The troough down the center of the ccattle barn is 40 cm wwide at thhe top and 20 cm at the botttom .it is 30 cm deep and 8 m long . It iis filled at a rate of 10 dm3/min. What is the rate of rise of water level .
I tried it and me ans is pretty close but not exact so just wanted if somebidy had any ideas?
look at a cross-section of the front of the trough
which would be a parallelogram.
draw verticals from the base to create two equal right-angles triangles, of height 30 cm and width 10 cm, with a rectangle of 20 by 30 between them.
draw an arbitrary water level, with r cm the value within each of the triangles, (length of water level = 20+2r)
let the water level be h cm
by ratio: r/h = 10/30 ---> r = h/3
V = 800(2 triangles + rectangle)
= 800(2(1/2)rh + 20h)
= 800((1/3)h^2 + 20h)
dV/dt = 800((2/3)h dh/dt + 20dh/dt)
1000 = 800(dh/dt)(2h/3 + 20)
1.25/(2h/3 + 20) = dh/dt
dh/dt = 3.75/(2h+60) ----- I multiplied top and bottom by 3
In most of these cases they would ask for the rate for some specific height given.
Are you sure there was not a height given?
If there was, just plug that in for h in the last part.
Notice I changed everything to cm
8 m = 800 cm and
since 1 dm = 10 cm
(1dm)^3 = (10 cm)^3 = 1000 cm^3
yea i was wondering that too but there isn't any height given so i think i will leave the answer with the height !thanks <3
Reiny the answer is suppose to be 65/(120+4h) . I dunno i m not geting itt
I mean 75 not 65
To find the rate of rise of the water level in the trough, we need to consider the volume of the trough and the rate at which it is being filled.
First, let's calculate the average width of the trough. We have the top width (40 cm) and the bottom width (20 cm), so we can calculate the average width using the formula:
Average Width = (Top Width + Bottom Width) / 2 = (40 cm + 20 cm) / 2 = 30 cm
Next, we can calculate the cross-sectional area of the trough using the average width and the depth:
Cross-sectional Area = Average Width x Depth = 30 cm x 30 cm = 900 cm²
The trough is 8 m long, which is equivalent to 800 cm. So the volume of the trough is given by:
Volume of Trough = Cross-sectional Area x Length = 900 cm² x 800 cm = 720,000 cm³
Now, we need to calculate the time it takes to fill the trough. We are given that the trough is filled at a rate of 10 dm³/min. This is equivalent to 10,000 cm³/min, since 1 dm³ = 1,000 cm³.
Finally, to find the rate of rise of the water level, we can divide the rate at which the trough is being filled by the cross-sectional area:
Rate of Rise of Water Level = Rate of Filling / Cross-sectional Area = 10,000 cm³/min / 900 cm² = 11.11 cm/min (rounded to two decimal places)
So, the rate of rise of the water level in the trough is approximately 11.11 cm/min.