Romanian NAAO 2014 Problems

8/10/2019 Romanian NAAO 2014 Problems

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Romanian National Astronomyand Astrophysics Olympiad2014

Theoretical and Data AnalysisProblems with Solutions


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Romania Astronomy and Astrophysics Olympiad, 2014 Theoretical Problems and Solutions

Compiled ByScience Olympiad Blog


3/25

National Astronomy and Astrophysics Olympiad 2014Seniors

Theoretic Phase

PART I - Short Problems

Problem 1

The height of the sky! The ancient Greeks knew that the diameter of the Earth issmall compared to the distance to the stars. For example, there is a legend thatthe god Hephaestus accidentally dropped his anvil on Earth. It took =9 days forthe anvil to eventually hit the ground.Estimate the height of the sky, in accordance to the beliefs of the ancientGreeks. It is known that the rotation of the Moon around the Earth is T L=27,3 daysand the radius of the Moon is a L= 384400 km.

Problem 2

The density of exoplanet X. A radio source is situated on the surface of exoplanetXs satellite. The source constantly emits radio waves but an observer from Earthcant always record the emitted signals because the satellite is occasionally

eclipsed by the planet.Making use of a graphic which illustrates the frequency of the registered signalrelative to the time, find the density of the exoplanet. The orbit of the satellite iscircular and the observer is situated within the same plane as the satellite s orbit.The following data is known: the radius of the exoplanet, R; the universalgravitational constant, K ; the speed of light in vacuum, c. It is also known that thesatellite revolves very close to the surface of the exoplanet.

Problem 3

Angular height of a star. The radar of an astronomical observatory is installed ona rocky plateau near the seashore, at the height h which is above the sea level.The receiver of the observatory records only the electromagnetic signals comingfrom the star . The vector E (the intensity of the electric field of these signals)oscillates parallel with the plane and horizontal surface of the sea, independentfrom the propagation direction of the electromagnetic wave. The intensity of anyrecorded signal is proportional with E 2. When the indicator of wavelengths showthe value , the radar receiver records minimum and maximum values.


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Determine the angular heights of the the star above sea level, at which theradar receiver records electromagnetic signals of maximum and minimumintensities.

Problem 4

Visible star? The Sun parallax is p sun = 8,8 and the parallax of a star, , which hasthe same absolute brightness (luminosity) as the Sun, is p star =0,022. Reveal whether this particular star can be observed on the night sky with thenaked eye. The following data is known: the distance between Earth and the Sun,rES = 149000000 km; Earth radius, R E= 6380 km.

Problem 5

Star Wars. In Star Wars, a star with the apparent magnitude of m initial =3m was

cut into four identical smaller stars that had the same density and temperaturelike the initial one.Determine the magnitude of the resulting quadruple star and compare it to themagnitude of the initial star.

PART II - Long Problems

Problem 1

Neutronic star. It is well known that many stars form binary systems. One type ofbinary system consists of a normal star (with the mass m 0 and radius R) and aneutronic star (much more compact and with a larger mass), which revolve arundtheir own centre of mass. In the following problem the Earths movement isneglected.

Based on terrestrial observations of this type of binary system, the followinginformation is known:- the maximum angular displacement of the normal star is and the maximum

angular displacement of the neutronic star is , as indicated in Figure 1;- the necessary time for this kind of maximum displacements is ;- the radiation attributes of the normal star show that its surface temperature is T

and the incident radiant energy per area unit of the Earths surface, per timeunit, is P;


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- the Calcium (Ca) spectrum line of this radiation has a wavelength which differsfrom the normal one ( 0) by , due only to the normal stars gravitational field.

a) Find the distance r between Earth and the binary system presented above byusing only the values of the observed units and the universal physical constantsinvolved.

b) Now, lets suppose that M>>m 0, so that the normal star revolves around theneutronic star on a circular orbit with the radius r 0. The normal star starts toemit gas twards the neutronic one with a relative speed of v 0 (relative to thenormal star) as indicated in Figure 2. Admitting that the neutronic star is thedominant source of the gravitational action and neglecting the orbit changes ofthe normal star, you are asked to determine the minimum distance r min atwhich the gas gets close to the neutronic star. It is known that the universalgravitational constant is K.

c) Determine the maximum distance r max at which the gas reaches close to theneutronic star.


