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### ch0223

Course: CHEM 101, Fall 2008
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2 Describing Chapter Motion: Kinematics in One Dimension Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, and so the downhill putt is more difficult. 68. (a) The train's constant speed is vtrain function of time is given by xtrain a 2 6 .0 m s , and the location of the empty box car as a v t ra i n t 6.0 m s t . The fugitive has v0 0 m s...

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Coursehero >> Pennsylvania >> Arcadia University >> CHEM 101

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2 Describing Chapter Motion: Kinematics in One Dimension Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, and so the downhill putt is more difficult. 68. (a) The train's constant speed is vtrain function of time is given by xtrain a 2 6 .0 m s , and the location of the empty box car as a v t ra i n t 6.0 m s t . The fugitive has v0 0 m s and 4 .0 m s until his final speed is 8.0 m s . The elapsed time during acceleration is v v0 8 .0 m s tacc 2.0 s . Let the origin be the location of the fugitive when he starts to a 4 .0 m s 2 run. The first possibility to consider is, "Can the fugitive catch the train before he reaches his maximum speed?" During the fugitive's acceleration, his location as a function of time is given by xfugitive x0 v0 t 1 at 2 0 0 1 4.0 m s 2 t 2 . For him to catch the train, we must have 2 2 x t ra i n x f u g i t i ve 6 .0 m s t 1 2 4.0 m s 2 t 2 . The solutions of this are t 0 s , 3 s . Thus the fugitive cannot catch the car during his 2.0 s of acceleration. Now the equation of motion of the fugitive changes. After the 2.0 s acceleration, he runs with a constant speed of 8.0 m s . Thus his location is now given (for times t 2 s ) by the following. xfugitive 1 2 4 .0 m s 2 2 .0 s 2 8 .0 m s t 2 .0 s 8 .0 m s t 8 .0 m . So now, for the fugitive to catch the train, we again set the locations equal. x t ra i x n f u g it i ve 4 .0 s 6 .0 m s t 8 .0 m s 8 .0 m s t 8 . 0 m 4 .0 s 8 .0 m 24 m . t 4 .0 s (b) The distance traveled to reach the box car is given by x fu g i t i v e t 69. Choose downward to be the positive direction, and y 0 0 to be at the roof from which the stones are dropped. The first stone has an initial velocity of v0 0 and an acceleration of a g . Eqs. 211a and 2-11b (with x replaced by y) give the velocity and location, respectively, of the first stone as a function of time. v v0 at v1 gt1 y y0 v0 t 1 at 2 y1 1 gt12 . 2 2 The second stone has the same initial conditions, but its elapsed time t 1.50 s , and so has velocity and location equations as follows. v2 g t1 1.50 s v2 y2 1 2 g t1 1.50 s 2 The second stone reaches a speed of v2 12.0 m s at a time given by 2 g 9 .8 0 m s The location of the first stone at that time is y1 1 2 t1 1 .5 0 s 1 .5 0 s 1 2 .0 m s 2 .7 2 s . gt12 1 2 9.80 m s 2 2 2.72 s 2 3 6 .4 m . 2 The location of the second stone at that time is y2 1 2 g t1 1.50 s 1 2 9.80 m s 2 .7 2 1 .5 0 s 2 7.35 m . Thus the distance between the two stones is y1 y2 3 6 .4 m 7 .3 5 m 2 9 .0 m . 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 34
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Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition70. To find the average speed for the entire race, we must take the total distance divided by the total time. If one lap is a distance of L , then the total distance will be 10 L . The time elaps
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension73. For the car to pass the train, the car must travel the length of the train AND the distance the train 95 km h t or travels. The distance the car travels can thus be written as either d car vcar
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThus the total time to reach the maximum altitude is t2 7 s 8 .9 4 s236 s .(e) For this part of the problem, the rocket has v0 0 m s , a 9.8 m s , and a displacement of 1600 m (it falls from
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension(a) The position vs. time graphs would qualitatively look like the graph shown here.x(b) The time to overtake the speeder occurs when the speeder Police car has gone a distance of 750 m. The time
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionand it takes 4.5 s for the stone to reach the final location y velocity, v0 B .y v0 B y0 v0 t1 20 . Use Eq. 2-11b to find the starting9 .8 0 m s 22at 201 6 .6 mv0 B 4 .5 s1 24 .5 s18
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One DimensionThe intent of the method was probably to use the change in air pressure between the ground level and the top of the building to find the height of the building. The very small difference in time mea
