F10P1AChap1Sol

# F10P1AChap1Sol - Introduction and Vectors SOLUTIONS TO...

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1 Introduction and Vectors SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time *P1.4 For either sphere the volume is V = 4 3 ! r 3 and the mass is m = V = 4 3 " r 3 . We divide this equation for the larger sphere by the same equation for the smaller: m l m s = 4 r l 3 3 4 r s 3 3 = r l 3 r s 3 = 5 . Then r l = r s 3 = 4.50 cm 1.71 ( ) = 7.69 cm . Section 1.2 Dimensional Analysis P1.7 (a) This is incorrect since the units of ax [ ] are m 2 2 , while the units of v [ ] are . (b) This is correct since the units of y [ ] are m, and cos kx ( ) is dimensionless if k [ ] is in m ! 1 . Section 1.3 Conversion of Units *P1.14 N atoms = m Sun m atom = 1.99 ! 10 30 kg 1.67 ! 10 " 27 kg = 1.19 ! 10 57 atoms

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2 Introduction and Vectors Section 1.4 Order-of-Magnitude Calculations P1.19 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 ! 4 ! 3 = 48 m 3 , while the volume of one ball is 4 ! 3 0.038 m 2 " # \$ % & ' 3 = 2.87 ( 10 ) 5 m 3 . Therefore, one can fit about 48 2.87 ! 10 " 5 ~ 10 6 ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is 1 6 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the
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## This note was uploaded on 11/27/2011 for the course PHYS 1A taught by Professor Onuchic during the Fall '07 term at UCSD.

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F10P1AChap1Sol - Introduction and Vectors SOLUTIONS TO...

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