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Antwoorden Natuurkunde (Newton) 5 VWO Hoofdstuk 11 $3.33   Add to cart

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Antwoorden Natuurkunde (Newton) 5 VWO Hoofdstuk 11

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Antwoorden Newton Natuurkunde 5 VWO Hoofdstuk 11

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  • March 12, 2017
  • 15
  • 2015/2016
  • Answers
  • Unknown
  • Secondary school
  • 5

2  reviews

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By: JanDirx • 6 year ago

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By: gijs16 • 7 year ago

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11 Vaardigheden
Wiskunde in de natuurkunde | vwo



Uitwerkingen basisboek
11.1 REKENVAARDIGHEDEN

1
18 7 5 18 28 60 106 53
a
120
+
30
+ 10 = 120 + 120 + 120 = 120 (= 60).
4 2 3 4∙2∙3 8∙3 3 1 1
b
9
× 7 × 8 = 9∙7∙8 = 9∙7∙8 = 9∙7 = 3∙7 = 21.
4 2 4 15 4∙15 2∙15 30
c
11
÷ 15 = 11 × 2
= 11∙2 = 11
= 11.

2
2
𝑙 𝑙 𝑙
a 2𝜋√9,81 = 0,452  (2𝜋√9,81) = 0,4522  4 ∙ 𝜋 2 ∙ 9,81 = 0,4522 
0,4522 ∙9,81
𝑙= 4∙𝜋 2
= 0,0508.
1 2∙4,7 2∙4,7
b × 10,5 × 𝑢2 = 4,7  𝑢2 = 𝑢 =√ = 0,95.
2 10,5 10,5
(3,156∙107 )2 4𝜋
c =  𝑟3 =
𝑟3 6,67384∙10 −06 ×1,9884∙1030
(3,156∙107 )2 ×6,67384∙10−6 ×1,9884∙1030
= 1,0518 ∙ 1039 (in rekenmachine laten
4𝜋
3
staan!)  𝑟 = √1,0518 ∙ 1039 = (1,0518 ∙ 1039 )1/3 = 1,017 ∙ 1013 .

3
a De remkracht blijft gelijk en de beginsnelheid is 90/50 = 9/5e keer zo groot. De
90 2
remweg is evenredig met 𝑣begin 2 , dus is 𝑠rem = 13,8 × ( ) = 45 m.
50
b De beginsnelheid blijft gelijk en de remkracht wordt 6,5/8,4 keer zo groot. De remweg
8,4
is omgekeerd evenredig met de remkracht, dus is 𝑠rem = 13,8 × 6,5 = 18 m.

4
2,07
a De omtrek is 2𝜋 ∙ 𝑟 dus 2,07 = 2𝜋 ∙ 𝑟  𝑟 = = 0,3295 m (in
2𝜋
rekenmachine laten staan!). De oppervlakte is 𝜋 ∙ 𝑟 2 dus
𝐴 = 𝜋 ∙ 𝑟 2 = 𝜋 ∙ 0,32952 = 0,341 m2.
b Maak een tekening:
Pythogoras: 𝑎2 + 4,52 = 5,82  𝑎 = √5,82 − 4,52 = 3,66 m.
De omtrek is 5,8 + 4,5 + 3,66 = 14,0 m.
1
De oppervlakte is ∙ 4,5 ∙ 3,66 = 8,2 m2 .
2
c De straal van de bal is 14 cm.
4 4
Het volume van de bal is 𝜋 ∙ 𝑟 3 = 𝜋 ∙ 143 = 1,1 ∙ 104 cm3 = 0,011 m3 .
3 3
De oppervlakte van de bal is 4𝜋 ∙ 𝑟 2 = 4𝜋 ∙ 142 = 2,5 ∙ 103 cm2 = 0,25 m2 .
d Het volume van een cilinder is π ∙ r 2 ∙ h dus 340 ∙ 10−3 = π ∙ r2 ∙ 1,20 
340∙10−3
r=√ π∙1,20
= 0,300 m. De diameter is dus 60,0 cm.

5 De springdrum wordt in de hand gehouden, dus de hoogte zal ongeveer 10 cm zijn. Op de
foto is de golflengte ongeveer 2 keer zo groot als de springdrum. We schatten de
golflengte dus op 2 x 10 = 20 cm. De frequentie is dan 𝑣 = 𝜆 ∙ 𝑓 


© ThiemeMeulenhoff bv CONCEPT Pagina 1 van 15

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