answer:
Ostrovityanka [42]3 years ago
7 0

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}

And the raising speed <em>v </em>of the water is given by:

v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}

where <em>q</em> is the water flow (1 cubic foot per minute).

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Semmy [17]

Answer:

Kinetic energy of diver at 90% of the distance to the water is 9000 J

Explanation:

Let d is the distance between the position of the diver and surface of the pool.

Initially, the diver is at rest and only have potential energy which is equal to 10000 J.

As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.

Energy before diving = Energy during diving

(Potential Energy + Kinetic Energy) =  (Kinetic Energy +  Potential Energy)

When the diver reaches 90% of the distance to the water, its kinetic energy

is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,

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we are given that Work = 1,500J and Distance = 20m

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assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

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