QUESTION 5 The uniform lamina shown in the diagram is formed by removing a square DEFG of side length a from the equilateral triangle ABC of side length Sa. The centre of mass of DEFG lies on the perpendicular bisector of AC Given that AC and DG are parallel and a distance a apart, work out the dist of the centre of mass of the whole lamina from B. O 0.898 O 1.89a O 289a O 3.89a W Sa

QUESTION 5 The uniform lamina shown in the diagram is formed by removing a square DEFG of side length a from the equilateral triangle ABC of side length Sa. The centre of mass of DEFG lies on the perpendicular bisector of AC Given that AC and DG are parallel and a distance a apart, work out the dist of the centre of mass of the whole lamina from B. O 0.898 O 1.89a O 289a O 3.89a W Sa

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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