Abschlussprüfung Telekolleg Multimedial Lehrgang 15

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Abschlussprüfung Telekolleg Multimedial Lehrgang 15

Fach: Physik

Lösungsvorschlag

Termin: 12. Dezember 2009 Arbeitszeit: 150 Minuten

Bewertungsschlüssel

BE | 60 – 52

|

51 – 43 | 42 – 34 | 33 – 25 | 24 – 13 | 12 – 0

Note | 1 | 2 | 3 | 4 | 5 | 6

(2)

- 2 -

Aufgabe 1 BE

1.1 = ⋅ ⋅

⋅ ⋅

= = =

2

h 1 g t 2

2 h 2 20,0m

t ; t 2,02 s

g 9,81 m s

3 1.2 ( )

( )

= ⋅ ⋅ = ⋅ ⋅ −

= ⋅ ⋅ − =

1 ges 1

1 2 1

v 2 g h 2 g h h

m m

v 2 9,81 50m 30m ; v 19,8

s s

3 1.3.1 =

⋅∆ = ⋅

⋅ ⋅

∆ = = ∆ =

D G

2

F F

D m g

75,0kg 9,81 m

m g s ; 8,66m

D 85,0 N m l

l l

4 1.3.2 Das Bezugsniveau für E

L

befindet sich in der Höhe 21,3 m über dem Boden.

( ) ( )

( )

+ = +

⋅ ⋅ + ⋅ ⋅∆ = ⋅ ⋅ ∆ + ⋅ ⋅

= + ⋅ ⋅∆ − ⋅ ∆

 

=     + ⋅ ⋅ − ⋅ =

1 2

k L S k

2 2 2

1 2

2 2

2 1

2

2 2 2

E E E E

1 1 1

m v m g D m v

2 2 2

v v 2 g D

m

85,0 N

m m m m

v 19,8 2 9,81 8,66m 8,66m ; v 21,8

s s 75,0kg s

l l

6

1.4.1 = ⋅π⋅ m = ⋅π⋅ 75,0kg =

T 2 2 ; T 5,90 s

D 85,0 N

m

2 1.4.2 ( ) ( )

( )

( ) ( )

= ⋅ω⋅ ω⋅

⋅π  ⋅π 

= ⋅ ⋅   ⋅  

 

⋅π ⋅π

= ⋅ ⋅  ⋅  = −

 

v t s ˆ cos t

2 2

v t s ˆ cos t

T T

2 2 m

v 10,0 s 1,00m cos 10,0 s ; v 10,0 s 0,361

5,90 s 5,90 s s

Die Bewegung ist nach unten gerichtet.

4

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- 3 -

Aufgabe 2 BE

2.1

−

=

⋅ = ⋅ ⋅

= ⋅ ⋅

⋅ ⋅ ⋅

= = ⋅

⋅

el k

2

B e

19

6 31

W E

e U 1 m v 2 2 e U

v m

2 1,60 10 As 250 V m

v ; v 9,37 10

9,11 10 kg s

4 2.2 Auf die Elektronen wirkt im Magnetfeld die Lorentzkraft ^F ^uur

^L

^{= ⋅ ×} ^{e v B} ( ) ^{r ur} ^.

Da v B r ur ⊥

ist, wirkt die Lorentzkraft senkrecht zur Elektronenbahn, was bedeutet, dass F r

L

keine Arbeit verrichtet, sondern nur die Richtung von v r

ändert, aber nicht den Betrag. Die Lorentzkraft wirkt als Zentralkraft einer Kreisbewegung.

3

2.3

−

− −

=

⋅ = ⋅ ⋅

= ⋅

⋅

⋅ ⋅ ⋅

= = ⋅

⋅ ⋅ ⋅

Z L

2 e

e

31 6

3

19 2

F F

m v e v B r

B m v e r

9,11 10 kg 9,37 10 m

B s ; B 1,07 10 T

1,60 10 As 5,00 10 m

4 2.4 Der Bahnradius der Protonen würde sich vergrößern, da

⋅ ⋅

= = ⋅ ⋅ = >

⋅ ⋅

2 2

Z L P e

m v m v

F F ; e v B; r ; r m und m m

r e v B : gilt.

3 2.5.1 2

2.5.2

−

=

⋅ = ⋅ ⋅

= ⋅

= ⋅ ⋅ ⋅ =

el L

6 3

2

F F

e E e v B E v B

m Vs kV

E 9,37 10 1,07 10 ; E 10,0

s m m

4 E r

B r

(4)

- 4 -

Aufgabe 3 BE

3.1

123 1 2 3

123

1 1 1 1 1 1 1

R R R R 220 330 470

R 103

= + + = + +

Ω Ω Ω

= Ω

2

3.2

2 2

R A

R A 103 0,100mm 0,500 mm

m 20,6m

= ρ ⋅

⋅ Ω ⋅

= =

Ω ⋅ ρ

= l l

l

3

3.3

123 4

4 123

123

4 4

123 4

U U U 12,0 V 5,91V 6,09 V R R

U U

R 103

R U 5,91V

U 6,09 V

R 100

= − = − =

=

= ⋅ = ⋅ Ω

= Ω

5 3.4 ^L

^FE ⁰

^N

²

^A ^{380 4} ¹⁰

⁷

_{Am 0,100m} ^Vs ³²⁸

²

( ^{1,00 10 m}

²

)

²

L 161mH

− −

= µ ⋅µ ⋅ ⋅ = ⋅ π ⋅ ⋅ ⋅ ⋅ ⋅ π

=

l 3

3.5

L L

eff eff

eff L

eff

X L und 2 f

X 2 f L

U U 12,0 V

I X 2 f L 2 50 1 0,161 Vs

s A

I 237 mA

= ω⋅ ω = π ⋅

= π ⋅ ⋅

= = =

π ⋅ ⋅ π ⋅ ⋅

=

5