16_PartUniversity Physics Solution

16_PartUniversity Physics Solution - 1-16 Chapter 1 " j A B...

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1-16 Chapter 1 () && 4.00 3.00 5.00 2.00 ×= + × = AB i j i j " " 20.0 8.00 15.0 6.00 × −× ii i jj i But && && 0 ×=×= ii jj and && & , i j k , × =− j ik so ( ) &&& 8.00 15.0 23.0 . + −= k k k " " The magnitude of × " " is 23.0. EVALUATE: Sketch the vectors A " and B " in a coordinate system where the xy -plane is in the plane of the paper and the z -axis is directed out toward you. Figure 1.58 By the right-hand rule × " " is directed into the plane of the paper, in the -direction. z This agrees with the above calculation that used unit vectors. 1.59. IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude. SET UP: 120.0 φ = ° . EXECUTE: (a) The direction of A&B " " is into the page (the -direction z ). The magnitude of the vector product is 2 sin 2.80 cm 1.90 cm sin120 4.61 cm AB == # . (b) Rather than repeat the calculations, Eq. (1.23) may be used to see that B&A " " has magnitude 4.61 cm 2 and is in the -direction z + (out of the page). EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is ( ) ( ) 2.80 cm cos60.0 1.90 cm sin60 zx yy x CA BA B =−= ° ° ( ) 2 2.80 cm sin60.0 1.90 cm cos60.0 4.61 cm −°° = . This gives the same result. 1.60. IDENTIFY: Area is length times width. Do unit conversions. SET UP: 1 mi 5280 ft = . 3 1 ft 7.477 gal = . EXECUTE: (a) The area of one acre is 2 11 1 8 80 640 mi mi mi , so there are 640 acres to a square mile. (b) 2 2 2 1 mi 5280 ft 1 acre 43,560 ft 640 acre 1 mi ⎛⎞ ××= ⎜⎟ ⎝⎠ (all of the above conversions are exact). (c) (1 acre-foot) 35 3 7.477 gal 43,560 ft 3.26 10 gal, 1 ft =×= × which is rounded to three significant figures. EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-foot is much larger than a gallon. 1.61. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from this the radius. SET UP: The earth has mass 24 E 5.97 10 kg m and radius 6 E 6.38 10 m r . The volume of a sphere is 3 4 3 Vr π = . 33 1.76 g/cm 1760 km/m ρ . EXECUTE: (a) The planet has mass 25 E 5.5 3.28 10 kg mm × . 25 22 3 3 1.86 10 m 1760 kg/m m V × = × . 1/3 22 3 74 [ 1 . 8 6 1 0 m ] 1.64 10 m 1.64 10 km 44 V r ππ × = = (b) E 2.57 rr = EVALUATE: Volume V is proportional to mass and radius r is proportional to V , so r is proportional to m . If the planet and earth had the same density its radius would be EE (5.5) 1.8 = . The radius of the planet is greater than this, so its density must be less than that of the earth.
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Units, Physical Quantities and Vectors 1-17 1.62. IDENTIFY and SET UP: Unit conversion. EXECUTE: (a) 9 1.420 10 cycles/s, f so 10 9 1 s7 . 0 41 0 s 1.420 10 × for one cycle. (b) 12 10 3600 s/h 5.11 10 cycles/h 7.04 10 s/cycle × (c) Calculate the number of seconds in 4600 million 9 years 4.6 10 y and divide by the time for 1 cycle: 97 6 10 (4.6 10 y)(3.156 10 s/y) 2.1 10 cycles 7.04 10 s/cycle 2 ×× × (d) The clock is off by 1 s in 5 100,000 y 1 10 y, so in 9 4.60 10 y × it is off by 9 4 5 4.60 10 (1 s) 4.6 10 s 11 0 ⎛⎞ × ⎜⎟ × ⎝⎠ (about 13 h).
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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16_PartUniversity Physics Solution - 1-16 Chapter 1 " j A B...

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