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Unformatted text preview: h ic e8 e k# i b iC a aha 1 Problem 11: We have a mass of m = 60 kg = 60 kg 1000 g 1 kg = 60 000 g . The volume of a sphere of radius r is V = 4 3 r 3 . If it has a density in g / cm 3 , then the total mass is m = V = 4 3 r 3 . Solving for the radius, we get = 3 radicalbigg 3 m 4 = 3 radicalBigg 3(60 000 g) 4 (19 . 5 g / cm 3 ) 9 cm . 1 Problem 21: Lets turn to a random page (I chose page 509) and count the number of words in the first inch, which is the first 5 lines. These lines contain 72 words (not counting the equations). The text is 9 1 4 inches high, but there are some equations, so lets round down to about 600 words per page. The book is about 1300 pages long, so this is about 800 000 words. 1 Problem 25: Lets say we have a cylinder about 2 m high and 0 . 5 m wide. (It would actually be shaped like some sort of conelike mound, but were only going for a very rough estimate.) Then the radius is r = 0 . 25 m and the volume is V = r 2 h = 0 . 4 m 3 = 4 10 5 cm 3 . To get the dollar value we multiply (4 10 5 cm 3 ) 19 . 3 g 1 cm 3 $10 1 g $80 000 000 . 1 Problem 30 ( a ): 1.8 m 2.4 m ( b ): 1.8 m 2.4 m ( c ): 2.4 m 1.8 m 1 1 Problem 41: The total displacement is (3 . 25 km 1 . 50 km) = 1 . 75 km to the North and 4.75 km to the West. The magnitude of the total displacement is radicalbig (1 . 75 km) 2 + (4 . 75 km) 2 = 5 . 06 km . The angle is Tan 1 parenleftbigg 1 . 75 km 4 . 75 km parenrightbigg = 20 . 2 North of West (clockwise from West). 3.25 km 1.5 km 4.75 km 1 Problem 42 ( a ): We have A = (1 . 30 cm , 2 . 25 cm) and B = (4 . 10 cm , 3 . 75 cm). Then A + B = ( (1 . 30 + 4 . 10) cm , (2 . 25 3 . 75) cm ) = (5 . 40 cm , 1 . 50 cm) . ( b ): The magnitude of A + B is  A + B  = radicalbig (5 . 40 cm) 2 + ( 1 . 50 cm) 2 = 5 . 60 cm ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
 Mass, Work

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