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Unformatted text preview: homework 01 – FONTENOT, BRIAN – Due: Jan 21 2008, 4:00 am 1 Question 1, chap 1, sect 1. part 1 of 2 10 points Two points in the xy plane have cartesian coordinates ( x 1 , y 1 ) and ( x 2 , y 2 ), where x 1 = 9 . 3 m, y 1 = − 10 m, x 2 = − 10 m, and y 2 = 5 . 4 m. Determine the distance between these points. Correct answer: 24 . 6911 m (tolerance ± 1 %). Explanation: By simple geometry of triangle, the dis tance between two points is d = radicalBig [ x 2 − x 1 ] 2 + [ y 2 − y 1 ] 2 = braceleftbigg bracketleftBig ( − 10 m) − (9 . 3 m) bracketrightBig 2 + bracketleftBig (5 . 4 m) − ( − 10 m) bracketrightBig 2 bracerightbigg 1 2 = 24 . 6911 m . Question 2, chap 1, sect 1. part 2 of 2 10 points What is the angle between the line con necting the two points and xaxis (measured counterclockwise, within the limits of − 180 ◦ to +180 ◦ )? Correct answer: 141 . 413 ◦ (tolerance ± 1 %). Explanation: The angle between the line connecting the two points and xaxis (positive in counter clockwise direction) is θ = arctan bracketleftbigg y 2 − y 1 x 2 − x 1 bracketrightbigg = arctan bracketleftbigg (5 . 4 m) − ( − 10 m) ( − 10 m) − (9 . 3 m) bracketrightbigg = arctan bracketleftbigg (15 . 4 m) ( − 19 . 3 m) bracketrightbigg = 2 . 46812 rad = 141 . 413 deg .1010998877665544332211 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 ( − 10, 5 . 4) (9 . 3, − 10) 1 4 1 . 4 1 3 d e g Question 3, chap 1, sect 6. part 1 of 1 10 points The velocity of a transverse wave traveling along a string depends on the tension of the string, F , which has units of force, and its mass per unit length, μ . Assume v = F x μ y . The powers of x and y may be determined based on dimensional analysis. The values of x and y for this problem are 1. x = − 1 2 , y = 1 2 2. x = − 1 2 , y = − 1 2 3. x = 1 , y = − 1 4. x = 1 2 , y = 1 2 5. x = − 1 , y = − 1 6. x = 0 , y = 1 7. x = 1 , y = 0 8. x = − 1 , y = 1 9. x = 1 2 , y = − 1 2 correct homework 01 – FONTENOT, BRIAN – Due: Jan 21 2008, 4:00 am 2 10. x = 1 , y = 1 Explanation: Recall that the dimensions of velocity, ten sion, and mass per unit length are [ v ] = L T , [ F ] = ML T 2 , [ μ ] = M L hence L T = [ F x μ y ] = parenleftbigg ML T 2 parenrightbigg x parenleftbigg M L parenrightbigg y = M x + y L x − y T − 2 x Equating the powers of M yields 0 = x + y . Equating the powers of L yields 1 = x − y . Equating the powers of T yields − 1 = − 2 x . Then, − 1 = − 2 x ⇒ x = 1 2 and 0 = x + y ⇒ y = − 1 2 . These values leads to the equation v = radicalBigg F μ , which is the correct expression for the velocity of a traveling wave. Note that dimensional analysis has led to the correct equation, up to an overall dimensionless constant. For the present case the constant happens to be 1....
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This note was uploaded on 03/22/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas.
 Fall '06
 MCCANN

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