{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution_pdf4

# solution_pdf4 - magalhaes(bam2734 Homework#3 Erskine(58200...

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

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 θ = 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) θ 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 , θ 1 = 37 . 9 , v 2 = 640 m / h , and θ 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 α = θ 2 θ 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 α . Since 2 d 1 d 2 cos α = 2 (1972 m) (1856 m) cos 145 . 1 = 6 . 00357 × 10 6 m 2 , then

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

View Full Document
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. 0 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 . Correct answer: 17 . 9629. Explanation: Let : a = 14 , b = 8 . 2 , and θ = 75 . b γ r a γ = 180 75 = 105 , so applying the law of cosines, r 2 = a 2 + b 2 2 a b cos γ = (14) 2 + (8 . 2) 2 2 (14) (8 . 2) cos 105 = 322 . 665 r = 322 . 665 = 17 . 9629 . 006 (part 2 of 2) 10.0 points Find the angle between the direction of the resultant vector A + B and the direction of the vector A . Correct answer: 26 . 164 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

solution_pdf4 - magalhaes(bam2734 Homework#3 Erskine(58200...

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

View Full Document
Ask a homework question - tutors are online