Question

university 1st year math question.

Answer 1

i looked up torricelli's law and only found a formula for a cylindrical tank..

Answer 2

@ganeshie8

Answer 3

@IrishBoy123

Answer 4

@ganeshie8

Answer 5

i am finding it very difficult

Answer 6

@ParthKohli try this

Answer 7

are you mathmath333?

Answer 8

what ?

Answer 9

No, he is not

Answer 10

I haven't studied fluids yet, but I looked at the Wikipedia page and it sorta makes sense.

Answer 11

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

Answer 12

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

Answer 13

noo \[v=\sqrt{2gh}\]

Answer 14

What does that k represents?

Answer 15

i was thinking of an approach where i find A(h) and use integration

Answer 16

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

Answer 17

yes 100% because we havent used other v=sqrt2gh yet and my teacher wouldnt expect us to use that

Answer 18

the only problem i am having is how to find A(h)

Answer 19

That wuld mean that the area of cross section varies with height will that make any sense?

Answer 20

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.

Answer 21

If my eyes work, we're looking at a rectangular cross-section, right?

Answer 22

yes

Answer 23

and yes i wan to find the area as a function of height plz

Answer 24

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...

Answer 25

\[\frac{ 10h }{ 8 }\]

Answer 26

noooo I didn't write h/8
h/4+4 means (1/4)*h + 4

Answer 27

\[\frac{ 10(h+16) }{ 4 }\]

Answer 28

40----> 50

A(y) = 40 + (y/4)*10

does that work ?

Answer 29

looks good

Answer 30

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

Answer 31

yeah 10cm is fixed, only the other dimension is changing, so it is linear

Answer 32

just a question srry interpreting ur thought process how did u get width as (1/4) times h +4

Answer 33

\[A(h)\frac{ dh }{ dt }=-k \sqrt{h}\]

plugin \(A(h)\), the de becomes

\[(40+(\frac{h}{4})*10)\frac{ dh }{ dt }=-k \sqrt{h}\]

Answer 34

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

Answer 35

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 ?