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\begin_body
\begin_layout Title
Centripetalni pospešek
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\begin_layout Author
\noun on
Anton Luka Šijanec
\end_layout
\begin_layout Date
14.
december 2022
\end_layout
\begin_layout Abstract
Poročilo pete vaje pri predmetu
\noun on
F41
\noun default
na Gimnaziji Bežigrad v 4.
letniku.
Vaja je potekala 10.
novembra 2022.
\end_layout
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LatexCommand tableofcontents
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\begin_layout Section
Potrebna oprema
\end_layout
\begin_layout Itemize
kos plastične cevi
\end_layout
\begin_layout Itemize
vrvica
\end_layout
\begin_layout Itemize
zamašek z luknjico
\end_layout
\begin_layout Itemize
štoparica
\end_layout
\begin_layout Itemize
ravnilo
\end_layout
\begin_layout Itemize
tehtnica
\end_layout
\begin_layout Itemize
uteži
\end_layout
\begin_layout Section
Cilj naloge
\end_layout
\begin_layout Standard
S poskusom preveriti veljavnost drugega Newtonovega zakona za kroženje.
\end_layout
\begin_layout Section
Potek meritve
\end_layout
\begin_layout Enumerate
Izmeri maso uteži in maso zamaška.
\begin_inset Newline newline
\end_inset
\begin_inset Formula
\[
m_{\text{{uteži}}}=\SI{30,5}{\gram}
\]
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\begin_inset Formula
\[
m_{\text{zamaška}}=\SI{3,5}{\gram}
\]
\end_inset
\end_layout
\begin_layout Enumerate
Primi za cev in jo vrti tako, da zamašek kroži v vodoravni ravnini s stalno
hitrostjo.
Pri tem pazi, da je razdalja od vrha cevi do zamaška ves čas enaka izbranemnu
polmeru.
Izmeri čas desetih obhodnih časov.
Meritev ponovi za deset različnih polmerov.
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Meritve
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\begin_layout Section
Naloge
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\begin_layout Enumerate
Za vsako meritev določi obhodni čas in izračunaj centripetalni pospešek.
Določi povprečni pospešek in njegovo napako.
\begin_inset Formula
\[
\overline{a_{r}}=\SI{99,970594}{\meter\per\second\text{\squared}}
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\[
\sigma_{a_{r}}=\SI{20,487365}{\meter\per\second\squared}
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\end_inset
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\begin_layout Enumerate
Centripetalni pospešek izračunaj tudi iz mas zamaška in uteži.
Ali se dobljena rezultata ustrezno ujemata?
\begin_inset Formula
\[
a_{r}=\frac{{F_{c}=gm_{\text{{uteži}}}}}{m_{\text{{zamaška}}}}=\frac{\SI{9,81}{\meter\per\second\squared}\cdot\SI{0,0305}{\kilo\gram}=\SI{0,299205}{\newton}}{\SI{0,0035}{\kilo\gram}}=\SI{85,487}{\meter\per\second\squared}
\]
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\begin_inset Newline newline
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Upoštevajoč napako merjenja se izračunana vrednost ustrezno ujema z izmerjeno.
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\begin_layout Enumerate
Izpelji izraz, ki povezuje kvadrat nihajnega časa
\begin_inset Formula $t^{2}$
\end_inset
z radijem
\begin_inset Formula $r$
\end_inset
!
\begin_inset Formula
\[
t^{2}=\frac{{4\pi^{2}r}}{a_{r}}
\]
\end_inset
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Nariši graf
\begin_inset Formula $t^{2}(r)$
\end_inset
.
Skozi narisane točke lahko narišemo premico skozi izhodišče.
Določi naklon te premice, pri tem pazi na enote rezultata!
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LatexCommand label
name "fig:graf"
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Graf kvadrata nihajnega časa v odvisnosti od polmera.
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begin{lstlisting}
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fit (x*p)
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podatki.tsv" using 2:6 via p
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\begin_layout Enumerate
Naklon premice izrazi z maso zamaška in uteži in ga tako tudi izračunaj.
\begin_inset Formula
\[
k=\frac{{t^{2}}}{r}=\frac{{4\pi^{2}}}{a_{r}}=\frac{{4\pi^{2}m_{\text{{uteži}}}}}{gm_{\text{{uteži}}}}=\SI{0,461805}{\second\squared\per\meter}
\]
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Izračunana vrednost je izven standardne napake, vendar ne pretirano.
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Uporabljen program
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lstinputlisting[language=Python]{meritev.py}
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