anonymous
  • anonymous
university 1st year math question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
welshfella
  • welshfella
i looked up torricelli's law and only found a formula for a cylindrical tank..
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@IrishBoy123

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anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
i am finding it very difficult
ganeshie8
  • ganeshie8
@ParthKohli try this
ParthKohli
  • ParthKohli
are you mathmath333?
anonymous
  • anonymous
what ?
ganeshie8
  • ganeshie8
No, he is not
ParthKohli
  • ParthKohli
I haven't studied fluids yet, but I looked at the Wikipedia page and it sorta makes sense.
ParthKohli
  • ParthKohli
This is what we're really looking at though. https://en.wikipedia.org/wiki/Torricelli%27s_law#Application_for_time_to_empty_the_container
ganeshie8
  • ganeshie8
same here, it seems torcelli's law gives the speed of fluid of height \(h\), leaving tank from bottom : \[v = \sqrt{2gh}\]
anonymous
  • anonymous
noo \[v=\sqrt{2gh}\]
ytrewqmiswi
  • ytrewqmiswi
What does that k represents?
anonymous
  • anonymous
i was thinking of an approach where i find A(h) and use integration
ParthKohli
  • ParthKohli
Is that the area of cross-section as a function of height? And are you sure this is the equation?\[A(h) \frac{dh}{dt} = -k \sqrt{h}\]
anonymous
  • anonymous
yes 100% because we havent used other v=sqrt2gh yet and my teacher wouldnt expect us to use that
anonymous
  • anonymous
the only problem i am having is how to find A(h)
ytrewqmiswi
  • ytrewqmiswi
That wuld mean that the area of cross section varies with height will that make any sense?
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
If my eyes work, we're looking at a rectangular cross-section, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and yes i wan to find the area as a function of height plz
ParthKohli
  • ParthKohli
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...
anonymous
  • anonymous
\[\frac{ 10h }{ 8 }\]
ParthKohli
  • ParthKohli
noooo I didn't write h/8 h/4+4 means (1/4)*h + 4
anonymous
  • anonymous
\[\frac{ 10(h+16) }{ 4 }\]
ganeshie8
  • ganeshie8
40----> 50 A(y) = 40 + (y/4)*10 does that work ?
ParthKohli
  • ParthKohli
looks good
ParthKohli
  • ParthKohli
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
ganeshie8
  • ganeshie8
yeah 10cm is fixed, only the other dimension is changing, so it is linear
anonymous
  • anonymous
just a question srry interpreting ur thought process how did u get width as (1/4) times h +4
ganeshie8
  • ganeshie8
\[A(h)\frac{ dh }{ dt }=-k \sqrt{h}\] plugin \(A(h)\), the de becomes \[(40+(\frac{h}{4})*10)\frac{ dh }{ dt }=-k \sqrt{h}\]
ytrewqmiswi
  • ytrewqmiswi
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<
ganeshie8
  • ganeshie8
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 ?
ParthKohli
  • ParthKohli
|dw:1443179501695:dw|
ytrewqmiswi
  • ytrewqmiswi
well...that thing can be used to find time tho..?
anonymous
  • anonymous
@ytrewqmiswi we cant use v=sqrt2gh because well i want the area as a function of h
ytrewqmiswi
  • ytrewqmiswi
A = cross sectional area of tank H=height till water is filled a=cross sectional area of hole
ParthKohli
  • ParthKohli
but didn't you assume A to be constant?
ytrewqmiswi
  • ytrewqmiswi
yes i took it constant
ParthKohli
  • ParthKohli
yes that solution matches ours so far
anonymous
  • anonymous
good because i am not sure about my final answer it doesnt make sense so i need to be assured
ParthKohli
  • ParthKohli
you shouldn't have any problem with integration
anonymous
  • anonymous
noo
anonymous
  • anonymous
but would love it if you solve it step by step plzzz
anonymous
  • anonymous
it will make my life so easy
ganeshie8
  • ganeshie8
you don't have teamviwer ?
anonymous
  • anonymous
again what is teamviewer?
anonymous
  • anonymous
do i have to buy it ?
