Recit%20Ch%201(1) - PHY2053 RECITATIONS SPRING 2011...

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Unformatted text preview: PHY2053 RECITATIONS, SPRING 2011 Recitation 1 (Chapter 1) Jan 17-21 Ch. 1 # 15. The corners of a square lie on a circle of diameter D = 0.35 m. The side of the square has a length L . Find L . REASONING: Using the Pythagorean theorem (Equation 1.7), we find that the relation between the length D of the diagonal of the square (which is also the diameter of the circle) and the length L of one side of the square is 2 2 2 D L L L = + = . SOLUTION: Using the above relation, we have 0.35 m 2 or 0.25 m 2 2 D D L L = = = = Ch. 1 # 21. Three deer, A , B , and C , are grazing in a field. Deer B is located 62 m from dear A at an angle of 51° north of west. Deer C is located 77° north of east relative to dear A . The distance between deer B and C is 95 m. What is the distance between deer A and C ? ( Hint : Consider the law of cosines given in Appendix E .) REASONING: The drawing shows the location of each deer A, B, and C. b = 62 m c = 95 m γ = 180 ° - 51 ° - 77 ° = 52 ° γ α β A B C 77° 51° b a c Applying the law of cosines (given in Appendix E) to the geometry in the figure, we have a 2- 2 ab cos γ + (b 2- c 2 ) = which is an expression that is quadratic in a. It can be simplified to...
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This note was uploaded on 09/21/2011 for the course PHY 1500 taught by Professor Staff during the Spring '09 term at University of Central Florida.

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Recit%20Ch%201(1) - PHY2053 RECITATIONS SPRING 2011...

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