Arcadia University - CHEM - 101
CHAPTER 3: Kinematics in Two Dimensions; Vectors Answers to Questions1. Their velocities are NOT equal, because the two velocities have different directions. 2. (a) During one year, the Earth travels a distance equal to the circumference of its orbit, bu
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors10. To find the initial speed, use the slingshot to shoot the rock directly horizontally (no initial vertical speed) from a height of 1 meter. The vertical displacement of the rock can be related to the time
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editioncrossing the river. For the upstream rower, some of his rowing effort goes into battling the current, and so his &quot;cross river&quot; speed will be only a fraction of his rowing speed. 17. The baseball
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectorsy4.Given that Vx given by V6.80 units and V yVx2 V y2 6 .8 0 2tan17.40 units, the magnitude of V isVx Vy7 .4 02x10.0 units . The directionis given by an angle of axis.7.40 6.8047 o , or 47 o b
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition10. AxBx Cx4 4 .0 co s 2 8 .0 o2 6 . 5 c o s 5 6 .0 o 3 1 .0 c o s 2 7 0 o3 8 .8 51 4 .8 2 0 .0AyBy Cy4 4 .0 s in 2 8 .0 o2 6 .5 s i n 5 6 . 0 o 3 1 .0 si n 2 7 0 o2 0 .6 62 1 .9 7 3 1
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors(b )ABCx3 8. 851 4 .8 20 .02 4 .0 3ABCABCy2 0 .6 6 2 1 .9 72 4 .0 323 1 .027 3 .6 3t an17 3 .6 37 7 .573.63 24.037 1 .9 o(c)CABx0 . 0 3 8 .8 51 4 .8 22 4 .0 3CABy3 1 .0 2 0 . 6
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionyy0v y 0t1 2ayt 26 .5 m001 29 .8 m s 2 t 2t2 6 .5 m 9 .8 m s 21 .1 5 s ecThe horizontal displacement is calculated from the constant horizontal velocity. x vx t 3 .5 m s 1 . 1 5 s e c
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectorsyy0v y 0t1 2ayt 24 5 .0 m1 29 .8 0 m s 2 t 2t2 4 5 .0 m 9 .8 0 m s 23 .0 3 secThe horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity: x vx t vx
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThus on the Moon, the person can jump 6 times farther . 26. (a) Choose downward to be the positive y direction. The origin is the point where the bullet 2 leaves the gun. In the vertical directio
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; VectorsSince the ball s height is less than 3.00 m, the football does not clear the bar . It is 0.76 m too low when it reaches the horizontal location of the goalposts. 30. Choose the origin to be where the project
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionvyvy0atv0 s i n0gt6 5 .0 m s s i n 3 7 . 0 o9 .8 0 m s 21 0 .4 s6 3 .1 m s(d) The magnitude of the velocity is found from the x and y components calculated in part c) above.v2 vx2 vy
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors34. Choose the origin to be the location from which the balloon is fired, and choose upward as the positive y direction. Assume the boy in the tree is a distance H up from the point at which the balloon is f
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(c) The horizontal component of the speed of the supplies upon landing is the constant horizontal speed of 69.4 m/s. The vertical speed is found from Eq. 2-11a. vy vy 0 ayt 8 .3 7 m s 9 . 8 0 m s
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors(b) The position of the boat after 3.00 seconds is given by d vboat rel.t 1.20, 2.30 m s 3.00 secshore3.60 m downstream, 6.90 m across the riverAs a magnitude and direction, it would be 7.8 m away from th
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition44. Call the direction of the boat relative to the water the x direction, and upward the y direction. Also see the diagram. v passenger v passenger v boat rel.rel. water rel. boat waterv passen
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors(b) The time is found from the constant velocity relationship for either the x or y directions. y 75 m y vyt t 17 0 s vy 0 .4 5 m s 48. (a) Call the direction of the flow of the river the x direction, and th
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionSince the police car doesn t accelerate until t1.00 s , the correct answer is t 15.8 s .51. Take the origin to be the location at which the speeder passes the police car. The speed of the speed
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectorsd xyd x2d y25022525 5 .9 md xy2For the vertical motion, consider another right triangle, made up of d xy as one leg, and the vertical displacement d z as the other leg. See the second figure, and t
