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Unformatted text preview: miller (zdm77) Homework02 Chiu (60180) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The square of the speed of an object undergo ing a uniform acceleration a is some function of a and the displacement s , according to the expression given by v 2 = ka m s n , where k is a dimensionless constant. Using dimensional analysis 1. v 2 = k a 2 s 2. v 2 = k radicalbigg a s 3. v 2 = k s a 4. v 2 = k a s correct 5. v 2 = k a s 6. v 2 = k a 2 s 2 7. v 2 = k a 2 s 8. v 2 = k a s 9. v 2 = k s 3 a 10. v 2 = k a 3 s Explanation: Solution: If the dimensions (length, mass, time) of the right hand side of a formula do not equal those of the left hand side, you know right off that formula is NOT correct. The dimensions of velocity are [ v ] = L T = L T 1 . So the dimensions of the left hand side of the formula in question are [ v 2 ] = L 2 T 2 = L 2 T 2 . On the right hand side of the formula, there is k , which is dimensionless, multiplied by the acceleration [ a ] = L T 2 = L T 2 and the displacement [ s ] = L . What exponent will give the dimensions of v 2 ? [ v 2 ] = [ k a m s n ] . L 2 T 2 = ( L T 2 ) m L n = L m + n T 2 m . For the dimensions of the two sides of the formula to be the same, the exponent of L has to be equal to 2 and the exponent of T has to be 2 m + n = 2 (1) 2 m = 2 . (2) Equation 2 implies that m = 1 . Substitute this into Eq. 1 to get 1 + n = 2 n = 1 . Therefore v 2 = k a s . 002 10.0 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. Which one is dimensionally incorrect ? 1. t = k radicalbigg s g + a v correct 2. t = v a + x v 3. a = g + k v t + v 2 s 4. s = s + v t + v 2 a miller (zdm77) Homework02 Chiu (60180) 2 5. v 2 = 2 a s + k s v t 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 = LT 1 LT 2 + L LT 1 = T + T = T It is consistent. (2) a = g + k v t + v 2 s [ a ] = LT 2 bracketleftbigg g + kv t + v 2 s bracketrightbigg = LT 2 + LT 1 T + L 2 T 2 L = LT 2 It is also consistent. (3) t = k radicalbigg s g + a v [ t ] = T bracketleftbigg k radicalbigg s g + a v bracketrightbigg = radicalbigg L LT 2 + LT 2 LT 1 = T + T 1 This is not dimensionally consistent. (4) v 2 = 2 a s + k s v t [ v 2 ] = L 2 T 2 bracketleftbigg 2 a s + k s v t bracketrightbigg = LT 2 L + LLT 1 T 1 = L 2 T 2 This is also consistent....
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This note was uploaded on 09/30/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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