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Title: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
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Tom Obrien
HNC Level 4
Unit Number and title: Unit 1
Analytical Methods for Engineers
Assignment Title: Algebraic
Methods
Assignment Number: 1
Date: October 2016
1
Tom Obrien
HNC Level 4
Contents Page:
Page 1:
Page 2:
-
Title Page
-
Contents Page
-
Question 1
Question 2a
-
Question 2a
Question 2b
Question 2c
-
Question 2c
Question 3
-
Question 4ai
Question 4aii
-
Question aii
Question 4aiii
Question 4aiv
-
Question 5
Question 6
Question 7a
-
Question 7b
Question 7c
Question 8a
-
Question 8b
Question 9
Page 3:
Page 4:
Page 5:
Page 6:
Page 7:
Page 8:
Page 9:
Page 10:
2
Tom Obrien
HNC Level 4
3
1)
Divide 6
2
− 6
−4
−4
+2
+0
by 2
−
+2
and state the remainder as a single term
...
72 kΩ and t is 1710*C
...
72 3
...
088 × 10
1710
For this conductor at what temperature will the resistance be 3
...
=
3
...
41 ×
3
...
41
...
026 =
...
×
×
×
×
×
ln 1
...
088 × 10
0
...
088 × 10
=
0
...
088 × 10
×
×
×
= 511
...
7939 + 0
...
2596 =
27 =
1
×
(27)
1
...
2596
1
×
(27)
1
+1=
(27)
1
×
(27)
3
...
852 = 4
...
4132 + 0
...
0001
1+
1 + 1
...
4203 = 2
...
7041 + 1 =
7
1
+1 =
(130)
(130)
1
×
(130)
1
(130)
1
...
7041
4aiii)
242 +
242 = 1
0
...
7796 = 1
1+
1 + 3
...
537 = 4
...
2827 + 1 =
1
+1 =
(242)
(242)
1
(242)
1
×
(242)
1
...
2827
4aiv)
328°15′ +
1
(242)
328°15′ = 1
0
...
2769 = 1
1+
1 + 2
...
383 = 1
...
852 + 1 =
(1 + 2
2+2
5)
) + (1 −
) =2
(1 +
=
2+2
=2
) + (1 − 2
+
= (1 +
2+2
1
(328°15′)
3
...
6114
4b)
(1 +
8
= 2+2
+
)
) + (1 +
= 2 sinh /2
= 2 × 50 sinh 40/2 × 50
= 100 × sinh 40/100
= 100 ×
ℎ 0
...
075
6)
= sinh
ℎ =2
2 = argsinh 2 = 1
...
3 + (23 − 1) × 0
...
3 + 24 × 0
...
5
= 1
6 = 100
6 = 100
6
=
100
1200
100
=
2
3
4 ℎ
5 ℎ
8a)
1
1+
2
= 1
...
6438 = 164
3 = 100 × 1
...
6438 = 444
5 = 100 × 1
...
5% is equivalent to 0
...
5% is equivalent to 0
...
=
N
= [ (1 + 0
...
025)]
=
=
(1 + 0
...
025)
N (1 + 0
...
025)
(1 + 0
...
0773
(1 − 0
...
9506
1
...
9506 = 1
...
41%, which is how much the flywheels energy increases by
Title: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers
Description: HNC HND in Mechanical Electrical Electronic Engineering Unit 1 - Analytical Methods for Engineers