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition1 2 5 k m h c o s 4 5 o 0 v w in d x v w in d x 8 8 . 4 k m h . From the y components of the above equation: 1 2 5 sin 4 5 o 155 vwind-y vwind-y 155 125 sin 45oThe magnitude of the wind velocity
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors62. The minimum speed will be that for which the ball just clears the fence; i.e., the ball has a height of 7.5 m when it is 95 m horizontally from home plate. The origin is at home plate, with upward as the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionTo find the full time of flight of the ball, set the final y location to be y = 0, and again use Eq. 2-11b. y y0 v y 0 t 1 a y t 2 0.0 m 2.50 m 1 9.80 m s 2 t 2 2 20 .7 1 4 3 0 .7 1 4 s 9.80 m s
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectorsthe river is t1D2 v2 u2D 2 v2 u2. The same relationship would be in effect for crossing. v u2 The speed v must be greater than the speed u . The velocity of the boat relative to the shore when going upst
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFor the vertical motion, applying Eq. 2-11b. y y0 v y 0 t 1 a y t 2 v0 sin t 1 gt 2 2 2 Substitute the expression for the time of flight and solve for the initial velocity.y v0 s in t1 2gt2v
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectorsv2 vx2 vy2 .3 1 m s tan121 0 .2 m s 77 o21 0 .4 5 8 m s10 m st an1vy vx1 0 .2 m s 2 .3 1 m sbelow the horizontal71. (a) Choose the origin to be the location where the car leaves the ramp, and
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition28 m s sin 61o28 m s 21 22sin 2 61o 9 .8 0 m s421 29 .8 0 m s 20 .9 m5 .0 3 4 s , 0 .0 3 6 5 s Choose the positive time, since the ball was hit at t 0 . The horizontal displacement of t
Arcadia University - CHEM - 101
Chapter 3Kinematics in Two Dimensions; Vectors(b) Since the horizontal velocity is known and the horizontal distance is known, the time of flight can be found from the constant velocity equation for horizontal motion. x 55 m x vx t t 4 .1 1 1 s v x 1 3.
Arcadia University - CHEM - 101
CHAPTER 4: Dynamics: Newton s Laws of Motion Answers to Questions1. The child tends to remain at rest (Newton s 1st Law), unless a force acts on her. The force is applied to the wagon, not the child, and so the wagon accelerates out from under the child,
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion9.When giving a sharp pull, the key is the suddenness of the application of the force. When a large, sudden force is applied to the bottom string, the bottom string will have a large tension in it. Because of t
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThe free body diagram below illustrates this. The forces are FT G , the force on team 1 from the1ground, FT G , the force on team 2 from the ground, and FTR , the force on each team from the ro
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion3.Use Newton s second law to calculate the tension.F FT ma 9 60 k g 1 .2 0 m s 2 1 .1 5 1 0 3 N4.In all cases, W (a) WEarth (b) WMoon (c) WMars (d) WSpacemg , where g changes with location.7 6 k g 9 .8 m s
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition8.We assume that the fish line is pulling vertically on the fish, and that the fish is not jerking the line. A free-body diagram for the fish is shown. Write Newton s 2nd law for the fish in the
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion13. In both cases, a free-body diagram for the elevator would look like the adjacent diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition17. (a) There will be two forces on the skydivers their combined weight, and the upward force of air resistance, FA . Choose up to be the positive direction. Write Newton s 2nd law for the skydiv
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion(b) While the player is in the air, the only force on the player is their weight. See the second diagram. 21. (a) Just as the ball is being hit, if we ignore air resistance, there are two main forces on the ball
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionF1 FT1 F2 FT2FT1 mg FT2 FT1mg0 31 N3.2 k g 9.8 m s 2 FT1 mg mg 2 mg 02 3.2 kg 9.8 m s 263 N(b) Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchang
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion28. Since all forces of interest in this problem are horizontal, draw the free-body diagram showing only the horizontal forces. FT1 is the tension in the coupling between the locomotive and the first car, and it
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionhorizontal force. Using Newton s 2nd law, we have the following. 720 N cos 22o FP cos 22o Fx FP cos 22o max ax m 65 kg10.27 m s 21.0 101 m s 2(b) Eq. 2-11a is used to find the final speed. The
