Lösung: Abschlussprüfung Telekolleg MultiMedial/Lehrgang 11 BE Aufgabe I:

- 1 -

Lösung: Abschlussprüfung Telekolleg MultiMedial/Lehrgang 11 BE Aufgabe I:

5 1.1

E s = W

R

+ E

k

; 0,5 D BA ² = µ g m AC + 0,5 m v o ²; v o ² = ⋅ − m

BA

D

²

2 µ g m AC;

v o = 0 , 80 m

² s 81 m , 9 61 , 0 kg 2

030 , 0

)² m 10 , 0 m ( 400 N

⋅

⋅

⋅

⋅ −

= 11 s m

2 1.2.1

x w = v o

g h

⋅ 2 ^; x w =

² s 81 m , 9

m 0 , 5 2 s

11 m ⋅ ⋅ = 11 m

4 1.2.2

v = v

²_o

+ ( gt )² ; t² = g

h

2 ; v = v

²_o

+ 2 gh ^; v = 5 , 0 m

² s 81 m , 9 2 s )² 11 m

( + ⋅ ⋅ = 15

s m

4 1.2.3

tan α =

w o y

x

² gt v

v = ; wobei v o = t x

_w

; t² = g

tan x

_w

⋅ α

; t =

g tan x

_w

⋅ α

4 1.2.4

tan α x

w

²

= gt ; t =

o w

v

x ; tan α =

² v

x g

o

⋅

w

; tan α = 0 , 89

s )² 11 m (

m

² 11 s 81 m ,

9 ⋅ =

; α = 42°

Aufgabe II:

4 2.1 Bahn Nr. 1 2 3 Mittelwert

³ r

² T in

³ m

²

10

⁻¹³

s 9,29 9,25 9,28 9,27 ⇒ T² = r ³

³ m

² 10 s 27 ,

9 ⋅

⁻¹³

⋅

4 2.2

Z

g

F

F = ; r

² T

² m 4

² r

m

f ⋅ m

^s

⋅

^M

=

_s

⋅ π ⋅ ; T² = r ³ m f

² 4

M

⋅ ⋅

π ;

M

M

f m

² C 4

⋅

= π = const.

3 2.3

v = T r 2 π

; v =

s 71 km , s 1 3600 24 621 , 0

m 10 6 , 14

2

⁶

⋅ =

⋅

⋅

⋅ π 4 2.4

² r T

² m 4

² r

m

f ⋅ m

^s

⋅

^M

=

_s

⋅ π ⋅ ;

f

² T

³ r

² m

_M

4

⋅

⋅

= π ;

² s kg

³ 10 m 67 , 6 )² s 3600 24 621 , 0 (

)³ m 10 6 , 14 (

² m 4

11 6 M

⋅ ⋅

⋅

⋅

⋅

⋅

⋅

= π

−

= 6 , 40 ⋅ 10

²³

kg

5 2.5

W = W kin + W h W = )

r 1 r ( 1 m m f

² v 2 m 1

2 M M s

s

+ ⋅ ⋅ ⋅ −

s )² 10 m 71 , 1 ( kg 10 4 , 2 30

W = 1 ⋅ ⋅

³

⋅ ⋅

³

+

+ ₎

m 10 14,6

1 m

10 3,4 ( 1 kg 10 6,40 10³kg s² 30,4

kg 10 m

6,67

²³ _{6 6}

3 11

− ⋅

⋅ ⋅

⋅

⋅

⋅

⋅ ⋅

⋅

⁻

= 3,37 ⋅ 10

¹¹

J

- 2 - Aufgabe III:

5 3.1 Φ

_max

= b ² ⋅ B ;

Φ

max

= (0,030 m)² 0,10

² m Vs

Φ

max

= 9,0 ⋅ 10

⁻⁵

Vs

5 3.2 I : 0 s ≤ t ≤ 1,5 s ; II: 3,0 s ≤ t ≤ 4,5 s;

U

i

= N

_i

t

∆

⋅ ∆Φ ; |U i | =

s 5 , 1

Vs 10 0 , 9 10 ⋅ ⋅

⁻⁵

= 0,60 mV I = R

U

_i

; I =

Ω

⋅

⁻

010 , 0

V 10 60 ,

0

³

= 6 , 0 ⋅ 10

⁻²

A

3 3.3 Nach dem lenzschen Gesetz bewirkt die Bewegung nach rechts einen Induk- tionsstrom entgegen dem Uhrzeigersinn. Die Folge ist eine nach rechts gerichtete Lorentzkraft, die durch die Zugkraft F r

_z

kompensiert werden muss.

5 3.4

B b I N F

F

_z

=

_L

=

_i

⋅ ⋅ ⋅ ; mit I = R U

_i

und U

_i

= N

_i

⋅ B ⋅ b ⋅ v folgt

I = R

v b B N

_i

⋅ ⋅ ⋅

; R

v )² b B N

F

_z

(

ⁱ

⋅ ⋅ ⋅

= 3 3.5

f 2 N

U

_m

=

_i

⋅ Φ

_m

⋅ π ^{; f =}

π

⋅ Φ

⋅ 2

N U

m i

m

; f =

π

⋅

⋅

⋅ 9 , 0 10

⁻

Vs 2 10

V 566 , 0

5

= 100 Hz

60 Summe