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> Opštinsko Takmičenje Iz Fizike 2007.

Iva
post Feb 10 2007, 09:00 AM
Post #41





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S ovim postom ni ti nisi daleko!


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^_NiN0_^
post Feb 10 2007, 11:44 AM
Post #42


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Ova tema ce biti LOCK, jer takmicenje je tek sutra , kada ce neko i da otvori, a mi smo naspamovali 3 strane dry.gif


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Iva
post Feb 11 2007, 12:07 PM
Post #43





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Dakle, da vas čujem..
Ja sam malopre došla sa takmičenja...
Ne želim ništa da kažem u vezi poena jer još ne znam kako glase rešenja.
Jedino, sad sam skontala kako sam zeznula 3. zadatak ohmy.gif unsure.gif wacko.gif


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pyost
post Feb 11 2007, 12:52 PM
Post #44


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Verovatno cu imati izmedju 85 i 90. zbog jedne glupe greske i necega sto nisam znao smile.gif Neka, ja sam zadovoljan.. Sad cu da okacim zadatke.


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Anchi
post Feb 11 2007, 01:07 PM
Post #45





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Ja sam uradila svih 5. Peti mi je verovatno netacan, ali nadam se oko 80 poena smile.gif
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pyost
post Feb 11 2007, 01:08 PM
Post #46


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Zadaci (II razred)

1. U sistemu prikazanom na slici 1 poznate su mase tela 1 i 2, a koeficijent trenja izmecu tela 1 i horizontalne ravni je k. Masa kotura je m i mozemo ga smatrati homogenim diskom. Nit po koturu ne proklizuje. U pocetnom trenutku telo 2 pocinje da se spusta. Zanemarujuci masu niti i trenje u osi bloka, naci:

1) ubrzanje tela 2
2) rad sile trenja, koja deluje na telo 2 u toku prvih 10 sekundi od pocetka kretanja

m_1=400g
m_2=200g
m=100g
k=0,1
g=9,81{{m}\over{s^2}}

(20 poena)


2. Zapremina neke mase gasa se, pri zagrevanju za 1^o C i pri nepromenjenom pritisku, poveca za 1\over335 deo svoje prvobitne zapremine. Izracunajte kolika je pocetna temperatura gasa.

(15 poena)


3. U posudi se nalazi gas u stanju "1" u kome je pritisak p_1 a zapremina V_1. Gas se rasiri do zapremine V_2 i pritiska p_2. Pri tome se pritisak linearno smanjuje sa povecanjem zapremine po zakonu p=-aV+b, (a,b\not=0). Masa gasa je m a molarna masa je M. Nadjite opsti izraz zavisnosti parametara a i b od datih vrednosti pritisaka i zapremina, kao i opsti izraz za zavisnost apsolutne temperature T od zapremine V u ovom slucaju. Univerzalna gasna konstanta je R.

(20 poena)

4. Pomocu kompresora se zahvata vazduh temperature T_1=300K, koji se nalazi pod atmosferskim pritiskom p_1=10^5Pa, i adijabatski se sabija do p_2=2\times10^5Pa. Kolicina vazduha koji se ubacuje iznosi {{\Delta V}\over{\Delta t}}=3{{dm^3}\over{s^2}}. Izracunajte korisnu snagu kompresora i temperaturu posle sabijanja. Koeficijent adijabate je \gamma=1,4.

(20 poena)

5. Kruzni ciklus, prikazan na slici 2, u kome ucestvuje n molova nekog gasa, sastoji se iz dve izohore i dve izobare. U stanju "1" temperatura gasa iznosi T_1, a u stanje "3" temperatura gasa iznosi T_3. Tacke 2 i 4 nalaze se na istoj izotermi. Pokazati da je rad koji se izvrsi u ovom ciklusu jednak A=nR(\sqrt{T_3}-\sqrt{T_1})^2, gde je R univerzalna gasna konstanta.

