HW1 - homework 01 – GAUTHIER, JOSEPH – Due: Sep 7 2007,...

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Unformatted text preview: homework 01 – GAUTHIER, JOSEPH – Due: Sep 7 2007, 1:00 am 1 Question 1, chap 1, sect 6. part 1 of 1 10 points A piece of pipe has an outer radius, an inner radius, and length as shown in the figure below. 3 9 c m 4 . 6 cm 2 . 9 cm b The density is 7 . 8 g / cm 3 . What is the mass of this pipe? Correct answer: 12 . 1848 kg (tolerance ± 1 %). Explanation: Let : r 1 = 4 . 6 cm , r 2 = 2 . 9 cm , ℓ = 39 cm , and ρ = 7 . 8 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross- sectional area times the length. Solution: V = ( π r 2 1 − π r 2 2 ) ℓ = π [ r 2 1 − r 2 2 ] ℓ = π [(4 . 6 cm) 2 − (2 . 9 cm) 2 ] (39 cm) = 1562 . 16 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 − r 2 2 ] ℓ = (7 . 8 g / cm 3 ) π [(4 . 6 cm) 2 − (2 . 9 cm) 2 ] (39 cm) = 12184 . 8 g = 12 . 1848 kg . Question 2, chap 1, sect 6. part 1 of 2 10 points Consider the dimension content of the cen- tripetal force. The force should depend on the mass m , the tangential speed v and the radius r . We write it in the form F = m x v y r z . Based on dimensional analysis, determine the powers x , y and z . 1. x = 1, y = 2, z = 1. 2. x = − 1, y = − 2, z = 1. 3. x = 1, y = − 2, z = 1. 4. x = 1, y = 2, z = − 1. correct 5. x = − 1, y = 2, z = − 1. 6. x = − 1, y = − 2, z = − 1. 7. x = 1, y = − 2, z = − 1. 8. x = − 1, y = 2, z = 1. Explanation: In order that F = m x v y r z to be dimen- sionally correct: [ F ] = M 1 L 1 T − 2 [ m x v y r z ] = M x parenleftBig L T parenrightBig y L z = M x L y + z T − y . By equating the powers of M , L , and T x = 1, y + z = 1, and − y = − 2. Therefore, x = 1, y = 2, and z = − 1. Question 3, chap 1, sect 6. part 2 of 2 10 points Consider the following set of equations, where s , s , x and r have units of length, t has units of time, v has units of velocity, g and a have units of acceleration, and k is dimensionless. homework 01 – GAUTHIER, JOSEPH – Due: Sep 7 2007, 1:00 am 2 Which one is dimensionally incorrect ? 1. a = g + k v t + v 2 s 2. v 2 = 2 a s + k s v t 3. t = k radicalbigg s g + a v correct 4. t = v a + x v 5. s = s + v t + v 2 a Explanation: For an equation to be dimensionally cor- rect, all its terms must have the same units. (1): t = v a + x v [ t ] = T bracketleftBig v a bracketrightBig + bracketleftBig x v bracketrightBig = L T − 1 L T − 2 + L L T − 1 = T + T = T is consistent. (2): a = g + k v t + v 2 s [ a ] = L T − 2 bracketleftbigg g + k v t + v 2 s bracketrightbigg = L T − 2 + L T − 1 T + L 2 T − 2 L = L T − 2 is also consistent. (3): t = k radicalbigg s g + a v [ t ] = T bracketleftbigg k radicalbigg s g + a v bracketrightbigg = radicalbigg L L T − 2 + L T − 2 L T − 1 = T + T − 1 is not dimensionally consistent....
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This homework help was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.

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HW1 - homework 01 – GAUTHIER, JOSEPH – Due: Sep 7 2007,...

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