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anonymous

  • one year ago

university 1st year math question.

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  1. welshfella
    • one year ago
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    i looked up torricelli's law and only found a formula for a cylindrical tank..

  2. anonymous
    • one year ago
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    @ganeshie8

  3. anonymous
    • one year ago
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    @IrishBoy123

  4. anonymous
    • one year ago
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    @ganeshie8

  5. anonymous
    • one year ago
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    i am finding it very difficult

  6. ganeshie8
    • one year ago
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    @ParthKohli try this

  7. ParthKohli
    • one year ago
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    are you mathmath333?

  8. anonymous
    • one year ago
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    what ?

  9. ganeshie8
    • one year ago
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    No, he is not

  10. ParthKohli
    • one year ago
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    I haven't studied fluids yet, but I looked at the Wikipedia page and it sorta makes sense.

  11. ParthKohli
    • one year ago
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    This is what we're really looking at though. https://en.wikipedia.org/wiki/Torricelli%27s_law#Application_for_time_to_empty_the_container

  12. ganeshie8
    • one year ago
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    same here, it seems torcelli's law gives the speed of fluid of height \(h\), leaving tank from bottom : \[v = \sqrt{2gh}\]

  13. anonymous
    • one year ago
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    noo \[v=\sqrt{2gh}\]

  14. ytrewqmiswi
    • one year ago
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    What does that k represents?

  15. anonymous
    • one year ago
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    i was thinking of an approach where i find A(h) and use integration

  16. ParthKohli
    • one year ago
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    Is that the area of cross-section as a function of height? And are you sure this is the equation?\[A(h) \frac{dh}{dt} = -k \sqrt{h}\]

  17. anonymous
    • one year ago
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    yes 100% because we havent used other v=sqrt2gh yet and my teacher wouldnt expect us to use that

  18. anonymous
    • one year ago
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    the only problem i am having is how to find A(h)

  19. ytrewqmiswi
    • one year ago
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    That wuld mean that the area of cross section varies with height will that make any sense?

  20. ParthKohli
    • one year ago
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    But in your equation, \(\sqrt{2g}\) or \(g\) won't appear. We can find the area of the cross-section as a function of height.

  21. ParthKohli
    • one year ago
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    If my eyes work, we're looking at a rectangular cross-section, right?

  22. anonymous
    • one year ago
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    yes

  23. anonymous
    • one year ago
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    and yes i wan to find the area as a function of height plz

  24. ParthKohli
    • one year ago
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    Width increases in a linear manner from 4 to 5. At h = 0, it is 4 and at h = 4, it is 5. Thus in general it should be \(w(h) = h/4 + 4\). Meanwhile the length remains 10 so \(l(h) = 10\). Multiply...

  25. anonymous
    • one year ago
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    \[\frac{ 10h }{ 8 }\]

  26. ParthKohli
    • one year ago
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    noooo I didn't write h/8 h/4+4 means (1/4)*h + 4

  27. anonymous
    • one year ago
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    \[\frac{ 10(h+16) }{ 4 }\]

  28. ganeshie8
    • one year ago
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    40----> 50 A(y) = 40 + (y/4)*10 does that work ?

  29. ParthKohli
    • one year ago
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    looks good

  30. ParthKohli
    • one year ago
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    it only works because one of them remains constant can't do the same thing when both width and length increase which is why it's a good practice to do them separately

  31. ganeshie8
    • one year ago
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    yeah 10cm is fixed, only the other dimension is changing, so it is linear

  32. anonymous
    • one year ago
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    just a question srry interpreting ur thought process how did u get width as (1/4) times h +4

  33. ganeshie8
    • one year ago
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    \[A(h)\frac{ dh }{ dt }=-k \sqrt{h}\] plugin \(A(h)\), the de becomes \[(40+(\frac{h}{4})*10)\frac{ dh }{ dt }=-k \sqrt{h}\]

