Homework02-solutions

# Homework02-solutions - miller (zdm77) Homework02 Chiu...

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Unformatted text preview: miller (zdm77) Homework02 Chiu (60180) 1 This print-out should have 26 questions. Multiple-choice 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.

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Homework02-solutions - miller (zdm77) Homework02 Chiu...

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