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Unformatted text preview: oldmidterm 01 NOORANI, ZOHEB Due: Feb 11 2008, 4:00 am 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered Jupiterlike planet has an average radius 11 . 1times that of the Earth and a mass 316 times that of the Earth. Calculate the ratio of new planets mass density to the mass density of the Earth. Correct answer: 0 . 231056 (tolerance 1 %). Explanation: Let : R np = 11 . 1 R E and M np = 316 M E . The volume of a sphere of radius R is V = 4 3 R 3 . The (average) density of a body is the ratio of the bodys mass to its volume, = M V . Planets are spherical, so the (average) density of a planet of a given mass M and a given radius R is = M 4 3 R 3 . Comparing the newly discovered planet to the Earth, we have np E = M np 4 3 R 3 np M E 4 3 R 3 E = M np M E parenleftbigg R np R E parenrightbigg 3 = 316 (11 . 1) 3 = . 231056 . Question 2, chap 1, sect 5. part 1 of 1 10 points A piece of wire has a density of 9 . 4 g / cm. What is the mass of 11 . 2 cm of the wire? Correct answer: 105 . 28 g (tolerance 1 %). Explanation: Let : = 9 . 4 g / cm and = 11 . 2 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 = = (9 . 4 g / cm) (11 . 2 cm) = 105 . 28 g . Question 3, chap 1, sect 6. 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. has units of inverse time. t is time, m is mass, V is volume, is density, and F is force. 1. F = m 2 r 2. v = radicalbig A 2 x 2 3. t = v + radicalBig v 2 + 2 ax a + v 2 a 2 t 4. v = radicalbig 2 g y + radicalbigg r F m 5. F = mg bracketleftbigg 1 + v r g bracketrightbigg correct 6. v = radicalbig v 2 2 ax 7. g = F m + m g V 8. y = A cos( k x t ) oldmidterm 01 NOORANI, ZOHEB Due: Feb 11 2008, 4:00 am 2 Explanation: Check y = A cos( k x t ): Since k is an inverse length, k x is dimension less, and similarly, t is also dimensionless, so k x t is dimensionless, as an argument of a trigonometry function must always be. The value of a trigonometry function ( e.g. cos( k x 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 parenleftbigg 1 T parenrightbigg 2 L = M L T 2 ....
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This note was uploaded on 09/16/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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