1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

HELP me please

  1. Nov 29, 2004 #1
    HELP me please!!!

    Hi, I am having trouble figuring out how to tackle this problem, it would be great if anyone can help! :eek:

    Here is the problem...
    The three containers shown below (sorry, i can only describe them and not draw them here, i'll describe em at the bottom) are all initially empty. Water is simultaneously poured into the containers at the constant rate of 20 cm^3/ sec. Water leaks out of a hole in the bottom of the second container at a constant rate. at some point in time, the water levels in the three containers are all rising at the same time. At what rate is the water leaking out of the second container?

    Container 1: A cylinder with radius of 10cm.
    Container 2: Cone shape with 60 degrees point
    Container 3: Another cone with a 90 degrees point.

    It would be awesome if i could get any help!! THANKS!!
     
  2. jcsd
  3. Nov 29, 2004 #2

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hint : Write V = f(h), (where V : volume, h : height) for all three containers. Find the time at which the levels rise at the same rate by comparing 1 and 3. From this determine tha leak rate of 2.
     
  4. Nov 29, 2004 #3
    can you give me another hint?
     
  5. Nov 30, 2004 #4
    Ok, do i set this up by using the volume equation, f(V)= 1/3 *pi *r^2* h......

    this is also given in the problem..

    [tex] \frac {dV_3} {dt} = 20, V_3 = 20t, and since \frac {dV_2}{dt} = r(in) - r(out) = 20 - r (out), V_2 = (20-r (out))t [/tex]

    sorry i didnt include this before...but yeah, another push in the right direction would be great, thanks.
     
  6. Dec 4, 2004 #5
    can anyone help me on this problem?
     
  7. Dec 6, 2004 #6

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [tex]V_1 = \pi R^2 h_1 => \frac{dV_1}{dt} = \pi R^2 \frac{dh_1}{dt} = 20 [/tex]

    [tex]=> \frac{dh_1}{dt} = \frac {20}{100 \pi} = \frac{1}{5 \pi} [/tex]

    [tex]V_3 = \frac{1}{3} \pi r_3^2 h_3 [/tex]

    [tex]But~\frac{r_3}{h_3} = tan 45 = \frac{1}{\sqrt{2}} => V_3 = \frac{1}{6} \pi h_3^3 [/tex]

    [tex]So~ \frac{dV_3}{dt} = \frac {\pi}{2} h_3^2 \frac {dh_3}{dt} = 20 [/tex]

    [tex]=> \frac{dh_3}{dt} = \frac {40}{\pi h_3^2} [/tex]

    [tex]But~at~some~t,~ \frac{dh_3}{dt} = \frac{dh_1}{dt} = \frac{1}{5 \pi} [/tex]

    From this, you can get h3, and from that, V3. Dividing V3 by 20 gives you the time in seconds. Can you take it from there ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: HELP me please
  1. Please Help me (Replies: 1)

Loading...