This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: hinojosa (jlh3938) oldmidterm1 02 Turner (58185) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A piece of wire has a density of 5 . 6 g / cm. What is the mass of 11 . 7 cm of the wire? Correct answer: 65 . 52 g. Explanation: Let : = 5 . 6 g / cm and = 11 . 7 cm . Since g cm cm = g, we must multiply the den- sity by the length of the wire to obtain the mass. This can also be determined by the definition of linear mass density: = m so that m = = (5 . 6 g / cm) (11 . 7 cm) = 65 . 52 g . 002 10.0 points Identify the equation below which is dimen- sionally incorrect . A, x, y and r have units of length. k here has units of inverse length. v and v have units of velocity. a and g have units of acceleration. has units of inverse time. t is time, m is mass, V is volume, is density, and F = ma is force. 1. v = radicalbig A 2- x 2 2. y = A cos( kx- t ) 3. F = mg bracketleftbigg 1 + v r g bracketrightbigg correct 4. v = radicalbig 2 g y + radicalbigg r F m 5. g = F m + m g V 6. t =- v + radicalBig v 2 + 2 a x a + v 2 a 2 t 7. F = m 2 r 8. v = radicalbig v 2- 2 a x Explanation: Check y = A cos( kx- t ): Since k is an inverse length, kx is dimension- less, and similarly, t is also dimensionless, so kx- t is dimensionless, as an argument of a trigonometry function must always be. The value of a trigonometry function ( e.g. cos( kx- t )) is also dimensionless. So the right hand side has the dimension of length. [ y ] = L [ A cos( k x )] = L This equation is dimensionally correct. Check F = m 2 r : The dimension of force is [ F ] = ML T 2 . The dimension of right hand side is bracketleftbig m 2 r bracketrightbig = M (1 /T ) 2 L = M L T 2 So it is also dimensionally correct. Check v = radicalbig 2 g y + radicalbigg r F m : Since bracketleftbigg F m bracketrightbigg = L T 2 = [ g ], and [ y ] = [ r ] = L , bracketleftBig radicalbig 2 g y bracketrightBig = bracketleftBigg radicalbigg r F m bracketrightBigg = radicalbigg L 2 T 2 = L T The left hand side of the equation is [ v ] = L T . So this equation is also dimensionally correct. Check F = mg bracketleftbigg 1 + v r g bracketrightbigg : hinojosa (jlh3938) oldmidterm1 02 Turner (58185) 2 As mentioned earlier bracketleftbigg F m bracketrightbigg = [ g ], so 1 + v r g should be dimensionless. However, bracketleftbigg v r g bracketrightbigg = L/T L 2 /T 2 = T L Therefore the right hand side of this equation is not dimensionally correct because quanti- ties of different dimensions cannot be added up together. Check t =- v + radicalBig v 2 + 2 a x a + v 2 a 2 t : Since [ a x ] = L 2 T 2 , we have bracketleftbigg radicalBig v 2 + 2 a x bracketrightbigg = L T = [ v ], therefore - v + radicalBig v 2 + 2 a x a = L/T L/T 2 = T and bracketleftbigg v 2 a 2 t bracketrightbigg = ( L/T ) 2 ( L 2 /T ) 2 T = T So this equation is dimensionally correct.So this equation is dimensionally correct....
View Full Document
This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08