Homework 1 - homework 01 BAUTISTA, ALDO Due: Sep 7 2006,...

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Unformatted text preview: homework 01 BAUTISTA, ALDO Due: Sep 7 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 part 1 of 3 10 points Stokes law says F = 6 rv. F is a force r the radius and v the velocity. The parameter has the dimension of 1. [ ] = T 2 L M 2. [ ] = M T L correct 3. [ ] = T M 4. [ ] = M T 2 L 2 5. [ ] = T L M 6. [ ] = T 2 L 2 M 7. [ ] = M T L 2 8. [ ] = M T 2 L 9. [ ] = M T 10. [ ] = T L 2 M Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 is dimensionless. The dimen- sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 r v ] M L T 2 = L L T [ ] Solving for [ ] yields [ ] = M LT . Question 2 part 2 of 3 10 points Consider a simple pendulum which consists of a string with a bob attached to its end. Its period, ( i.e. , the time interval taken for the bob to complete one cycle of motion), may be written in the form T = k m x g y b z , where k is a dimensionless constant, g the magnitude of the gravitational acceleration, m the mass of the bob, and b the length of the string. The appropriate x , y , and z values are given respectively by 1. ( x, y, z ) = 1 ,- 1 2 ,- 1 2 2. ( x, y, z ) = (1 ,- 1 , 1) 3. ( x, y, z ) = ,- 1 2 , 1 2 correct 4. ( x, y, z ) = 1 , 1 2 , 1 2 5. ( x, y, z ) = (0 ,- 1 , 1) 6. ( x, y, z ) = 1 ,- 1 2 , 1 2 7. ( x, y, z ) = , 1 2 , 1 2 8. ( x, y, z ) = ,- 1 2 ,- 1 2 9. ( x, y, z ) = , 1 2 ,- 1 2 10. ( x, y, z ) = 1 , 1 2 ,- 1 2 Explanation: [ T ] = [ k m x g y b z ] T = M x L y T 2 y L z T = M x L y + z T- 2 y homework 01 BAUTISTA, ALDO Due: Sep 7 2006, 4:00 am 2 Hence x = 0, y + z = 0,- 2 y = 1 y =- 1 2 , z = 1 2 ( x, y, z ) = ,- 1 2 , 1 2 . Question 3 part 3 of 3 10 points Consider a piece of string which is placed along the x-axis. Let m be the mass of a segment of the string and x the length of this segment. The linear mass density, , of a piece of string is defined as = m x . Denote to be its mass density defined as = mass volume and A its cross sectional area. Let us write = x A y . Use dimensional analysis to determine the equations for x and y . 1. x =- 1 ,- 2 y- 3 x =- 1 2. x = 1 , 2 y + 3 x =- 1 3. x = 1 , 2 y- 3 x =- 1 correct 4. x =- 1 , 3 x + 2 y = 1 5. x =- 1 , 2 y + 3 x =- 1 6. x = 1 , 2 y- 3 x = 1 7. x =- 1 , 2 y- 3 x =- 1 8. x =- 1 , 2 y- 3 x = 1 9. x = 1 ,- 2 y- 3 x =- 1 10. x = 1 , 3 x + 2 y = 1 Explanation: [ ] = [ x A y ] M L- 1 = M x L- 3 x L 2 y = M x L 2 y- 3 x Equating both sides yields the equations x = 1 , 2 y- 3 x =- 1 . Question 4 part 1 of 1 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....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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Homework 1 - homework 01 BAUTISTA, ALDO Due: Sep 7 2006,...

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