Exam (elaborations)
Alle antwoorden Sattelieten en Aardobservatie 5VWO NLT
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In dit document staan alle antwoorden van de module Sattelieten en Aardobservatie.
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1.1 -1.2
Functie satelliet Kijkrichting: Baan: polair / Baanhoogte:
geostationair/
naar aarde/ van aarde hoog/laag
Lagrangepunt
af
Observatie naar aarde polair laag
luchtvervuiling
Wetenschappelijk van aarde af Lagrangepunt hoog
onderzoek zonnewind
Weerobservatie naar aarde polair laag
Spionage naar aarde polair laag
Communicatie (TV) naar aarde geostationair hoog
Navigatie (GPS) naar aarde geostationair hoog
1.3 a. 2/3 niet operationeel, 1/3 operationeel
b. NASA en Rusland
c. Ruimteparaplu. Zo klein mogelijk bouwen en zo min mogelijk uitstekende delen eraan zetten.
1.4 veeteeltboeren.
1.5 a. 17 pixels = 243km
= 14.29 km
b. Die zijn langer dan de verticale afmeting. Dit komt doordat de satelliet beweegt in de richting
zuidoost naar noordwest, en dus meer kortere stukjes kan zien in die richting.
1.6 a. Mitochondrium
b.
1.7 a. CO2
, b. Ja, alleen met grote maatregelen vanuit de overheid gaat iedereen meehelpen aan het klimaat
redden. Zonder wetten tegen klimaatverandering zullen gigantische bedrijven rustig door blijven
gaan met alles vervuilen.
1.8 2 NH3 + 2 O2 -> 2 NO + 3 H2O
2 NO + O2 -> 2 NO2
3 NO2 + H2O -> 2 HNO3 + NO
1.9 a. Vlees eten, (benzine)autorijden, ademen.
b. Elektrisch rijden, minder vlees eten.
1.10 a. O = 32 amu, N = 28 amu, Ar = 36 amu
32*0.21 + 28*0.78 + 36*0.01 = 28.92 amu = 29 gr/mol
b. 1.225 * avogadro/28.92 = 2.55*1025 moleculen/m3
c. 56%
1.11 a. De 3e wet van Kepler
b.
T2 = (4π2*r3) / GM
T = √((4π2*r3) / GM)
T = √((4π2*67710003) / 6.6726*10-11*5.972*1024) = 5545.6396 s ~=~ 1.5 uur
c. T = 1 draaiing aarde ~=~ 24 uur
1.12 a. chemische energie in hoogte-energie en kinetische energie.
b. vo = √((2GM)/r)
c. vo = √((2*6.6726*10-11*5.972*1024)/ 6371000) = 11184.55 m/s = 11.18 km/s
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