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Problem 2

A. Sun dusk. The dusk and dawn are two events of lengths that depend solely onthe place and time of the observation.a) Determine the duration of the dusk/dawn for an observer situated in a place

with the latitude on equinox days;b) Localise the observer so that during the equinox days, the duration of the

dusk/dawn is maximum/minimum;c) Determine the duration of the dusk/dawn for an observer situated in a place

with the latitude on solstice days;d) Localise the observer so that during the solstice days, the duration of the

dusk/dawn is maximum/minimum.The following data is known: the apparent angular diameter of the Sun, =3159,3; the rotation period of the Earth around the Sun T E=24h, the anglebetween the equator plane and the ecliptic plane =23 27. The effects ofatmospheric refraction are neglected.

B. The third cosmic speed. You are asked to determine the approximate minimumvalue of escape velocity that a body must have so that when launched from Earth,it would escape the Solar System forever (third cosmic speed).The following data is known: V 0 30km/s, the speed of Earth around its circularorbit around the Sun; v 0 7,9km/s, the speed of a low orbit satellite that revolvesaround the Earth (first cosmic speed).

It is also known that

. The bodys kinetic energy relative to the Sun is

neglected from the moment of launch until reaching the limit of the Earthsgravitational field.

C. Fall from the Earth on the Sun! You are asked to determine the minimumspeed that is needed for a spaceship to escape Earths gravity and fall on theSuns surface. The following data is known: the distance between Earth and theSun, r ES= 1,5 10

11 m; the rotation period of the Earth around the Sun,TE=3,15 10

7s.

Prof. dr. Mihail SanduLiceul Tehnologic de TurismCa lima nes


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Romania Astronomy and Astrophysics Olympiad, 2014 Data Analysis Problems and Solutions

Compiled ByScience Olympiad Blog


9/25

National Astronomy and Astrophysics Olympiad 2014Seniors

Data analysis Phase

Problem 1

Speed of light. Lets imagine that in a distant future, the Solar System will be occupied byour descendants. A small mining robot installed on the SALTIS asteroid is supervised byCelesta Spacedigger, who happens to also be a passionate amateur astronomer. During thelong nights of Saltis, Celesta (character from the Greek mythology) observes the stars andplanets, particularly the beautiful planet Saturn. An old but trustworthy astronomicalalmanac helps her follow certain celestial events such as Titans eclipses due to Saturnsmovement. To her astonishment, Celesta discovers large differences between the timevalues she noticed while observing Titans eclipses and the existent values from thealmanac. After years of careful observation (as she was detached to stay on SALTIS for along time), Celesta eventually finds an explanation. The differences are the largest whenSaturn is close to the opposition or conjunction with the Sun, both seen from Saltis. Celestafigures out that this is because the speed of light is finite. Also, she discovers that a sketchfrom the almanac confirms the fact that the synchronizations from its tables are heliocentric(relative to the Sun and not to Saltis). Very satisfied with her discovery, Celesta used theseobservations to calculate the speed of light.In the following problem you have to repeat the computation done by Celesta using herobservations. The units of time and length used by Celesta are fairly different than the ones

we use. The unit of time which is called pinit is defined so that Saltiss synodic period ofrotation is T synodic Saltis = 1000 pinit. The unit of length called seter is defined so that 1 seter isequal to 10 -9 the mean distance from the Sun to Saltis. In other words, r Saltis-Sun =10

9 seter.a) Six (6) records made by Celesta on Titans eclipses when Saturn was close to the

opposition or conjunction are represented below. The columns are described asfollows:

I) the values from the almanac table regarding the moment when an observersituated on the Sun could see the beginning of the eclipse;

II) the values of Celestas observations regarding the beginning of the eclipse asseen from Saltis. The accuracy of the synchronizations is = 0,03 pinit;

III) Saturns position during Titans eclipse (close to the opposition or conjunction ).