  34. ytrewqmiswi
    • one year ago
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    the liquid escapes with a velocity =sqrt(2gh) which varies with the level of liquid. Hence, we have to use integral. Let y be the height of the liquid at an instant. This height changes by dy in time dt. Volume of water leaving out per secound=-Ady.dt at the hole volume escaping per second is av=a[sqrt(2gy)] a[sqrt(2gy)] =-Ady/dt ---->\[\int\limits_{H}^{0}\frac{ -dy }{ \sqrt{y} } =\frac{ a \sqrt{2g} }{ A }\int\limits_{t}^{0}dt\] \[t=\frac{ A }{ a \sqrt{2g}}2\sqrt{H}\] a<<A? what did i do o.o

  35. ganeshie8
    • one year ago
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    maybe think of it in reverse : A(h) = 40 + (h/4)*10 plugin h=0, what do you get ? plugin h=4, what do you get ?

  36. ParthKohli
    • one year ago
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    |dw:1443179501695:dw|

  37. ytrewqmiswi
    • one year ago
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    well...that thing can be used to find time tho..?

  38. anonymous
    • one year ago
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    @ytrewqmiswi we cant use v=sqrt2gh because well i want the area as a function of h

  39. ytrewqmiswi
    • one year ago
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    A = cross sectional area of tank H=height till water is filled a=cross sectional area of hole

  40. ParthKohli
    • one year ago
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    but didn't you assume A to be constant?

  41. ytrewqmiswi
    • one year ago
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    yes i took it constant

  42. ParthKohli
    • one year ago
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    yes that solution matches ours so far

  43. anonymous
    • one year ago
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    good because i am not sure about my final answer it doesnt make sense so i need to be assured

  44. ParthKohli
    • one year ago
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    you shouldn't have any problem with integration

  45. anonymous
    • one year ago
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    noo

  46. anonymous
    • one year ago
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    but would love it if you solve it step by step plzzz

  47. anonymous
    • one year ago
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    it will make my life so easy

  48. ganeshie8
    • one year ago
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    you don't have teamviwer ?

  49. anonymous
    • one year ago
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    again what is teamviewer?

  50. anonymous
    • one year ago
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    do i have to buy it ?

  51. ganeshie8
    • one year ago
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    no, it is a free screen sharing app

  52. anonymous
    • one year ago
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    for the solution u an see i am stuck in the end

  53. anonymous
    • one year ago
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    ahm do i have to get it for this question

  54. ParthKohli
    • one year ago
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    \[A(h) \frac{dh}{dt} = -k\sqrt{h}\]We've got all messy differentials here. What we do is we take all the \(h\)-terms to to one side and \(t\)-terms to the other. Then integrate both sides with proper limits.

  55. anonymous
    • one year ago
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    ok

  56. ParthKohli
    • one year ago
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    \[\frac{A(h)}{\sqrt h} dh=-kdt\]

  57. anonymous
    • one year ago
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    \[\frac{ A(h) }{ \sqrt{h} } dh=-k dt\]

  58. ParthKohli
    • one year ago
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    Great. Now when \(t= 0\), \(h=4\). When \(t = t_0\), \(h = 0\). So you integrate both sides with those limits corresponding to each other.

  59. anonymous
    • one year ago
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    how?

  60. ParthKohli
    • one year ago
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    OK, first put the value of \(A(h) \) in the equation and simplify. You don't have to type it out here. Keep doing that on paper. I'll guide you through each step.

  61. anonymous
    • one year ago
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    which A(h) value must i put?

  62. ParthKohli
    • one year ago
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    The one we found out earlier?

  63. ParthKohli
    • one year ago
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    We found the value of area of the cross-section in terms of height earlier, yes?

  64. anonymous
    • one year ago
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    \[\frac{ 10(h+16) }{ 4 }\ /\sqrt{h} =-kdt\]

  65. ParthKohli
    • one year ago
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    Good. I know you're having trouble with typing everything out. So don't. Just keep doing it on paper alright?

  66. anonymous
    • one year ago
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    no i need to show u , i can be writting jubrish

  67. ganeshie8
    • one year ago
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    //ahm do i have to get it for this question Absolutely not, but considering the fact that you already have a detailed solution which you don't understand it, the jibber jabber that I do here only confuses you more. If you have teamviewer, we could use voice also and engage more efficiently.

