Finding h in the Fluid Dynamics Lab

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SUMMARY

The discussion focuses on calculating the height 'h' in a fluid dynamics lab experiment, where 'h1' is the initial height before a hole is opened. The equation provided, x = 2√(h(h1-h)), is central to the calculations. Participants are tasked with rearranging this equation to express 'h' in terms of 'x' and 'h1'. A derived equation, h1 = (1/4)h x² - (h²/h), is also mentioned, indicating the need for a linear relationship between functions f(x) and g(h1) for further analysis.

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belinda
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Homework Statement


okay, so the lab that we did is called fluid dynamics. where we open a hole but before we are supposed to record h1 before the hole is open and also measure the height of the hole. h1 is higher than h. so every 30 seconds, we have to record h2, h3 ,h4 corresonding with their range, thus their x values until we reach close to h. an equation that x can also be calculated as x = 2√(h(h1-h). so we are told to rearrange the equation to deduce it to where h is either the slope or the intercept and then to find h and compare it with the recorded h. My question is how do i find h from this equation. and what is going to be the x value, is the x value the recorded value that corresponds to h recorded


Homework Equations



x = 2√(h(h1-h).

The Attempt at a Solution


so far, this is the equationh1=1/4h x^2-h^2/h that i got but i have no idea how to find h
 
Last edited:
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Welcome to PF!

Hi belinda! Welcome to PF! :smile:

(try using the X2and X2 tags just above the Reply box :wink:)

You have x2 = 4(h1h - h2).

You need to find functions f(x) and g(h1) such that the f,g graph is linear (ie, f = Ag + B). :smile:
 

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