Answer:
Following are the proving to this question:
Explanation:
using the energy equation for entry and exit value
:
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%20%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D)
where
![= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\](https://tex.z-dn.net/?f=%3D%20%28%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%20%29%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%29%5E4%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C%5C%5C)
![= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%29%20%20%29%7D%20%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C)
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D%20%5C%5C%5C%5C)
![\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})](https://tex.z-dn.net/?f=%5Cto%200%2B0%2BZ_0%20%3D%200%20%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%20%7D%7B2g%7D%20%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%29%7D%5C%20V%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%20%20%5C%5C%5C%5C%5Cto%20Z_0%20-Z_1%20%3D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%5C%20%281%2Bf%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%20%29%7D%20%29%20%20%5C%5C%5C%5C%5Cto%20H%3D%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%28%5Cfrac%7B3%7D%7B2%7D%29%20%5C%5C%5C%5C%5Cto%20%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%3D%20H%28%5Cfrac%7B3%7D%7B2%7D%29)
L.H.S = R.H.S
Answer:
The publication of a parody for commercial gain does not fall within the protection afforded by Section 107, as it is used for commercial gain.
Explanation:
<h2><u><em>
PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
Answer:
I am not sure I am understanding plz more context
Explanation:
Answer:
there's no photo? but I'm willing to help
Answer:
Ususushehehehhuuiiïbbb
Explanation:
Yyshehshehshshsheyysysueueue