Abschlussprüfung Telekolleg Multimedial Lehrgang 13 Fach: Physik

Abschlussprüfung Telekolleg Multimedial Lehrgang 13

Fach: Physik

Lösungsvorschlag

Termin: 13.01.07 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-

Aufgabe 1 BE

1.1

= = = = = =

⋅ ⋅

m m

(8,3 )² 8,3

v² s v s

s ; s 69m ; t ; t 17 s

m m

2 a 2 0,50 a 0,50

s² s²

4

1.2 = ⋅ =

=µ ⋅ ⋅ =µ ⋅ = ⋅ =

= = = + =

⋅

Re ak

R gleit gleit

Brems Anhalt

s 8,3m 1,0 s 8,3 m ; s

m m

F m g; a g; a 0,20 9,81 2,0

s² s²

(8,3m)²

s s 17m ; s 17m 8,3m 25m;

2 2,0m s²

Die Katze kommt mit dem Schrecken davon.

6

1.3

= = =

8,3m

v s

t ; t 4,2 s

a 2,0m

s²

4

1.4 = + µ ⋅ ⋅ ⋅ = ⋅ + ⋅ ⋅

= ⋅ ⋅ − = = °

= ⋅ ⋅ ° − ° =

haft

haft

F F F ; m g cosα m a m g sinα

haft a H

a g (μ cosα sinα) ; tanα 0,10 ; α 5,7

m m

a 9,81 (0,25 cos 5,7 sin5,7 ) 1,5

s² s²

6 0,0 17 21 25 t in s

8,3

v in m/s

-3-

Aufgabe 2 BE

2.1

( )

−

= ⋅ ⋅ +

⋅ ⋅ ⋅ ⋅

= ⋅ =

⋅ + ⋅

E 2 E

11 24

6

F f m m r h

6,67 10 m³ 5,98 10 kg 75,0kg kg s²

F 651 N

(6,40 10 m 380 10³ m)²

Die Gravitationskraft wird als Zentralkraft gebraucht, so dass die resultierende Kraft, die die Astronauten auf den Boden der ISS drückt, null ist.

4

2.2

−

π ⋅ π ⋅

⋅ ω ⋅ = ⋅ ⋅ ⋅ = =

⋅

= + = ⋅ + ⋅ = ⋅

π ⋅ ⋅

= = ⋅ =

⋅ ⋅ ⋅

⋅

E 2

E

E

6 6 6

E

6

3

2 11 24

1 4 ² f m 4 ² r³

m ² r f m m ; ; T ;

r ² T² r³ f m

r r h 6,40 10 m 0,38 10 m 6,78 10 m 4 ² (6,78 10 m)³

T 5,55 10 s 92,6 min

6,67 10 m³ 5,98 10 kg kg s²

6

2.3

−

= + ∆ = ⋅ + ⋅ ⋅ ⋅ − = ⋅

= ⋅ ⋅ = ⋅ = ⋅ ⋅ ⋅ + ⋅ −

= ⋅ ⋅ ⋅ ⋅ ⋅ +

⋅

+ ⋅ ⋅ ⋅

⋅

E ISS ISS

E

2 2 2

E E E

E 2 ISS E ges E E

ISS ISS ISS E ISS

6 3

6

11 2

1 1 1

Eges Ekin W 2m v² f m m (r r); mit v g r folg t

r r 1 r 1 1

v² g r g ; E m [ g f m ( )]

r r 2 r r r

1 m (6,40 10 m)² Eges 183 10 kg (2 9,81s² 6,78 10 m

6,67 10 m³ 5,98 10

kg s² ⋅ − = ⋅

⋅ ⋅

4 12

6 6

1 1

kg ( ) 6,06 10 J

6,40 10 m 6,78 10 m

6

2.4

−

π ⋅ ⋅ ⋅

⋅ ω ⋅ = ⋅ ⋅ ⋅ = =

π

⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ = ⋅

π

E 3 E

E

11 24

3 7

1 4 ² f m f m T²

m ² r f m m ; ; r ;

r² T² r³ 4 ²

6,67 10 m³ 5,98 10 kg (24,0 3600 s)² kg s²

r 4,23 10 m

4 ²

4

-4-

Aufgabe 3 BE

3.1 Br

-Feldlinien senkrecht von oben nach unten einzeichnen.

Vorderansicht: Nach der UVW-Regel wirkt auf einen Strom durchflos- senen Leiter (U: Daumen nach links) in einem magnetischen Feld Br (V: Zeigefinger nach unten) eine Kraft Fr

(W: Mittelfinger aus der Zeichenebene heraus), die den Leiter aus der Zeichenebene herauslenkt, also in der Seitenansicht nach rechts auslenkt.

3

3.2

−

⋅ ⋅ ⋅

α = = = ⋅ α

⋅ ⋅

⋅ ⋅

= ⋅ ° =

⋅

mag G

3

F I b B m g

tan ; B tan

F m g I b

5,0 10 kg 9,81m

B s² tan11,5 50 mT

10A 0,020m

4

3.3.1 = ⋅ ω = α = ⋅ ° ⋅ π =

°

ω = π = ⇒ ω = ω = =

π

= ⋅ ⋅

l

l l

ˆ ˆ ˆ 2

s(t) s cos t ; s ; s 0,40 m 11,5 0,080m 360

1 g g 9,81m 1

2 f ; f ; 5,0

2 0,40ms² s

s(t) 8,0cm cos(5,01 t) s

5

3.3.2

= ⋅ ω = ⋅ = = − ⋅ ω⋅

π π π ⋅

= − = ω⋅ ω⋅ = = = =

⋅ ω ⋅

1 m

ˆ ˆ ˆ ˆ

v s ; v 0,080m 5,0 0,40 ; v(t) v sin( t)

s s

ˆv 1 s

v(t) ; sin( t) ; t ; t ; t 0,10 s

2 2 6 6 6 5,0

4

3.4

−

−

= ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅

π π

= = =

ω

i

3 i

4 i

ˆ ˆ

U B b v ;

Vs m

Uˆ 50 10 0,020m 0,40

m² s

Uˆ 4,0 10 V

2 2

T ; T 1,3 s;

5,01 s

4