solution_pdf4 - magalhaes (bam2734) Homework #3 Erskine...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: magalhaes (bam2734) Homework #3 Erskine (58200) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Two points have cartesian coordinates (8 . 3 m , 10 m) and ( 10 m , 4 . 9 m). Find the distance between these points. Correct answer: 23 . 5987 m. Explanation: Let : x 1 = 8 . 3 m , y 1 = 10 m , x 2 = 10 m , and y 2 = 4 . 9 m . The distance is d = radicalBig ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 = { [ 10 m (8 . 3 m)] 2 + [4 . 9 m ( 10 m)] 2 } 1 / 2 = 23 . 5987 m . 002 (part 2 of 2) 10.0 points What is the angle between the line connect- ing the two points and the x-axis (measured counter-clockwise from the x-axis and within the limits of 180 to +180 )? Correct answer: 140 . 847 . Explanation: The angle is = arctan parenleftbigg y 2 y 1 x 2 x 1 parenrightbigg = arctan bracketleftbigg 4 . 9 m ( 10 m) 10 m (8 . 3 m) bracketrightbigg = 140 . 847 . ( 10, 4 . 9) (8 . 3, 10) 003 10.0 points Two airplanes leave an airport at the same time. The velocity of the first airplane is 680 m / h at a heading of 37 . 9 . The velocity of the second is 640 m / h at a heading of 183 . How far apart are they after 2 . 9 h? Correct answer: 3652 m. Explanation: Let : v 1 = 680 m / h , 1 = 37 . 9 , v 2 = 640 m / h , and 2 = 183 . Under constant velocity, the displacement for each plane in the time t is d = v t. These displacements form two sides of a tri- angle with the angle = 2 1 = 145 . 1 between them. The law of cosines applies for SAS, so the distance between the planes is d = radicalBig d 2 1 + d 2 2 2 d 1 d 2 cos . Since 2 d 1 d 2 cos = 2 (1972 m) (1856 m) cos 145 . 1 = 6 . 00357 10 6 m 2 , then magalhaes (bam2734) Homework #3 Erskine (58200) 2 d = bracketleftbig (1972 m) 2 + (1856 m) 2 6 . 00357 10 6 m 2 bracketrightbig 1 / 2 = 3652 m . 004 10.0 points A vector of magnitude 2 CANNOT be added to a vector of magnitude 3 so that the magnitude of the resultant is 1. 2 2. 3 3. more information is needed. 4. correct 5. 5 6. 1 Explanation: The resultant vector of two added vectors can be zero only when the two vectors have the same magnitudes and opposite directions or are both zeroes. 005 (part 1 of 2) 10.0 points Consider two vectors vector A and vector B and their resul- tant vector A + vector B . The magnitudes of the vectors vector A and vector B are, respectively, 14 and 8 . 2 and they act at 75 to each other. vector A vector B vector A + vector B Find the magnitude of the resultant vector vector A + vector B ....
View Full Document

This note was uploaded on 04/13/2010 for the course PHY phy303k taught by Professor Erskine during the Spring '09 term at University of Texas-Tyler.

Page1 / 10

solution_pdf4 - magalhaes (bam2734) Homework #3 Erskine...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online