ganeshie8
  • ganeshie8
no, it is a free screen sharing app
anonymous
  • anonymous
for the solution u an see i am stuck in the end
anonymous
  • anonymous
ahm do i have to get it for this question
ParthKohli
  • ParthKohli
\[A(h) \frac{dh}{dt} = -k\sqrt{h}\]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.
anonymous
  • anonymous
ok
ParthKohli
  • ParthKohli
\[\frac{A(h)}{\sqrt h} dh=-kdt\]
anonymous
  • anonymous
\[\frac{ A(h) }{ \sqrt{h} } dh=-k dt\]
ParthKohli
  • ParthKohli
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.
anonymous
  • anonymous
how?
ParthKohli
  • ParthKohli
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.
anonymous
  • anonymous
which A(h) value must i put?
ParthKohli
  • ParthKohli
The one we found out earlier?
ParthKohli
  • ParthKohli
We found the value of area of the cross-section in terms of height earlier, yes?
anonymous
  • anonymous
\[\frac{ 10(h+16) }{ 4 }\ /\sqrt{h} =-kdt\]
ParthKohli
  • ParthKohli
Good. I know you're having trouble with typing everything out. So don't. Just keep doing it on paper alright?
anonymous
  • anonymous
no i need to show u , i can be writting jubrish
ganeshie8
  • ganeshie8
//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.
anonymous
  • anonymous
\[\frac{ h+16 }{ \sqrt{h}} dh = - \frac{ 2 }{ 10 } \times 0.03 dt\]
anonymous
  • anonymous
@ganeshie8 i am on computer , will that work on this and how do i add uz
ganeshie8
  • ganeshie8
Hey it is really not required, let try and finish it off here
anonymous
  • anonymous
ok
ganeshie8
  • ganeshie8
have you deleted the attachment ?
anonymous
  • anonymous
noo
ganeshie8
  • ganeshie8
i couldn't find the solution above
ganeshie8
  • ganeshie8
your solution..
anonymous
  • anonymous
ok u see i did the A(h) wrong then towards the end i thought when t=0 , h=3
anonymous
  • anonymous
my solution is wrong
anonymous
  • anonymous
i did A(h) wrong
ganeshie8
  • ganeshie8
your A(h) looks fine to me
anonymous
  • anonymous
and i dont seem to get what that refers to as 3 quarter filled
anonymous
  • anonymous
so plz ignore the solution
ganeshie8
  • ganeshie8
dude you're good, don't you see what you have is same as `A(h) = 40 + (h/4)*10`
anonymous
  • anonymous
ok so my solution makes sense all of it ok what about part c
anonymous
  • anonymous
i just gave up on that
anonymous
  • anonymous
i mean iii
anonymous
  • anonymous
and also i dunno how to go furthur from here
ganeshie8
  • ganeshie8
so whats you final answer for part i
anonymous
  • anonymous
i dunno what is h and t if i dunno t how am i suppose to find h
anonymous
  • anonymous
what about h
ganeshie8
  • ganeshie8
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 \(t\)
anonymous
  • anonymous
is this what u saying?
ganeshie8
  • ganeshie8
I am asking you to simply replace \(h\) by \(0\) because, empty tank means that the depth of water is 0
anonymous
  • anonymous
ohhh
ganeshie8
  • ganeshie8
\[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\] \[0=-0.012t+58.89\] solve \(t\) and you're done!
anonymous
  • anonymous
t=4907.5
anonymous
  • anonymous
just one question did uz consider the three quarter filled thing
ganeshie8
  • ganeshie8
Oops! we missed that, my mistake, sorry, lets start over..
anonymous
  • anonymous
ok so part i is finding an expression we have done it
anonymous
  • anonymous
ur being sarcastic right
ganeshie8
  • ganeshie8
I'm never sarcastic
ganeshie8
  • ganeshie8
we need to start over, our work for all parts is wrong because we assumed that the tank was full
anonymous
  • anonymous
ok
ganeshie8
  • ganeshie8
gimme some time, il type in the complete solution
anonymous
  • anonymous
ok
Jhannybean
  • Jhannybean
nice comforter I mean. Lol.