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion(b) All of the vertical forces on each block add up to zero, since there is no acceleration in the vertical direction. Thus for each block, FN mg . For the horizontal direction, we haveF F F12 F21 F23 F32 F m1
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFT FT a2m1 g m2 g m1 m2m1a2 m2 a2 gFTm1 g m1a2 m2 g 9.8 m s 2 m2 a2 m1 g m2 g m1a2 m2 a2m1 g m1a2 3.2 kg 2.2 kgm1 m2 3.2 kg 2.2 kg The lighter block starts with a speed of 0, and moves a di
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion(a) To start the box moving, the pulling force must just overcome the force of static friction, and that means the force of static friction will reach its maximum value of Ffr F . Thus we have for the starting m
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionmg sin FfrFfrma x ma x 15.0 kg 9.80 m s 2 s in 3 2 o 0.30 m s 2 73.40 N 73 Nmg sinNow solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from t
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion45. A free-body diagram for the bobsled is shown. The acceleration of the sled is found from Eq. 2-11c. The final velocity also needs to be converted to m/s. 1m s v 60 km h 16.667 m s 3.6 km h2 v 2 v02ax x x0
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFN m1 m2 g , and so Ffr F kN horizontal direction. Fx FP Ffr m1 m2 aa FP m1 Ffr m2 FPkkm1m2 g . Write Newton s 2nd law for them1 m2m2 g620 N0 . 1 5 1 8 5 k g 9 .8 m s 2 185 kgm11 .9 m
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion51. We assume that the child starts from rest at the top of the slide, and then slides a distance x x0 along the slide. A force diagram is shown for the child on the slide. First, ignore the frictional force and
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionFxmg sinmaag sinUse Eq. 2-11c with v0 3.0 m s and v stopping. 2 v 2 v0 2a x x0x x0 v22 v00 m s to find the distance that it slides before03.0 m s22a2 9.8 m s 2 sin 22.0o1.2 mThe ne
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of MotionFfrm g si n 1 0 1 .3 Nma 101 Nkm g sina18.0 kg9.80 m s 2sin 37.0o0.270 m s 2FfrkFNmg cosFfrk101.3 N 18.0 kg 9.80 m s 2 cos 37.0omg cos0.71956. Consider a free-body diagram for the box, showing
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionvertical velocity is (5.66 m/s) sin 30o = 2.83 m/s, the vertical acceleration is 9.8 m/s2, and the vertical displacement is 10.0 m. Use Eq. 2-11c to find the final vertical speed. 2 2 v y v y 0 y
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of MotionWrite Newton s 2nd law for both the x and y directions. Note that the net force in the y direction is zero, since the block does not accelerate in the y direction. Fy FN mg cos 0 FN mg cosm Now equate the two e
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionacceleration of the box, it is apparent that FN mg , and so Ffr the horizontal direction to find the acceleration.smg . Use Newton s 2nd law inFFfrsmgmaasg0. 60 9 . 8 m s 25.88 m s 2
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion2 2 (b) To have an acceleration of zero, the expression for the acceleration must be zero. 1 sin cos k ag 0 1 sin cos 0 k 2ag1 sinkcos9.8 m s 21 sin 25o0.15 cos 25o2.2 m s 21 sink1 sin 25o cos 25oco
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition68. (a) Assume that the earthquake is moving the Earth to the right. If an object is to FN hold its place , then the object must also be accelerating to the right with the Earth. The force that w
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of Motion70. Assume that kinetic friction is the net force causing the deceleration. See the free-body diagram for the car, assuming that the right is the positive direction, and the direction of motion of the skidding c
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition73. Consider the free-body diagram for the cyclist in the mud, assuming that the cyclist is traveling to the right. It is apparent that FN mg since there is no vertical acceleration. Write Newton
Arcadia University - CHEM - 101
Chapter 4Dynamics: Newton s Laws of MotionUse Eq. 2-11a with v00 to find the final velocity (takeoff speed).v v0atvv0at04 .6 m s 2 1 8 s82 m s76. (a) Consider the free-body diagrams for both objects, initially stationary. As sand is added, the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition78. (a) To find the minimum force, assume that the piano is moving with a constant velocity. Since the piano is not accelerating, FT 4 Mg . For the lower pulley, since the tension in a rope is th