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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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In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir
Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be S_{1}

and the distance covered by Sir Alfred be S_{2}

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

S = ut + \frac{1}{2}at^{2}

Where S is the distance traveled

u is the initial velocity

a is the acceleration

and t is the time

For Sir George,

S = S_{1}

u = 0 m/s (Since they start from rest)

a =0.21 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{1}  = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1}  = 0.105 t^{2}\\

t^{2} = \frac{S_{1}}{0.105}

Now, for Sir Alfred

S = S_{2}

u = 0 m/s (Since they start from rest)

a =0.26 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{2}  = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2}  = 0.13 t^{2}\\

t^{2} = \frac{S_{2}}{0.13}

Since, they traveled for the same time, t just before collision, we can write

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

S_{1} + S_{2} = 86 m

Then, S_{2} = 86 - S_{1}

Then,

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13} becomes

\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}

0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}

S_{1} = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

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