ganeshie8
  • ganeshie8
Hey, it turns out our previous solution works out perfectly fine because we have worked it from bottom.
ganeshie8
  • ganeshie8
No need to redo anything. we are good.
ganeshie8
  • ganeshie8
@ParthKohli please double check if psble
anonymous
  • anonymous
srry last thing what about part b
anonymous
  • anonymous
as u said make h=0 and solve for T
anonymous
  • anonymous
thnx guys cheers i got it
ganeshie8
  • ganeshie8
try http://www.wolframalpha.com/input/?i=plot+%5Cfrac%7B+2+%7D%7B+3+%7D%5Csqrt%7Bh%7D+%28h%2B48%29%3D-0.012t%2B58.89%2C+0%3Ct%3C5000
ganeshie8
  • ganeshie8
deleted, let me know if you want me delete any other stuff
anonymous
  • anonymous
no its ok thak uso much
ganeshie8
  • ganeshie8
np
anonymous
  • anonymous
how can i manupilate above equation so the subject is h because thats what it is ultimately asking
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@pooja195
anonymous
  • anonymous
but the question is asking me to make h the subject
dan815
  • dan815
ok
dan815
  • dan815
it says you can use wolfrad alpha so use that
anonymous
  • anonymous
noo i cant because its for graph only h vs t
dan815
  • dan815
yeah?
anonymous
  • anonymous
when u read the question b (i) last line determine the depth of the milk at any instant after the drain has opened
anonymous
  • anonymous
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
anonymous
  • anonymous
did u understand i am after i want an expression where h = .....
dan815
  • dan815
\[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\]
dan815
  • dan815
you just specify in wolfram plot h vs t
anonymous
  • anonymous
look i am not talking about ii use wolfram alpha to obtain plot h verus t
dan815
  • dan815
what do u want then
anonymous
  • anonymous
i am talking about question above (i)
dan815
  • dan815
oh i thought u are done that
anonymous
  • anonymous
nooo
dan815
  • dan815
do u want me to solve b(i)
anonymous
  • anonymous
how do i make h the subject
anonymous
  • anonymous
in this expression
anonymous
  • anonymous
how do i get rid of sqrt h
anonymous
  • anonymous
so i have an expression like h =..... when i manipulate above eq
dan815
  • dan815
it's not going to be pretty
dan815
  • dan815
yes.. this 0.03 for k means nothing to me
anonymous
  • anonymous
k is just viscosity its a constant
dan815
  • dan815
doesnt help
anonymous
  • anonymous
see the solution u will realise
dan815
  • dan815
k im assuming is a function of the radius of the opening
dan815
  • dan815
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
anonymous
  • anonymous
yea can u see any way where h is the subject
dan815
  • dan815
what do u mean h is the subject
anonymous
  • anonymous
ok h is the depth
anonymous
  • anonymous
at 4 sec
anonymous
  • anonymous
h= depth in equation and t = time
dan815
  • dan815
what you have is good enough
dan815
  • dan815
i dont see where it says to make h the subject in your question
anonymous
  • anonymous
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
anonymous
  • anonymous
substituing 10*
dan815
  • dan815
you gotta use nonlinear methods lol
anonymous
  • anonymous
how would i use non linear method
dan815
  • dan815
just use matlab or something for now
anonymous
  • anonymous
ok thnx for ur help btw before u go mind deleting this whole question plz
dan815
  • dan815
i have no powers
dan815
  • dan815
you know as far as applications and solutions go solving for t=f(h) is good enough
anonymous
  • anonymous
yea i hope so
dan815
  • dan815
as you can graph that, and rotating it returns h=f(t)
anonymous
  • anonymous
yea true
dan815
  • dan815
if you really must solve for h i can show u a way
anonymous
  • anonymous
can u plz delete the expression above
dan815
  • dan815
it's not pretty yo uwill need to use complex numbers
dan815
  • dan815
and it might be too early for you right now in first year
anonymous
  • anonymous
no dont i have solved its not pretttyyy
dan815
  • dan815
okay

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