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NLT Opdrachten kernfusie VWO alle antwoorden

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  • Course
  • NLT
  • Level
  • VWO / Gymnasium

Alle antwoorden van de NLT module 5/6 VWO. Let op! het zijn de antwoorden van de opdrachten, niet van de toets maar ik moest een documenttype aanklikken. Ik heb met dit dossier een 8,5 behaald.

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Document information

  • March 23, 2021
  • 17
  • 2020/2021
  • Exam (elaborations)
  • Questions & answers

Subjects

  • nlt
  • dossier
  • antwoorden
  • module kernfusie
  • vwo nlt

Written for

  • Secondary school
  • VWO / Gymnasium
  • NLT
  • 6
All documents for this subject (195)

1  review

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By: emmamulder3 • 1 year ago

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JohannaJanssen

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Content preview

1 meten van energie
1.1 Energie en vermogen

Opdracht 1.1




Opdracht 1.2

a.
E=F z⋅s=m⋅g⋅s

E= 60*9,81*4= 2354,4 Joule

P = E/t

P = 2354,4/8 = 294,3 Watt

b.
E=F z⋅s=m⋅g⋅s

500L water= 500 kg

E= 500*9,81*2= 9810 Joule

P = E/t

, P = 9810/3600 = 2,725 Watt




1.2 Energie-eenheden

Opdracht 1.3

100 kWh per jaar

Opdracht 1.4

apparaat kWh Tijd / dag

Koelkast * 2 1,096 * 2 24 uur

boiler 0,53 24 uur

Oven + magnetron 0,028 1 uur
Wasmachine+droger 0,28 4 uur
cv-ketel 0,79 24 uur
computer 0,29 4 uur
vaatwasser 0,086 3 uur
fornuis 0,11 1 uur
aquarium 2,4 24 uur
Tv plasma 0,92 4 uur
Stofzuiger 0,67 1 uur
fohn 0,31 30 minuten
lampen 1,38 5 uur
Koffiezetapparaat 0,11 10 minuten
waterontharder 0,085 2 uur
Totaal: 10,181 kWh

10,181 * 3600000 = 36651600 Joule



Opdracht 1.5

41.868 megajoule
a.

10231*106 toe * 41868*106 Joule = 4,28*1020 Joule

b. 4,28*1020/3600000 = 1,19*1014 kWh

, Opdracht 1.6

Plek Totaal in Toe Verbruik p.p. in Toe
6
India 539*10 5,14*10-1
6
Europa 1692*10 3,72
6
VS 2290*10 7,98
In India gebruiken ze p.p. minder toe dan in de EU en de VS, en in de VS gebruiken ze meer toe, ook
al zijn er minder mensen in de VS.



Opdracht 1.7

ΔE kin=−q⋅U ak
a.

ΔEkin = -1,602*10-19 * 500 = -8*10-17 Joule

b. -8*10-17/1,602*10-19 = -500 eV

Opdracht 1.8

a. Elektronen gaan dan naar de kathode toe, want ze worden weggedreven door de anode, doordat
deze negatief is.

b. Dan gaan de elektronen naar de anode toe, omdat ze worden aangetrokken door de positieve
lading.

c. ΔEkin=-q*Uak

ΔEkin=-1,6*10-19*100 = -1,6*10-17 J

-1,6*10-17/1,6*10-19= -100 eV



Opdracht 1.9

6,5*10-,6*10-19 = 4,063 eV



Opdracht 1.10

9,5*10-19 / (1,6*10-19 * 2) = 2,97 eV



Opdracht 1.11

7,0·1012 eV * 1,6*10-19 = 1,12*10-6 J

1,12*10-6 * 3,23·1014 = 361760000 = 3,62*108 Joule per bundel



Opdracht 1.12

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