Observation number Almanac table(pinit)I

Observations madeby CelestaII

Statement

III1 456,47 450,32 Opposition2 18,50 12,28 Opposition3 821,41 815,29 Opposition4 444,70 450,85 Conjunction5 615,43 621,52 Conjunction

6 791,94 798,02 Conjunction


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Carefully analyzing the data from the table, estimate the speed of light expressed inseter/pinit and state what is the possible error of the estimation.

b) During those days on Saltis when she feels lonely, Celesta likes to listen to the radiosignals coming from Earth. Now that she knows the speed of light, Celesta wants todetermine Earths radius expressed in seter. Her watch is accurately synchronizedwith the signals coming from Earth. The results from the measurements are shown inthe graphic below:

Estimate the radius of planet Earth expressed in meter using Celestas data from the chartabove.

c) Knowing that: 1 au=149,6 10 6 km; c=2,998 10 8 m/s, determine the equivalent inmeters for 1 seter; the equivalent in seconds for 1 pinit.

d) Estimate the sidereal orbital period of Saltis expressed in years using c) and thegraphic above.

Problem 2

The orbit of planets. It is accepted that Earths orbit around the Su n is a circle with theradius r E=1 au. In the following table the maximum Eastern and Western angularelongations of Mars and Venus are specified.

a) Admitting that the orbits of the three planets (Mercury, Venus and Earth) relative tothe Sun are described as concentric and coplanar circles, lets suppose thatVenus/Mercury are in the corresponding position for their maximum Easternelongation. Determine after what time Venus/Mercury will be:I) for the first time, in the corresponding position for the maximum Western

elongation;II) again in the corresponding position for their maximum Eastern elongation.


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b) According to the data from the table and using a piece of paper on which Earthscircular orbit around the Sun (which is in the middle) is represented, localise Mercuryand Venuss approximate position around he Sun and determine the mean distancebetween Mercury and the Sun and the mean distance between Venus and the Sun.

c) Determine the exact values of Mercury and Venus s orbit parameters , (a; b; e) ,expressed in au, knowing that:

18 max ,East Mercury

28 ; 18 max ,West Mercury

28 ;

45 max ,East Venus 48 ; 45 max ,West

Venus 48 .

Table - The maximum elongations of Mercur and Venus


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Problem 3The Great Opposition. On the 28 th August 2003, at 17:56 UT (Universal Time), the latestGreat Opposition of Mars took place, as the Earth passed between Mars and the Sun,aligning with them. The distance between Earth and Mars had the smallest possible value.The next Great Opposition of Mars will take place in 2018. Someone doesnt understand thespecial aspect of the Great Opposition and thinks that a Simple Opposition (Earth passingbetween Mars and the Sun and aligning with them) will happen in 2018, not a GreatOpposition. a) Determine the circular orbit parameters of the hypothetical planet Mars 2 which will be ina Simple Opposition to the Sun as seen from Earth in 2018. Earths orbit around the Sun iscircular.b) Estimate the apparent magnitude of Mars 2 during its opposition in 2018, as seen fromEarth. It is known that the apparent visual magnitude of Mars during the Great Oppositionin 2003 was m M,2003 = 2

m. The physical characteristics of Mars 2 are identical to those ofMars.c) The following table lists all the Oppositions of Mars from 1955 to 2037. Identify theOppositions taking place near the Perihelion, the Oppositions near the Aphelion and theGreat Oppositions. State the criteria used for identification.