(25 poena)


Resenja

1.
a=2,41{{m}\over{s^2}}
A=-47,28J

2.
T_0=335K

3.
a={{p_1-p_2}\over{V_2-V_1}}

a={{p_1V_2-p_2V_1}\over{V_2-V_1}}

T=VM{{b-aV}\over{mR}}

4.
T_2\approx366K

P=164.25W (nije provereno)

5. Treba izracunati rad gasa, nakon cega treba dokazati da je \sqrt{T_1T_3}=\sqrt{T_2T_2} - cista matematika.

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Cekaaa
post Feb 11 2007, 01:28 PM
Post #47





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Uradio sam sve i sve mi je tacno, sem sto sam napravio racunsku gresku u 5-om. Resenja koja sam dobio:

1. 19.xxxxxxx125
2. 20m/s,20m/s
3.24s,288m
4.t(sqrt(5)-1)
5.v/((cos a) na kvadrat), i lepo dobijem 2v rezultat, medjutim napisem na kraju da je to 10km/h, a treba 40km/h XD.gif XD.gif XD.gif

This post has been edited by Cekaaa: Feb 11 2007, 01:33 PM
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pyost
post Feb 11 2007, 01:34 PM
Post #48


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Vama su bili prilicno laki zadaci smile.gif


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Cekaaa
post Feb 11 2007, 01:36 PM
Post #49





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lol vazi,... to uvek kazes

btw zadaci su bili najtezi u poslednjih 6godina(te sam radio)... sve se uradi vrlo brzo... medjutim ovo takmicenje je bilo mnoooogo teze nego prethodna...

This post has been edited by Cekaaa: Feb 11 2007, 01:38 PM
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pyost
post Feb 11 2007, 01:38 PM
Post #50


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Pa dobro, nisu bas toliko laki.. Prva tri bih mogao da odradim opusteno, a za 4. i 5. bi se mozda malo pomucio.


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Cekaaa
post Feb 11 2007, 01:39 PM
Post #51





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4-ti bi verovatno uradio... posto smo mi imali srecu da ga radimo na casu u skoli.. biggrin.gif a 5-i je po meni bio zajeban do daske,,,,
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Iva
post Feb 11 2007, 01:40 PM
Post #52





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Da, a ja sam imala 47.
Uh! Uzas, sramota!!!






Ali, prva sam..


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pyost
post Feb 11 2007, 01:41 PM
Post #53


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QUOTE(Cekaaa @ Feb 11 2007, 01:39 PM)
4-ti bi verovatno uradio... posto smo mi imali srecu da ga radimo na casu u skoli.. biggrin.gif a 5-i je po meni bio zajeban do daske,,,,
*



Da imam 3 sata, verovatno bih sve nekako uspeo da uradim.. Ovako sam samo bacio pogled smile.gif


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Cekaaa
post Feb 11 2007, 01:42 PM
Post #54





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*** nisi za dzabe drugi razred... tongue.gif
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Anchi
post Feb 11 2007, 01:43 PM
Post #55





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[attachmentid=160]Evo zadataka sa prvi
mozda ima gresaka pisala sam na brzinu biggrin.gif
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pyost
post Feb 11 2007, 01:46 PM
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QUOTE(Cekaaa @ Feb 11 2007, 01:42 PM)
*** nisi za dzabe drugi razred... tongue.gif
*



Nemoj tako, te stvari se lako zaboravljao.. A i nikada nisam voleo rotaciju i kruzno kretanje dry.gif


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Iva
post Feb 11 2007, 01:51 PM
Post #57





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:bag:

kod nas je ovako:

1. 47
2. 40 i nesto
3. 19
4. 18
5. 2
6. 2
7. 0
8. 0

i to je to

za prvi razred


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NeverMore21
post Feb 11 2007, 01:55 PM
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Uhhhh, prokleti peti zadatak mad.gif mad.gif mad.gif !!! I shta kazesh - kosinus na kwadrat je u reshenju....kako si to dobio tongue.gif::tongue.gif Treba na kub tongue.gif tongue.gif tongue.gif


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Iva
post Feb 11 2007, 02:04 PM
Post #59





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wtf blink.gif ???


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smarac
post Feb 11 2007, 02:37 PM
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poyst jel mozes da postavis postupna resenja za 1, 3 i 4 zadatak za drugi razred sa sistemom bodovanja?
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