  68. anonymous
    • one year ago
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    \[\frac{ h+16 }{ \sqrt{h}} dh = - \frac{ 2 }{ 10 } \times 0.03 dt\]

  69. anonymous
    • one year ago
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    @ganeshie8 i am on computer , will that work on this and how do i add uz

  70. ganeshie8
    • one year ago
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    Hey it is really not required, let try and finish it off here

  71. anonymous
    • one year ago
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    ok

  72. ganeshie8
    • one year ago
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    have you deleted the attachment ?

  73. anonymous
    • one year ago
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    noo

  74. ganeshie8
    • one year ago
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    i couldn't find the solution above

  75. ganeshie8
    • one year ago
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    your solution..

  76. anonymous
    • one year ago
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    ok u see i did the A(h) wrong then towards the end i thought when t=0 , h=3

  77. anonymous
    • one year ago
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    my solution is wrong

  78. anonymous
    • one year ago
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    i did A(h) wrong

  79. ganeshie8
    • one year ago
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    your A(h) looks fine to me

  80. anonymous
    • one year ago
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    and i dont seem to get what that refers to as 3 quarter filled

  81. anonymous
    • one year ago
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    so plz ignore the solution

  82. ganeshie8
    • one year ago
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    dude you're good, don't you see what you have is same as `A(h) = 40 + (h/4)*10`

  83. anonymous
    • one year ago
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    ok so my solution makes sense all of it ok what about part c

  84. anonymous
    • one year ago
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    i just gave up on that

  85. anonymous
    • one year ago
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    i mean iii

  86. anonymous
    • one year ago
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    and also i dunno how to go furthur from here

  87. ganeshie8
    • one year ago
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    so whats you final answer for part i

  88. anonymous
    • one year ago
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    i dunno what is h and t if i dunno t how am i suppose to find h

  89. anonymous
    • one year ago
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    what about h

  90. ganeshie8
    • one year ago
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    Notice that the depth of water, h, is 0 when the tank is empty. so, for part iii, i think you simply plugin h=0 and solve \(t\)

  91. anonymous
    • one year ago
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    is this what u saying?

  92. ganeshie8
    • one year ago
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    I am asking you to simply replace \(h\) by \(0\) because, empty tank means that the depth of water is 0

  93. anonymous
    • one year ago
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    ohhh

  94. ganeshie8
    • one year ago
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    \[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\] \[0=-0.012t+58.89\] solve \(t\) and you're done!

  95. anonymous
    • one year ago
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    t=4907.5

  96. anonymous
    • one year ago
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    just one question did uz consider the three quarter filled thing

  97. ganeshie8
    • one year ago
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    Oops! we missed that, my mistake, sorry, lets start over..

  98. anonymous
    • one year ago
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    ok so part i is finding an expression we have done it

  99. anonymous
    • one year ago
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    ur being sarcastic right

  100. ganeshie8
    • one year ago
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    I'm never sarcastic

  101. ganeshie8
    • one year ago
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    we need to start over, our work for all parts is wrong because we assumed that the tank was full

  102. anonymous
    • one year ago
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    ok

  103. ganeshie8
    • one year ago
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    gimme some time, il type in the complete solution

  104. anonymous
    • one year ago
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    ok

  105. Jhannybean
    • one year ago
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    nice comforter I mean. Lol.

  106. ganeshie8
    • one year ago
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    Hey, it turns out our previous solution works out perfectly fine because we have worked it from bottom.

  107. ganeshie8
    • one year ago
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    No need to redo anything. we are good.

  108. ganeshie8
    • one year ago
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    @ParthKohli please double check if psble

  109. anonymous
    • one year ago
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    srry last thing what about part b

  110. anonymous
    • one year ago
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    as u said make h=0 and solve for T

  111. anonymous
    • one year ago
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    thnx guys cheers i got it

  112. ganeshie8
    • one year ago
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    deleted, let me know if you want me delete any other stuff

  113. anonymous
    • one year ago
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    no its ok thak uso much

  114. ganeshie8
    • one year ago
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    np

  115. anonymous
    • one year ago
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    how can i manupilate above equation so the subject is h because thats what it is ultimately asking

  116. anonymous
    • one year ago
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    @dan815

  117. anonymous
    • one year ago
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    @pooja195

  118. anonymous
    • one year ago
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    but the question is asking me to make h the subject

  119. dan815
    • one year ago
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    ok

  120. dan815
    • one year ago
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    it says you can use wolfrad alpha so use that

  121. anonymous
    • one year ago
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    noo i cant because its for graph only h vs t

  122. dan815
    • one year ago
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    yeah?

