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https://github.com/danbulant/notes
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vault backup: 2023-04-04 12:35:18
This commit is contained in:
parent
44df9bad80
commit
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5 changed files with 115 additions and 19 deletions
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@ -12,8 +12,8 @@
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@ -1154,7 +1154,11 @@
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@ -10,6 +14,11 @@
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@ -38,6 +47,12 @@
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@ -198,6 +214,22 @@
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22
notes/.obsidian/workspace.json
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22
notes/.obsidian/workspace.json
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@ -20,6 +20,18 @@
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@ -231,14 +243,17 @@
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"fyz/Jevy mezi pevnýma tělesama a kapalinama.md",
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@ -247,14 +262,11 @@
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"data/Screenshot_20230402_150549.png",
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"data/Screenshot_20230402_150005.png",
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"fyz/pohyb/pohyb.md",
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"fyz/Pevne blbosti/Vzorky.md",
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"fyz/Pevne blbosti/Priklady.md",
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"fyz/Pevne blbosti/Deformace.md",
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@ -2,10 +2,10 @@
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Zakřivení kapaliny u stěn nádoby (voda se “lepí na stěnu” a je tak dutá, rtuť se zase snaží stěny dotékat co nejméně)
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$\vartheta$ - stykový úhel
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$p_k=\frac{2\sigma}R$ - kapilární tlak
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$p_k=\frac{2\sigma}r$ - kapilární tlak
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$p_h=\rho gh$ - hydrostatický tlak
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$p_h=p_k$ - rovnováha v kapiláře
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$h=\frac{2\sigma}{\rho gR}$ -
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$h=\frac{2\sigma}{\rho gr}$ - výška kapiláry
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@ -53,21 +53,48 @@ Zvýšení volné hladiny vody v kapiláře.
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Snížení volné hladiny rtuti v kapiláře
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### Tlak
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$p_k=\frac{2\sigma}R$
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$p_k=\frac{2\sigma}r$
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$r$ … poloměr kapiláry
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Vyvolán výslednicí povrchových sil $F_V$ působící kolmo k obsahu průřezu $S$ kapiláry.
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### Výška hladiny
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Výška hladiny kapaliny v kapiláře $h$ je dána rovnováhou kapilárního a hydrostatického tlaku.
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$p_k=\frac{2\sigma}R$
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$p_k=\frac{2\sigma}r$
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$p_h=\rho gh$
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$p_k=p_h$
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$\frac{2\sigma}R=\rho gh$
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$h=\frac{2\sigma}{\rho gR}$
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$\frac{2\sigma}r=\rho gh$
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$h=\frac{2\sigma}{\rho gr}$
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Zvýšení hladiny je nepřímo úměrné poloměru kapiláry
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## Příklady
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Jaký je vnitřní průměr kapiláry, vystoupí-li v ní voda do výšky $2cm$ nad volnou hladinu vody v širší nádobě?
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$\sigma=73mNm^{-1}$
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$\frac{2\sigma}r=\rho gh$
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$r=\frac{2\sigma}{\rho gh}$
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$r=\frac{2*73*10^{-3}}{1*10^{6}*10*2*10^{-3}}$
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$r=\frac{146}{2*10^7}$
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$r=73*10^{-7}$
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$d=1.5mm$
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---
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Jaký tlak má vzduch v kulové bublině s průměrem $1\micro m$ v hloubce $5m$, je-li atmosferický tlak $1000hPa$
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$\sigma=73mNm^{-1}$
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$p=p_h+p_k+p_a$
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$p_h=h\rho g$
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$p_k=\frac{2\sigma}r$
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$p_a=1000hPa=10^5$
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$p=h\rho g+\frac{2\sigma}r+10^5$
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$p=5*10^3*10+\frac{2*73*10^{-3}}{0.5*10^{-6}}+10^5$
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$p=5*10^4+10^5+292*10^3$
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$p=442*10^3Pa$
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@ -8,6 +8,9 @@ $W=\Delta E$ - Práce vyk. působením vnějších sil je rovna přírůstku pov
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$2F\Delta x=\sigma2l\Delta x$
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$F=\sigma l$ [ref](#^fl)
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$l=\pi d$ u kruznice
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[Příklady](#Příklady)
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@ -49,11 +52,10 @@ $\Delta x=2cm=2*10^{-2}$
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$\sigma=\frac{F}l$
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$\Delta E=\sigma2l\Delta x$
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$\sigma=\frac{2*10^{-3}}{5*10^{-2}}=\frac{2*10^{-1}}5=\frac2{50}$
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$\Delta E=\frac1{25}*2*10^{-2}*2*10^{-2}$
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$\Delta E=W=F\Delta x*2$ - z hodiny
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$\Delta E=2*10^{-3}*2*10^{-2}*2$*
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$\Delta E=\frac{4*10^{-4}}{25}$
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Blbě - zeptat se na hodině
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$\Delta E=80\micro J$
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---
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@ -63,4 +65,23 @@ Kruhová smyčka z nitě o průměru $1cm$ je vytvořena na mydlinové bláně.
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$F=\sigma l$
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$\sigma=42mNm^{-1}=42*10^{-3}Nm^{-1}$
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???
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$l=\pi d$
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$F=\sigma \pi d2$ - dva povrchy
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$F=42*10^{-3}\pi 1*10^{-2}*2$
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$F=2.64*10^{-3}N$
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---
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Tlustostěnnou kapilárou vnějšího průměru $3,41mm$ odkapalo $100$ kapek vody teploty $15\degree C$ o celkové hmotnosti $8.11g$. Vypočtěte povrchové napětí vody se vzduchem při dané teplotě.
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$F_g=F_P$
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$m_1g=\sigma l$
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$m_1g=\sigma\pi d$
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$m_1$ … hmotnost jedné kapky
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$\sigma=\frac{m_1g}{\pi d}$
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$m_1=8.11/100$
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$\sigma=\frac{\frac{m}kg}{\pi d}=\frac{mg}{k\pi d}$
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$\sigma=\frac{8.11*10}{100\pi 3.41*10^{-3}}$
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$\sigma=74.3mNm^{-1}$
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