An upside down cone with diameter 10m and height 8m is being filled with water, 0.1 m^3 per minute.
What's the speed the surface raising when the water height in the cone is 4 meter
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An upside down cone with diameter 10m and height 8m is being filled with water, 0.1 m^3 per minute.
What's the speed the surface raising when the water height in the cone is 4 meter
I make it (48/(250*pi)) * ( ( 192 / (25*pi) )^(-2/3) ) /60
= 5.61296766986877E-04 m/s
about half a centimetre per second
Hey, I got 2.12E-5 m/s
That's 21 micrometers per second, about 1,2 millimeters/s
Something must have screwed up, either you or me or both got wrong answers, so
did you get this equation for height?
h=2t/(625pi)^(2/3)
and then got the derivate and put in t in it?
I got t=20000pi seconds for h=4
838 m^3 should go in the cone, and if you fill it 0,1m^3 /minute, should take up nearly six days to fill
my bad, I missed out a couple of constants
I got 2.97001026127106E-04 m/s this time, I've done it a completley different way though. I worked out the height in terms of the volume first then found the derivitive and divided by 600
Hey guys, we will never decide what is correct answer without showing how it was derived.
Check the following.
V = Pi*r^2*h/3 (Cone volume in terms of height & radius).
For cone in question, r = 5*h/8
Hence V = Pi*(5*h/8)^2*h/3
or V = 25*Pi*h^3/192
dV = (25*Pi*h^2/64)*dh
dV = (25*Pi*16/64)*dh
dV = (25*Pi/4)*dh
.1 = (19.635)*dh
dh = .1/19.635
dh = .00509 meters/second
I did not recheck all of the above. If you disagree, show me my error. There might be one.
I'm sorry there was a typo in the qwestion, the radius should be 10m not the diameter. But that won't affect the methods to solve it, thanks for your input Guv :)
the speed it takes is about the speed it takes arbiter to down a Pint hehehe. sorry just needed to put some humor into this post. back to the assylum I go.