Tutorial_1_Solution.. BENE1133 Electronic Engineering
2
Tutorial 1 Selected Answer – BENE1133 Chapter 1 10) a) 9 6 6 1 2 12 2 2 12 9 10 (10 10 )(5 10 ) 1.125 0.02 kQ Q F kN r 32 0.90 F kN Resultant force acting on Q2 is 12 23 0.225 F F F kN b) Resultant force acting on Q1 is 12 23 0.675 F F F kN 11) 2 2 9 19 5 2 31 ( )( ) 9 10 1.6 10 6.3 10 9.1 10 0.02 e k e e F ma r a ms 12) a) 8 8.2 10 F N b) 2 6 1 2.2 10 mv F r v ms 13) 14 24 34 24 34 0.2 0.054 0.084 4 cos 0.054 0.084 0.0408 5 x F N F N F N F F F N 24 14 sin 0.166 y F F F N 2 2 0.17 x Y F F F N Direction, 0 tan 4.069 76.2 y x F F
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Transcript of Tutorial_1_Solution.. BENE1133 Electronic Engineering
Tutorial 1 Selected Answer – BENE1133
Chapter 1
10)
a) 9 6 6
1 212 2 2
12
9 10 (10 10 )(5 10 )1.125
0.02
kQ QF kN
r
32 0.90F kN
Resultant force acting on Q2 is
12 23 0.225F F F kN
b) Resultant force acting on Q1 is
12 23 0.675F F F kN
11)
2
29 19
5 2
31
( )( )
9 10 1.6 106.3 10
9.1 10 0.02
e
k e eF m a
r
a ms
12) a) 88.2 10F N
b)
2
6 12.2 10
mvF
r
v ms
13)
14
24
34
24 34
0.2
0.054
0.084
4cos 0.054 0.084 0.0408
5x
F N
F N
F N
F F F N
24 14sin 0.166yF F F N
2 2 0.17x YF F F N
Direction,
0
tan 4.069
76.2
y
x
F
F
Chapter 2
2) 23000 /Wb m
3) 37.5 Wb
4) 56.4 Wb
13) 597r
14) 233333 /At Wb
15) 150mF At
16) 750 /H At m
18) 175indv V