Mars s Oppositions, 1955 - 2037Opposition date Date of maximum proximity Minimum distance between Earth

and Mars (ua/ millions of miles)12 th February 1995 11 th February 1995 0,67569/62,8 17 th March 1997 20 th March 1997 0,65938/61,324 th April 1999 1 st May 1999 0,57846/53,813 th June 2001 21 st June 2001 0,45017/41,828 th August 2003 27 th August 2003 0,37272/34,67 th November 2005 30 th October 2005 0,46406/43,124 th December 2007 18 th December 2007 0,58935/54,829 th January 2010 27 th January 2010 0,66398/61,73rd March 2012 5 th March 2012 0,67368/62,68 th April 2014 14 th April 2014 0,61756/57,422 nd May 2016 30 th May 2016 0,50321/46,827 th July 2018 31 st July 2018 0,38496/35,813 th October 2020 6 th October 2020 0,41492/38,68 th December 2022 1 st December 2022 0,54447/50,616 th January 2025 12 th January 2025 0,64228/59,719 th February 2027 20 th February 2027 0,67792/63,025 th March 2029 29 th March 2029 0,64722/60,24 th May 2031 12 th May 2031 0,55336/51,427 th June 2033 5 th July 2033 0,42302/39,315 th September 2035 11 th September 2035 0,38041/35,419 th November 2037 11 th November 2037 0,49358/45,9

d) Determine the ratio of the diurnal equatorial horizontal parallax of the Sun (p eS) and thediurnal equatorial horizontal parallax of Mars (p eM ).


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ONAA 2014SENIORI

Analiza Datelor - BARAJ

Problema 1

a)

;Opozitie

SoareSaltis

t r

c

; pinitseter 1060,1

pinitseter 10

22,610

pinit22,6seter 10 889 c

, pinitseter

1003,060,1 8 c 3p

reprezentnd viteza luminii, determinat de Celesta, exprimat n seter/pinit.

b) Utiliznd desenele din figurile alturate, rezult:

Fig.

t

mint

maxt


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Fig.

3,5minmax t t t pinit;;2 SoarePamantPamantSaltismin,PamantSaltismax, rrr

; pinitseter

1060,1 8 c

seter;105,8 pinitseter

106,1 pinit3,5 88 ct

.UA1seter 1025,421 8

SoarePamant ct r .2p

c)

m;352seter 1 188 pinit1 s................................................................. 2p

d)

;m10149m35210

an12/3

9

92/3

Pamant

SaltisPamantsideral,Saltissideral,

a

aT T

2/32/3

Saltissideral, 36,2149352an1

T ani 6,3 ani3p

SoareSaltisr PamantSaltismin, r

SoareleSaltis n Opoziie

Pmntul

Saltis n Conjuncie

PamantSaltismax, r

SoarePamantr


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Problema 2

Rezolvarea)

;27Mercur Vest,max,Mercur Est,max,

.47VenusVest,max,VenusEst,max,

1) Pentru planeta Venus:

zile;6,157zile7,234zile2,365

zile7,234zile2,36524,024,036086

VP

VP

VP

VP T T

T T T T

T T t .......... 1p

Pentru planeta Mercur:

zile;6,40zile88zile2,365

zile88zile2,36535,035,0

360126

MP

MP

MP

MP T T

T T T T

T T t ............... 1p

2) Pentru planeta Venus:

zile;8,656zile7,234zile2,365

zile7,234zile2,365VP

VP T T

T T t ........................... 1p

0P

Estmax,,0V

S

P

Vestmax,V

Estmax,

Vestmax,

V

P


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Pentru planeta Mercur:

zile;116zile88zile2,365

zile88zile2,365

MP

MP T T

T T t ................ 1p

b)

M 0,360 UAr .................................................. 2p

0P

Estmax,,0V

S

P

Estmax,V

Estmax,

Estmax, V

P

I1

II1

III1

IV1 V1

VI1

VII1

IX1

X1 XI1

XII1

VIII1

MERCUR

P

S


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Fig.

V 0,718 UAr .......................................... 2p

c)

Fig.

