Physics Solution Manual for 1100 and 2101

Solving this equation for d yields 2 d r h 2 r 2 2 2

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Unformatted text preview: onvert the distance from meters to miles, we use the relation 1609 m = 1 mi (see the page facing the inside of the front cover of the text). SOLUTION a. The Pythagorean theorem (Equation 1.7) states that the square of the hypotenuse is equal to the sum of the squares of the sides, or ( R + h ) = d + R . Solving this equation for d yields 2 d= ( R + h )2 − R 2 2 2 = R + 2 Rh + h − R ( 2 2 2 ) = 2 Rh + h 2 = 2 6.38 ×106 m (1.6 m ) + (1.6 m ) = 4500 m 2 b. Multiplying the distance of 4500 m by a factor of unity, (1 mi)/(1609 m) = 1, the distance (in miles) from the person's eyes to the horizon is 1 mi d = ( 4500 m )(1) = 4500 m = 2.8 mi 1609 m ______________________________________________________________________________ ( ) 69. REASONING AND SOLUTION If D is the unknown vector, then A + B + C + D = 0 requires that DE = –(AE + BE + CE) or DE = (113 u) cos 60.0° – (222 u) cos 35.0° – (177 u) cos 23.0° = –288 units The minus sign indicates that DE has a direction of due west. Also, DN = –(AN + BN + CN) or DN = (113 u) sin 60.0° + (222 u) sin 35.0° – (177 u) sin 23.0° = 156 units ______________________________________________________________________________ CHAPTER 2 KINEMATICS IN ONE DIMENSION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (b) Displacement, being a vector, conveys information about magnitude and direction. Distance conveys no information about direction and, hence, is not a vector. 2. (c) Since each runner starts at the same place and ends at the same place, the three displacement vectors are equal. 3. (c) The average speed is the distance of 16.0 km divided by the elapsed time of 2.0 h. The average velocity is the displacement of 0 km divided by the elapsed time. The displacement is 0 km, because the jogger begins and ends at the same place. 4. (a) Since the bicycle covers the same number of meters per second everywhere on the track, its speed is constant. 5. (e) The average velocity is the displacement (2.0 km due north) divided by the elapsed time (0.50 h), and the direction of the velocity is the same as the direction of the displacement. 6. (c) The average acceleration is the change in velocity (final velocity minus initial velocity) divided by the elapsed time. The change in velocity has a magnitude of 15.0 km/h. Since the change in velocity points due east, the direction of the average acceleration is also due east. 7. (d) This is always the situation when an object at rest begins to move. 8. (b) If neither the magnitude nor the direction of the velocity changes, then the velocity is constant, and the change in velocity is zero. Since the average acceleration is the change in velocity divided by the elapsed time, the average acceleration is also zero. 9. (a) The runners are always moving after the race starts and, therefore, have a non-zero average speed. The average velocity is the displacement divided by the elapsed time, and the displacement is zero, since the race starts and finishes at the same place. The average acceleration is the change in the velocity divided by the elapsed time, and the velocity changes, since the contestants start at rest and finish while running. 10. (c) The equations of kinematics can be used only when the acceleration remains constant and cannot be used when it changes from moment to moment. 11. (a) Velocity, not speed, appears as one of the variables in the equations of kinematics. Velocity is a vector. The magnitude of the instantaneous velocity is the speed. Chapter 2 Answers to Focus on Concepts Questions ( 43 ) 12. (b) According to one of the equation of kinematics v = v0 + 2 ax, with v0 = 0 m/s , the 2 2 displacement is proportional to the square of the velocity. ( ) 13. (d) According to one of the equation of kinematics x = v0t + 1 at , with v0 = 0 m/s , the 2 2 displacement is proportional to the acceleration. 14. (b) For a single object each equation of kinematics contains four variables, one of which is the unknown variable. 15. (e) An equation of kinematics ( v = v0 + at ) gives the answer directly, since the initial velocity, the final velocity, and the time are known. 16. (c) An equation of kinematics x = 1 2 ( v0 + v ) t gives the answer directly, since the initial velocity, the final velocity, and the time are known. ( ) 17. (e) An equation of kinematics v = v0 + 2ax gives the answer directly, since the initial 2 2 velocity, the final velocity, and the acceleration are known. 18. (d) This statement is false. Near the earth’s surface the acceleration due to gravity has the approximate magnitude of 9.80 m/s2 and always points downward, toward the center of the earth. 19. (b) Free-fall is the motion that occurs while the acceleration is solely the acceleration due to gravity. While the rocket is picking up speed in the upward direction, the acceleration is not just due to gravity, but is due to the combined effect of gravity and the engines. In fact, the effect of the engines is greater than the effect of gravity. Only when the e...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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