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fund_odd_final - FUNDAMENTALS A.1 A.3(a chemical(b...

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FUNDAMENTALS A.1 (a) chemical; (b) physical; (c) physical A.3 The temperature, the humidity, and the evaporation of water are physical properties. The ripening of oranges is a chemical change. A.5 (a) intensive; (b) intensive; (c) extensive; (d) extensive A.7 3 3 112.32 g 1 mL 29.27 mL 23.45 mL 1 cm 19.3 g cm = = ⎠ ⎝ = m d V A.9 3 3 , rearranging gives 0.750 carat 200 mg 1 g 1 carat 1000 mg 3.51 g cm 0.0427 cm = = ⎞ ⎛ ⎞ ⎛ = ⎟ ⎜ ⎟ ⎜ ⎠ ⎝ = m m d V V d A.11 3 22 3 10 4 3 3 3.95 10 g 1 pm (138 pm) 1 10 cm 35.9 g cm π = ⎞ ⎛ × = ⎟ ⎜ × = m d V Because the density of metallic uranium is much less than the density of a uranium atom, the metallic form of uranium must contain considerable empty space. SM-3
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A.13 (a) 3 3 0.213 g 1.100 cm 0.531 cm 0.212 cm 0.213 g 0.1238 cm 1.72 g cm = = × × = = m d V This determination is more precise because the volume is not limited to 2 significant figures as it is in part (b). (b) 3 3 41.003 g 39.753 g 20.37 mL 19.65 mL 1.250 g 1 mL 0.72 mL 1 cm 1.7 g cm = = ⎞⎛ = ⎟⎜ ⎠⎝ = m d V A.15 2 K 2 2 1 2 2 2 1 2 1 h 1000 m 1 (4.2 kg)(14 km h ) 2 3600 s 1 km 32 kg m s 32 J = ⎞ ⎛ = ⎟ ⎜ ⎠ ⎝ = = E mv A.17 -1 -1 =2.8 metric tons, 100 km hr , 50 km hr = = i f m v v 2 1 2 = K E mv ( ) 3 ( ) 2 2 -2 6 2 -2 1 10 kg 2.8 metric tons 2 1 metric ton 100 km 1 hr 1 min. 1 hr 60 min. 60 sec. 4.32 kg km s 4.32 10 kg m s 4,320 kJ = ⎞⎛ ⎞⎛ ⎟⎜ ⎟⎜ ⎠⎝ ⎠⎝ = = × = K init E SM-4
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( ) 3 ( ) 2 2 -2 6 2 -2 1 10 kg 2.8 metric tons 2 1 metric ton 50 km 1 hr 1 min. 1 hr 60 min. 60 sec. 0.27 kg km s 0.27 10 kg m s 270 kJ = ⎞⎛ ⎞⎛ ⎟⎜ ⎟⎜ ⎠⎝ ⎠⎝ = = × = K final E 3 ( ) ( ) (4,320 270) kJ=4,050 kJ= 4.0 10 kJ (2 SF) = × K init K final E E This amount of energy could have been recovered, neglecting friction and other losses, or used to drive the vehicle up a hill. -2 9.81 ms = = P E mgh g Setting potential energy equal to 4,050 kJ=4.05 kg m 2 s -2 and solving for height gives 6 2 -2 -2 4.05 10 kg m s 147 m=150 m (2 SF) (2800 kg)(9.81 ms ) × = = = p E h mg A.19 -2 2 -2 1 kg (40.0 g)(9.81 m s )(0.50 m) 1000 g 0.20 kg m s for one raise of a fork. = = P E mgh = For 30 raises, 2 -2 (30)(0.20 kg m s ) 6.0 J = A.21 (a) The energy is all potential energy before the ball is dropped. After the ball has fallen halfway, half of the energy has been converted to kinetic energy. -2 2 -2 1 2 1 (0.95 kg)(9.81 m s )(13.9 m) 2 65 kg m s 65 J = = = = K E mgh (b) When the ball hits the floor, all of the energy has been converted to kinetic energy. SM-5
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-2 2 2 -2 2 (0.95 kg)(9.81 m s )(13.9 m) 1.3 10 kg m s 1.3 10 J = = = × = × K E mgh A.23 We need to use the expansion given in Exercise A.22 to help solve this problem. We also need to recognize that = P E egh for the small difference in distance, h, can be represented by subtracting P E at distance r
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