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    Dwnload Full University Physics With Modern Physics 2nd Edition Bauer Solutions Manual PDF

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    Dwnload Full University Physics With Modern Physics 2nd Edition Bauer Solutions Manual PDF

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    Dwnload Full University Physics With Modern Physics 2nd Edition Bauer Solutions Manual PDF

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    University Physics with Modern Physics 2nd Edition Bauer Solutions Manual

    University Physics with Modern


    Physics 2nd Edition Bauer Solutions
    Manual
    Visit to get the accurate and complete content:

    https://testbankfan.com/download/university-physics-with-modern-physics-2nd-edition
    -bauer-solutions-manual/

    Visit TestBankFan.com to get complete for all chapters


    University Physics with Modern Physics 2nd Edition Bauer Solutions Manual

    Chapter 2: Motion in a Straight Line

    Chapter 2: Motion in a Straight Line

    Concept Checks
    2.1. d 2.2. b 2.3. b 2.4. c 2.5. a) 3 b) 1 c) 4 d) 2 2.6. c 2.7. d 2.8. c 2.9. d

    Multiple-Choice Questions
    2.1. e 2.2. c 2.3. c 2.4. b 2.5. e 2.6. a 2.7. d 2.8. c 2.9. a 2.10. b 2.11. b 2.12. d 2.13. c 2.14. d 2.15. a 2.16. c

    Conceptual Questions
    2.17. Velocity and speed are defined differently. The magnitude of average velocity and average speed are the
    same only when the direction of movement does not change. If the direction changes during movement, it
    is known that the net displacement is smaller than the net distance. Using the definition of average velocity
    and speed, it can be said that the magnitude of average velocity is less than the average speed when the
    direction changes during movement. Here, only Christine changes direction during her movement.
    Therefore, only Christine has a magnitude of average velocity which is smaller than her average speed.
    2.18. The acceleration due to gravity is always pointing downward to the center of the Earth.

    It can be seen that the direction of velocity is opposite to the direction of acceleration when the ball is in
    flight upward. The direction of velocity is the same as the direction of acceleration when the ball is in flight
    downward.
    2.19. The car, before the brakes are applied, has a constant velocity, v0 , and zero acceleration. After the brakes
    are applied, the acceleration is constant and in the direction opposite to the velocity. In velocity versus
    time and acceleration versus time graphs, the motion is described in the figures below.

    2.20. There are two cars, car 1 and car 2. The decelerations are a1 = 2a2 = −a0 after applying the brakes. Before
    applying the brakes, the velocities of both cars are the same, v=
    1 v=
    2 v0 . When the cars have completely
    v0
    stopped, the final velocities are zero, v f = 0 . v f =v0 + at =0 ⇒ t=− . Therefore, the ratio of time taken
    a
    time of car 1 −v0 / −a0 1
    to stop is Ratio = = = . So the ratio is one half.
    time of car 2  1  2
    −v0 /  − a0 
     2 

    45

    Visit TestBankFan.com to get complete for all chapters


    Bauer/Westfall: University Physics, 2E

    2.21. Here a and v are instantaneous acceleration and velocity. If a = 0 and v ≠ 0 at time t, then at that moment
    the object is moving at a constant velocity. In other words, the slope of a curve in a velocity versus time
    plot is zero at time t. See the plots below.

    2.22. The direction of motion is determined by the direction of velocity. Acceleration is defined as a change in
    velocity per change in time. The change in velocity, ∆v , can be positive or negative depending on the
    values of initial and final velocities, ∆v = v f − vi . If the acceleration is in the opposite direction to the
    motion, it means that the magnitude of the objects velocity is decreasing. This occurs when an object is
    slowing down.
    2.23. If there is no air resistance, then the acceleration does not depend on the mass of an object. Therefore,
    both snowballs have the same acceleration. Since initial velocities are zero, and the snowballs will cover the
    same distance, both snowballs will hit the ground at the same time. They will both have the same speed.
    2.24. Acceleration is independent of the mass of an object if there is no air resistance.

    Snowball 1 will return to its original position after ∆t , and then it falls in the same way as snowball 2.
    Therefore snowball 2 will hit the ground first since it has a shorter path. However, both snowballs have the
    same speed when they hit the ground.

    46
    Chapter 2: Motion in a Straight Line

    2.25. Make sure the scale for the displacements of the car is correct. The length of the car is 174.9 in = 4.442 m.

    Measuring the length of the car in the figure above with a ruler, the car in this scale is 0.80 ± 0.05 cm.
    Draw vertical lines at the center of the car as shown in the figure above. Assume line 7 is the origin (x = 0).

    Assume a constant acceleration a = a0 . Use the equations v= v0 + at and x = x0 + v0t + (1/ 2 ) at 2 . When
    the car has completely stopped, v = 0 at t = t 0 .
    0 =+ v0 at 0 ⇒ v0 = −at 0
    Use the final stopping position as the origin, x = 0 at t = t 0 .
    1
    0 =x0 + v0t 0 + at 02
    2
    Substituting v0 = −at 0 and simplifying gives
    1 1 2x
    x0 − at 02 + at 02 = 0 ⇒ x0 − at 02 = 0 ⇒ a = 2 0
    2 2 t0
    Note that time t 0 is the time required to stop from a distance x0 .First measure the length of the car. The
    length of the car is 0.80 cm. The actual length of the car is 4.442 m, therefore the scale is
    4.442 m
    = 5.5 m/cm . The error in measurement is (0.05 cm) 5.5 m/cm ≈ 0.275 m (round at the end).
    0.80 cm
    So the scale is 5.5 ± 0.275 m/cm. The farthest distance of the car from the origin is 2.9 ± 0.05 cm.
    =
    Multiplying by the scale, 15.95 m, t 0 ( =0.333 )( 6 s ) 1.998 s . The acceleration can be found using
    2(15.95 m)
    a ==
    2 x0 / t 02 : a = 7.991 m/s 2 . Because the scale has two significant digits, round the result to
    (1.998 s)2
    two significant digits: a = 8.0 m/s 2 . Since the error in the measurement is ∆x0 =
    0.275 m, the error of the
    acceleration is
    2∆x0 2 ( 0.275 m )
    =
    ∆a = ≈ 0.1 m/s 2 .
    ( )
    2
    t 02 1.998 s

    47
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