- 412-25 and B =-18.14A =-16.-1-64Find:e.) BB"; f.) A(B + A");g.) A? – 3A + 61.

To find (e) B.BT (f) A(B+AT) (g) A2-3A+6𝕀 Given A=01-4-16565-1B=2-24-18-6450

Answered: 1 -4 -2 and B =-1 4 A =-1 -6 6 -1 4…

Engineering

Mechanical Engineering

1 -4 -2 and B =-1 4 A =-1 -6 6 -1 4 Find : e.) BB"; f.) A(B + A"); g.) A? – 3A + 61.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provide resistance in changing its state of velocity or motion for coming to a state of rest. In other words, Newton's first law of motion describes the law of inertia of a rigid body.

Question

- 41
2
-2
5 and B =-1
8.
1
4
A =-1
6.
-1
-6
4
Find:
e.) BB"; f.) A(B + A");
g.) A? – 3A + 61.

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Q12. The simple connecting rod mechanism shown in Figure Q12 has a connecting rod length AB of 142mm and a crank radius OA of 46mm. The connecting rod has a centre of gravity C, which is located 32mm from end A. a) b) c) d) e) When the crank angle 0 is 305° and the crankshaft speed (w) is 6500rpm clockwise as shown, determine by kinematic analysis: The magnitude and direction of the angular velocity of the connecting rod. The magnitude and direction of the velocity of the piston. The magnitude and direction of the angular acceleration of the connecting rod. The magnitude and direction of the acceleration of the piston. The magnitude of the acceleration of the centre of gravity (C) in the X and Y directions. Bo a Fig. 12 Connecting Rod Mechanism A Ꮎ a) b) c) d) e) 131.2 rad/s (ACW) 30.6 m/s to the left 122782.7 rad/s² (ACW) 9953.8 m/s² to the right x: 11712.9 m/s² to the right y: 13524.2 m/s² upwards 4
9- O A) 3*F(AD)=(5.385)*F(AE) B) F(AC)=(5.385)*F(AB) C) 2*F(AD)=(5.385)*F(AC) O D) F(AE)=W O E) 4*F(AD)=(5.385)*F(AB)