hmwk-1-K09-soln - Physics 317K Homework 1 SOLUTIONS 1. B 1...

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Physics 317K Homework 1 SOLUTIONS 1. B 1 mi = 1.609 km; 1 gal = 3.785 L. So 30 . 2mi / gal = 30 . 2 mi gal ± 1 . 609 km 1mi ²± 1 gal 3 . 785 L ² =12 . 8km / L . 2. B A nanonsecond is 10 - 9 s by defnition. 3. C The kilogram is the SI unit oF mass. 4. C (5 . 0 × 10 4 ) × (3 . 0 × 10 6 )=15 × 10 10 =1 . 5 × 10 11 5. B In scientifc notation this is 3 . 30 × 10 - 4 . There are three signifcant fgures. 6. D In scientifc notation this is 6 . 500 × 10 2 . There are Four signifcant fgures. 7. B 27.3 has three signfcant fgures and the implied precision is about 0.05. So, the sum 27.3+7.513 cannot be more precisely known than 0.05. Hence, the resulting 34.813 should be reported as 34.8, since more signifcant fgures than this would imply unrealistic precision. 8. E 65 mi hr ± 1hr 3600 s 1609 m ² =29m / s 9. B The radius r =2 . 4cm=2 . 4 × 10 - 2 m and the surFace area is given by A =4 πr 2 . This gives A π (2 . 4 × 10 - 2 m) 2 =7 . 2 × 10 - 3 m 2 . 10. A L =0 . 75 Ft is the length oF each side oF the cube. Using 1 m = 3.281 Ft, the volume V oF the cube in m 3 is V = L 3 =(0 . 75 Ft) 3 ± 1m 3 . 281 Ft ² 3 . 2 × 10 - 2 m 3 . 1
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11. C Each term in the equation v = at + bt 3 + c/t must have the same units, m/s. Thus, the units of a must be m/s 2 , the units of b must be m/s 4 , and the units of c must be m. 12. D A = BC = ± L 2 M 3 ² = B ± L T ² so B = ± L 2 M 3 ²± T L ² = ± LT M 3 ² 13. E A = B m C n [ L 2 T ]= ± L T ² m [ L 2 T 2 ] n =[ L ] m +2 n [ T ] - m +2 n The units must be the same on both sides of an equation, so 2 = m +2 n and 1 = - m n . Solve these equations. For instance, add them (3 = 4 n ) and subtract them (1 = 2 m ). Thus, m = 1 2 and n = 3 4 . 14. A The displacement is given by Δ x = x f - x i . This quantity is negative only for case A, where Δ x = x f - x i = - 1m - 4m= - 5m. 15. A In the time interval Δ t , one car will cover (60 km/h)Δ t and the other will cover (40 km/h)Δ t . When the sum of these equals 250 km, Δ t = 250 km/(60 km/h + 40 km/h) = 250 km/100 km/h = 2.5 h. 16. C The car travels 60 km + 40 km = 100 km in total. It takes 3/4 h to travel 60 km at 80 km/h. It takes 2/3 h to travel 40 km at 60 km/h. So the total time duration of the trip is 0.75 h + 0.67 h = 1.42 h. The average speed, then, is v ave = 100 km/1.42 h = 70 km/h. 17. B The particle is at rest when it has zero velocity. Its position is x ( t )=5 . 5+8 . 0 t - 3 . 0 t 3 , so its velocity is v ( t )= dx dt =8 - 9 t 2 Thus, v ( t ) = 0 requires 0 = 8 - 9 t 2 . Solve for t . t = ³ 8 9 s=0 . 94 s 18. D If x 0 is the position of the rocket at t = 0, its position at a later time t is x = x 0 + ´ t 0 vdt . Thus, it travels a distance x - x 0 = µ t 0 vdt = µ t 0 ( bt ct 2 )=( 1 2 bt 2 + 2 3 ct 3 ) | t 0 = 1 2 bt 2 + 2 3 ct 3 . 2
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19. B Let t 1 =3s , v 1 = v ( t 1 ) = 20 cm / s, t 2 = 7 s, and v 2 = v ( t 2 )=2cm / s, . Then the average acceleration a ave is given by a = v 2 - v 1 t 2 - t 1 = 2cm / s - 20 cm / s 7s - 3s = - 18 4 cm / s 2 = - 4 . 5cm / s 2 . 20. C Since a = d 2 x dt 2 , take the second derivative with respect to time of each x ( t ) function. Particle 1 has a = - 2 . 7 and particle 2 has a =2 . 7 (i.e., constant values), while particle 2 has a . 7 t and particle 4 has a = - 2 . 7 t (i.e., not constants).
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This note was uploaded on 09/14/2009 for the course CH 310 N taught by Professor Blocknack during the Fall '08 term at University of Texas at Austin.

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hmwk-1-K09-soln - Physics 317K Homework 1 SOLUTIONS 1. B 1...

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