- #1

Danneskjöld

- 4

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of this subject. It is the first semester of Physics, but the end, and has become quite

complicated.

## Homework Statement

There is a 2.0 L soda bottle used, with a hole a certain height (h

_{hole}) from

the bottom (and a small mini-hose to assist in the motion of the exiting of the fluid from the

bottle). The bottle is aligned with the opening of a sink, and a hose is attached to the top

so as to fill the bottle to a certain height and remain at that height constantly (h

_{1}).

Then the distance the water travels out the hole is measured and recorded as a value

of x

_{1}.

Subsequently, this is done one more time with the water level at a new higher height (h

_{2}) and a new value of x

_{2}.

Okay, hopefully that makes sense.

Next, we are given two equations that are used to derive some other new equations

meant to discover the velocity (I'm uncertain if I should include the messy method of

proving these equations weren't pulled out of my ear, or just cut to the chase).

THEN... we are to solve for velocity.

Then we are asked to comment on our results, giving probable reasons for difference

in values obtained from equations 1 & 2.

## Homework Equations

Now for the equations.

__Equation 1__:

__Point 1_______

__Point 2_______

__Point 3__

___P

_{0}___= P

_{0}+ [tex]\rho[/tex]gh

_{1}= P

_{0}+([tex]\rho[/tex]v

_{H2O}

^{2})/2

OR

__[tex]\rho[/tex]v__= [tex]\rho[/tex]gh

_{H2O}^{2}_{1}; v

_{H2O}= [tex]\sqrt{(2gh[sub1])}[/tex]

___2

Equation 1: v

_{H2O}= [tex]\sqrt{(2gh[sub1])}[/tex]

__Equation 2__:

h

_{2}= ½ gt

^{2}

**OR**t = [tex]\sqrt{[2h(sub2)]/g}[/tex] ; X = v

_{H2O}t ;

**OR**t = x/v

_{H2O}= [tex]\sqrt{(2gh[sub1])}[/tex]

THEN

v

_{H2O}= X*[tex]\sqrt{g/(2h(sub2))}[/tex] ; v

_{H2O}= 2.21 * X/[tex]\sqrt{h(sub2)}[/tex]

Equation 2: v

_{H2O}= 2.21 * X/[tex]\sqrt{h(sub2)}[/tex]

## The Attempt at a Solution

h

_{hole}= h

_{2}= 6.9 cm

h

_{1}= 8.4 cm

x

_{1}= 8.6 cm

v

_{1H2O}=________?

h

_{hole}= h

_{2}= 6.9 cm

h

_{1}= 11.1 cm

x

_{2}= 9.8 cm

v

_{2H2O}= _______?

__Using Equation 1__:

v

_{1H2O}= [tex]\sqrt{2gh(sub1)}[/tex] = [tex]\sqrt{(2)(9.8)(0.084m)]}[/tex] = 1.6464 m/s

v

_{2H2O}= [tex]\sqrt{2gh(sub2)}[/tex] = [tex]\sqrt{(2)(9.8)(0.11m)]}[/tex] = 2.156 m/s

__Using Equation 2__:

v

_{1H2O}= (2.21) (X/[tex]\sqrt{h(sub2)}[/tex] = [(2.21)(0.086m)]/[tex]\sqrt{0.084m}[/tex] = 0.655769301 m/s

v

_{2H2O}= (2.21) (X/[tex]\sqrt{h(sub2)}[/tex] = [(2.21)(0.098m)/[tex]\sqrt{0.11m}[/tex] = 0.65302 m/s

So to discuss these differences and probably reasons, I came to the conclusion

that the 1st equation is lacking the value of X. I purport that because we normally

consider volume (not cross-sectional area) when we are dealing with pressure,

and also because the X value is not taken into consideration within the first equation,

that perhaps it is wrong for those reasons.

However, the truth is, I don't even know what I am talking about, and am simply

left in a complete state of perplexity. Any and all help is appreciated.

P.S.: I have even asked my class mate and my father for help, both of whom

are equally confused.

P.P.S.: My roommate mentioned something along the lines of the potential for

the two Bernoulli equations in question to be from two different relatives from

the Bernoulli family (the roommate has read many a Mathematician's biography,

which is how they know about the relations)... I believe this to be nonsense. ;)