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Solutions Manual for University Physics with Modern Physics 2nd Edition by Bauer

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Full download: http://goo.gl/6hLSot Solutions Manual for University Physics with Modern Physics 2nd Edition by Bauer,2nd Edition, Bauer, Solutions Manual, University Physics with Modern Physics, Westfall
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  Chapter 2 : Motion in a Straight Line  45 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, 0 v  , 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 = = − 120 2 a a a  after applying the brakes. Before applying the brakes, the velocities of both cars are the same, = = 120 v v v  . When the cars have completely stopped, the final velocities are zero, = f  0 v  . = + = ⇒ = − 0f0 0 v v v at t a . Therefore, the ratio of time taken to stop is − −= = − −   0000 /time of car 11Ratio = .1time of car 22/2 v av a So the ratio is one half. Solutions Manual for University Physics with Modern Physics 2nd Edition by Bauer Full Download: http://downloadlink.org/product/solutions-manual-for-university-physics-with-modern-physics-2nd-edition-by-ba Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org  Bauer/Westfall: University Physics, 2E   46 2.21.  Here a  and v   are instantaneous acceleration and velocity. If a  = 0 and ≠ 0 v   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, ∆ = − fi vvv  . 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 srcinal 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.  Chapter 2 : Motion in a Straight Line  47 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 srcin ( x   = 0). Assume a constant acceleration = 0 aa .   Use the equations = + 0 v v at   and ( ) = + + 200 1/2 x x v t at  .   When the car has completely stopped, = 0 v   at = 0 t t  . = + ⇒ = − 0000 0 vatvat   Use the final stopping position as the srcin,  = 0 x   at = 0 t t  . = + + 20000 102 x v t at   Substituting = − 00 vat   and simplifying gives − + = ⇒ − = ⇒ = 22200000020 2110 0 22 x xatatxatat   Note that time 0 t   is the time required to stop from a distance 0 x  .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 m5.5 m/cm0.80 cm . The error in measurement is (0.05 cm) 5.5 m/cm ≈ 0.275 m   (round   at   the   end). So the scale is 5.5 ± 0.275 m/cm. The farthest distance of the car from the srcin is 2.9 ± 0.05 cm. Multiplying by the scale, 15.95 m, ( )( ) = = 0 0.3336 s1.998 s t  . The acceleration can be found using = 200 2/ a x t  : = = 22 2(15.95 m)7.991 m/s(1.998 s) a .   Because the scale has two significant digits, round the result to two significant digits: = 2 8.0 m/s. a   Since the error in the measurement is ∆ = 0 0.275 m, x   the error of the acceleration is   ( )( ) ∆∆ = = ≈ 20220 20.275 m20.1 m/s.1.998 s x at     Bauer/Westfall: University Physics, 2E   48 2.26.  Velocity can be estimated by computing the slope of a curve in a distance versus time plot . Velocity is defined by    = ∆ ∆ /. vxt   If acceleration is constant, then − ∆= =− ∆ fifi v v  v at t t  . (a) Estimate the slope of the dashed blue line. Pick two points: it is more accurate to pick a point that coincides with horizontal lines of the grid.   Choosing points t   = 0 s, x   = 0 m and t   = 6.25 s, x   = 20 m:   −= =− 20. m0 m3.2 m/s6.25 s0 s v   (b) Examine the sketch. There is a tangent to the curve at t   = 7.5 s. Pick two points on the line. Choosing points: t   = 3.4 s, x   = 0 m and t   = 9.8 s, x   = 60 m: −=− 60. m0 m = 9.4 m/s9.8 s3.4 s v   (c) From (a), v   = 3.2 m/s at t   = 2.5 s and from (b), v   = 9.4 m/s at t   = 7.5 s. From the definition of constant acceleration, −= = =− 2 9.4 m/s3.2 m/s6.2 m/s1.2 m/s.7.5 s2.5 s5.0 s a  2.27.  There are two rocks, rock 1 and rock 2. Both rocks are dropped from height h . Rock 1 has initial velocity = 0 v  and rock 2 has = 0 v v   and is thrown at = 0 t t  . Rock 1: = ⇒ = 2 12 2 hhgtt  g   Rock 2: = − + − ⇒ − + − − = 22000000 11()() ()()022 h v t t g t t g t t v t t h  This equation has roots − ± +− = 2000 2 v v ght t  g  . Choose the positive root since − > 0 ()0. tt  Therefore − += + 2000 2. vvghtt  g   Substituting = 2 ht  g   gives: +  = + − + − +   2200000 2222 or . v ghv v v h h ht  g g g g g g g     Chapter 2 : Motion in a Straight Line  49 2.28.  I want to know when the object is at half its maximum height. The wrench is thrown upwards with an initial velocity = = 0 (0). v t v  , = + − 200 1,2 x x v t gt    = − 0 , v v gt    and = 2 9.81 m/s.  g   At maximum height, v   = 0. = − ⇒ = − ⇒ = 00max0max  0 . v v gt v gt v gt    Substitute = max0 / tvg   into ( ) = + − 200 1/2. x x v t gt           = − = − = − =              22222000000max0 11112222 v v v v v v x v g  g g g g g g   Therefore, half of the maximum height is = 2012 4 v x  g  . Substitute this into the equation for x  . = = − ⇒ − + = 222200120121/21/2012 11 04224 vv xvtgtgtvt  gg   This is a quadratic equation with respect to 1/2 t  . The solutions to this equation are:     ± − ±   ± −        = = = = ±       220220000000012 11142412211222 v vvg vv vvv  g v t  ggg  g    Exercises  2.29. THINK: What is the distance traveled,  p , and the displacement d   if = 1 30.0 m/s v   due north for = 1 10.0 min t   and = 2 40.0 m/s v   due south for = 2 20.0 min t  ? Times should be in SI units: ( ) = = ⋅ 12 10.0 min60 s/min6.0010 s, t    ( ) = = ⋅ 23 20.0 min60 s/min1.2010 s. t    SKETCH: RESEARCH:  The distance is equal to the product of velocity and time. The distance traveled is = + 1122  p v t v t   and the displacement is the distance between where you start and where you finish, = − 1122 d v t v t  .  SIMPLIFY: There is no need to simplify.
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