  123. anonymous
    • one year ago
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    when u read the question b (i) last line determine the depth of the milk at any instant after the drain has opened

  124. anonymous
    • one year ago
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    so we need to find expression where h is the subject and i did use wolfram alpha it didnt work there were always two h

  125. anonymous
    • one year ago
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    did u understand i am after i want an expression where h = .....

  126. dan815
    • one year ago
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    \[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\]

  127. dan815
    • one year ago
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    you just specify in wolfram plot h vs t

  128. anonymous
    • one year ago
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    look i am not talking about ii use wolfram alpha to obtain plot h verus t

  129. dan815
    • one year ago
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    what do u want then

  130. anonymous
    • one year ago
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    i am talking about question above (i)

  131. dan815
    • one year ago
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    oh i thought u are done that

  132. anonymous
    • one year ago
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    nooo

  133. dan815
    • one year ago
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    do u want me to solve b(i)

  134. anonymous
    • one year ago
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    how do i make h the subject

  135. anonymous
    • one year ago
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    in this expression

  136. anonymous
    • one year ago
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    how do i get rid of sqrt h

  137. anonymous
    • one year ago
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    so i have an expression like h =..... when i manipulate above eq

  138. dan815
    • one year ago
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    it's not going to be pretty

  139. dan815
    • one year ago
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    yes.. this 0.03 for k means nothing to me

  140. anonymous
    • one year ago
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    k is just viscosity its a constant

  141. dan815
    • one year ago
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    doesnt help

  142. anonymous
    • one year ago
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    see the solution u will realise

  143. dan815
    • one year ago
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    k im assuming is a function of the radius of the opening

  144. dan815
    • one year ago
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    yep see that equation u are using there is an application of the bernoullis principle, where that k in there would be some factor with the radius of the opening at the bottom accounted in for it

  145. anonymous
    • one year ago
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    yea can u see any way where h is the subject

  146. dan815
    • one year ago
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    what do u mean h is the subject

  147. anonymous
    • one year ago
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    ok h is the depth

  148. anonymous
    • one year ago
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    at 4 sec

  149. anonymous
    • one year ago
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    h= depth in equation and t = time

  150. dan815
    • one year ago
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    what you have is good enough

  151. dan815
    • one year ago
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    i dont see where it says to make h the subject in your question

  152. anonymous
    • one year ago
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    ok at 10 sec what would be the depth of tank can u give me answer for that by substituting 4 into the final eq

  153. anonymous
    • one year ago
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    substituing 10*

  154. dan815
    • one year ago
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    you gotta use nonlinear methods lol

  155. anonymous
    • one year ago
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    how would i use non linear method

  156. dan815
    • one year ago
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    just use matlab or something for now

  157. anonymous
    • one year ago
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    ok thnx for ur help btw before u go mind deleting this whole question plz

  158. dan815
    • one year ago
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    i have no powers

  159. dan815
    • one year ago
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    you know as far as applications and solutions go solving for t=f(h) is good enough

  160. anonymous
    • one year ago
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    yea i hope so

  161. dan815
    • one year ago
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    as you can graph that, and rotating it returns h=f(t)

  162. anonymous
    • one year ago
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    yea true

  163. dan815
    • one year ago
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    if you really must solve for h i can show u a way

  164. anonymous
    • one year ago
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    can u plz delete the expression above

  165. dan815
    • one year ago
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    it's not pretty yo uwill need to use complex numbers

  166. dan815
    • one year ago
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    and it might be too early for you right now in first year

  167. anonymous
    • one year ago
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    no dont i have solved its not pretttyyy

  168. dan815
    • one year ago
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    okay

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spraguer (Moderator)
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