This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: midterm 01 – ALIBHAI, ZAHID – Due: Feb 14 2007, 11:00 pm 1 Question 1 part 1 of 1 10 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. ω has units of inverse time. t is time, m is mass, V is volume, ρ is density, and F is force. 1. v = ± ω radicalbig A 2 x 2 2. v = radicalbig 2 g y + radicalbigg r F m 3. y = A cos( k x ω t ) 4. g = F m + ρ m g V 5. F = mg bracketleftbigg 1 + v r g bracketrightbigg correct 6. v = radicalbig v 2 2 ax 7. F = mω 2 r 8. t = v + radicalBig v 2 + 2 ax a + v 2 a 2 t Explanation: • Check y = A cos( k x ω t ): Since k is an inverse length, k x is dimension less, and similarly, ωt is also dimensionless, so k x ω t is dimensionless, as an argument of a trigonometry function must always be. The value of a trigonometry function ( e.g. cos( k x ω 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 = mω 2 r : The dimension of force is [ F ] = ML T 2 . The dimension of right hand side is bracketleftbig mω 2 r bracketrightbig = M parenleftbigg 1 T parenrightbigg 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 : 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 L T 2 = T L . Therefore the right hand side of this equation is not dimensionally correct because quanti ties of different dimensions can not be added up together. • Check t = v + radicalBig v 2 + 2 ax a + v 2 a 2 t : Since [ ax ] = L 2 T 2 , we have bracketleftbigg radicalBig v 2 + 2 ax bracketrightbigg = L T = [ v ], therefore  v + radicalBig v 2 + 2 ax a = L T L T 2 = T and bracketleftbigg v 2 a 2 t bracketrightbigg = parenleftbigg L T parenrightbigg 2 parenleftbigg L T 2 parenrightbigg 2 T = T . midterm 01 – ALIBHAI, ZAHID – Due: Feb 14 2007, 11:00 pm 2 So this equation is dimensionally correct. • Check v = radicalbig v 2 2 ax : As mentioned earlier, this one is correct. • Check v = ± ω radicalbig A 2 x 2 : bracketleftBig ω radicalbig A 2 x 2 bracketrightBig = 1 T √ L 2 = L T = [ v ] ....
View
Full Document
 Spring '08
 Turner
 Acceleration, Velocity, m/s, Fundamental physics concepts, Fnet

Click to edit the document details