Elongaia Estic

MINIM

Elongaia Vestic

MINIM

Elongaia Vestic

MAXIM

Elongaia Estic

MAXIM

S

P

minr maxr

Pr

ESTmin,

VESTmin, ESTmax,

VESTmax,

Mercur/Venus

I1

II1

III1

IV1 V1

VI1

VII1

IX1

X1 XI1

XII1

VIII1

VENUS

S

P


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- pentru planeta Mercur:

UA;389,02

Mercur max,Mercur min,Mercur

rra

;020,0Mercur max,Mercur min,

Mercur min,Mercur max,Mercur

rr

rre

;1 2Mercur

2Mercur

Mercur ab

e

UA;388,01 2Mercur Mercur Mercur eab ..................................................... 1p - pentru planeta Venus:

UA;725,02

Venusmax,Venusmin,Venus

rra

;024,0Venusmax,Venusmin,

Venusmin,Venusmax,Venus

rr

rre

;1 2Venus

2Venus

Venus ab

e

.UA724,01 2VenusVenusVenus eab ................................................... 1p

Problema 3

a)

;111

2Msideral,Psideral,2Msinodic, T T T

07,1an1ani15

ani15an1

Psideral,2Msinodic,

2Msinodic,Psideral,2Msideral,

T T

T T T

ani.

Fig.

2Ma

Pa

2MP,r

2M P

Orbita circu lar

a lui M 2

Soarele

Opoziia lui M 2

Orbitacircular aPmntului


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;3

P

2M2

Psideral,

2Msideral,

a

aT

T

047,132

Psideral,

2Msideral,P2M

T

T aa UA,...................................... 2p

b)

Fig. ;4,0log 2MM

Observator ,2Marte

Observator Marte, mm

;4,0

24

24log 2MM

22MP,

22M

22MS,

S2M

2MP,

2M

2MS,

SM

mm

r R

r L

r R

r L

;2MM ;2MM R R

;log5 2MM2M.P

MP,

2MS,

MS, mmr

r

r

r

MarteSoare,Soare, r

SoareleMarteincident,F

Martereflectat,F

Observator Marte,

Pmntul (observatorul)

M a r t e

M R

PMr

MarteSoare,r


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;6m2MM mm ;2

mM

m ,8m2M m ................................................................... 3p

reprezentnd magnitudinea aparent a planetei Marte 2, vzut de pe Pmnt.

c)Atunci cnd Opoziia lui Marte se ntmpl la Periheliul orbitei sale, sau foarte aproape

acesta, avem de a face cu o Mare Opoziie a lui Marte, aa cum indic desenul din figura altur

distana dintre Pmnt i Marte fiind minim posibil (60 milioane km). Aa s-a ntmplat la 28August 2003, aceasta fiind ultima Mare Opoziie a lui Marte. Atunci cnd Opoziia lui Marte se ntmpl la Apheliul orbitei sale, sau foarte aproape

acesta, distana dintre Pmnt i Marte este maxim posibil (100 milioane km).

Fig.

Tabelul de mai jos prezint o list a tuturor Opoziiilor lui Marte din 1955 i pn n 2037. Dinacest tabel rezult c Pmntul este relativ apropiat de Marte n anii 2001 i 2005, iar n 20Pmntul este foarte aproape de Marte. Apoi, n anii 2020 i 2033, Pmntul va fi din nou relapropiat de Marte, iar n anii 2018 i 2035 Pmntul va fi din nou foarte aproape de Marte, ca 2003.

P

Orbita eliptica lui Marte

M

Orbita circulara Pmntului

Marea O poziie a lui Marte

Periheliu

60 milioane km

100 milioane km

P M Apheliu

Opoziia lui Marte

150 milioane km

2 x 230 milioane km

M M25 log 1, 46 11,15 ;m m


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Opoziiile lui Marte, 1995 - 2037

Data O poziiei Data apropierii maxime Distana minim

(UA/milioane mile)

12 Februarie 199517 Martie 199724 Aprilie 199913 Iunie2001

28 August 2003 07 Noiembrie2005 24 Decembrie 2007

29 Ianuarie 201003 Martie 201208 Aprilie 201422 Mai 2016

27 Iulie 2018 13 Octombrie 2020

08 Decembrie 202216 Ianuarie 2025

19 Februarie 202725 Martie 2029

04 Mai 203127 Iunie 2033

15 Septembrie 2035 19 Noiembrie 2037


01 Mai 199921 Iunie 2001

27 August 2003 30 Octombrie 2005 18 Decembrie 2007

27 Ianuarie 201005 Martie 201214 Aprilie 201430 Mai 2016

31 Iulie 2018 06 Octombrie 2020

01 Decembrie 202212 Ianuarie 2025


12 Mai 203105 Iulie 2033

11 Septembrie 2035 11 Noiembrie 2037

0,67569/62,80,65938/61,30,57846/53,8

0,45017/41,8

0,37272/34,6 0,46406/43,1 0,58935/54,80,66398/61,70,67368/62,60,61756/57,4

0,50321/46,80,38496/35,80,41492/38,6 0,54447/50,60,64228/59,70,67792/63,00,64722/60,20,55336/51,4

0,42302/39,30,38041/35,4 0,49358/45,9

n tabel sunt indicate dou date:data O poziiei , cnd Pmntul trece printre Marte i Soare,aliniindu- se cu acetia;data apropierii maxime dintre Pmnt i Marte, care este cu cteva zile maidevreme dect

data Opoziiei, cnd Marte se deprteaz de Soare (apropiindu-se de Apheliu) i cu

cteva zile mai trziu dectdata Opoziiei , cnd Marte se apropie de Soare (apropiindu-se dePeriheliu), aa cum ilustreaz desenul din figura alturat.

Dac data Opoziiei este foarte aproape de Periheliu, atunci data apropierii maxime eaproximativ aceeai cu data Opoziiei (aa cum s-a ntmplat n 2003). Pmntul trece mai aproape deMarte, dac data Opoziiei este mai apropiat de Periheliu, aa cum se va ntmpla n anii 2012035.


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Fig.

Identificrile Opoziiilor ............................................................................................3p- Opoziiile marcate cu BOLD ITALIC se produc atunci cnd distana dintre Pmnt i Mar

este minim, ceea ce se ntmpl cnd Marte este la Periheliu sau foarte aproape de acesta. - Opoziiile marcate cu BOLD DREPT se produc atunci cnd distana dintre Pmnt i Ma

este mic, ceea ce se ntmpl cnd Marte este n apropierea Periheliului. - Opoziiile pentru care distanele dintre Pmnt i Marte sunt maxime se produc atunci c

Marte este la Apheliului.- Opoziiile pentru care distanele dintre Pmnt i Marte sunt apropiate de valorile maxime

produc atunci cnd Marte este aproape de Apheliul orbitei sale.

d)

.54,1113/2

P

M

Me,

Se,

T T

e p

p.2p

P e r i h e l i uApheliu

Soarele

Marte

Pmntul

Marte

Pmntul

Opoziie

minPM,d

Opoziie

minPM,d


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Problema 4A.

;

v

v1 2

02rad

2rad

0

min

rrr

;4,0log 00 mm E E

;

v

log52

0

2rad

00

p D

p D

mm

;61,05,189,03,05 mmmm0 mm ............................................. 2p

13

6

7

00 1015

10

km1015

p

Dr ;

Soarele

Altair

0A radv

v tgv

10 sinr

minr

B

min0 rr

A

Pmntul

S 0A 0r

d

0 p

90


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km;105,721km1015

vv

v 13132tg

2rad

tg0min rr ........................... 2p

;v radmin0 rr

95130s103v

12

rad

min0 rr

ani............................... 1p

B.

,4,0log SS

mm E

E

S Sk E ; k E

;10 5,2/S Smmk k ;kW/m37,1

2S k

;1m m ;8,26 m

S m .

mW

10 28 k

t S k W total J,3000s103m1000smJ

10 8228

300total W W J.

kg102 6

V M .;W Mc 8

6105,3

K kgJ

4200kg102

J300

McW

K.................................... 2p

C.

.log10log5stea

Soare

stea

SoaresteaSoare T

T mm

........................................... 2p

.195,25 m

Soare m ..1p

Romanian NAAO 2014 Problems

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