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ch0110

Course: CHEM 101, Fall 2008
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1 Introduction, Chapter Measurement, Estimating 47. A cubit is about a half of a meter, by measuring several people s forearms. Thus the dimensions of Noah s ark would be 150 m long , 25 m wide, 15 m high . The volume of the ark is found by multiplying the three dimensions. V 150 m 25 m 15 m 5 .6 2 5 1 0 4 m 3 6 10 4 m 3 48. The volume of the oil will be the area times the thickness. The area is r2 d 2 ,...

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1 Introduction, Chapter Measurement, Estimating 47. A cubit is about a half of a meter, by measuring several people s forearms. Thus the dimensions of Noah s ark would be 150 m long , 25 m wide, 15 m high . The volume of the ark is found by multiplying the three dimensions. V 150 m 25 m 15 m 5 .6 2 5 1 0 4 m 3 6 10 4 m 3 48. The volume of the oil will be the area times the thickness. The area is r2 d 2 , and so 2 V d2 t 2 d 2 V t 1000cm 3 2 1m 3 100 cm 2 10 10 m 3 103 m . 49. Consider the diagram shown. L is the distance she walks upstream, which is about 120 yards. Find the distance across the river from the diagram. d t an 6 0 o d L tan 60o 120 yd tan 60 o 210 yd L d 60o L 210 yd 3 ft 1 yd 0 .3 0 5 m 1 ft 190 m 50. 8s 1y 1y 3.156 10 s 7 100% 3 10 5 % 51. volume The of a sphere is found by V VMoon VEarth VMoon 4 3 4 3 4 3 4 3 3 r3 . 2.21 1019 m 3 3 3 RMoon 3 REarth 3 RMoon 4 3 1 .74 1 0 6 m REarth RMoon 3 6.38 106 m 1.74 106 m 49.3 . Thus it would take about 49.3 Moons to create a volume equal to that of the Earth. o o 52. (a) 1.0 A o 1.0 A o 10 10 o m 1 nm 10 9 m 1 fm 10 m 15 0.10 nm 1A (b) 1.0 A 1.0 A 10 10 o m 1.0 105 fm 1A o (c) 1.0 m 1.0 m 1A 10 m 10 o 1.0 1010 A o (d) 1.0 ly 1.0 ly 9.46 1015 m 1 ly 1A 10 10 o m 9.5 10 25 A 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. 10
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Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition53. (a) Note that sin 15.0o1000.259 and sin 15.5o 0.5o 100 3% 15.0o0.267 .sin sin1008 1030.2591003%(b) Note that sin 75.0o0.966 and sin 75.5o0.5o0.968 .sin 100 2 103100 0.2% 75.
Arcadia University - CHEM - 101
CHAPTER 2: Describing Motion: Kinematics in One Dimension Answers to Questions1. A car speedometer measures only speed. It does not give any information about the direction, and so does not measure velocity. By definition, if an object has a constant vel
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition7.If an object is traveling to the north but slowing down, it has a northward velocity and a southward acceleration. The velocity of an object can be negative when its acceleration is positive.
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimensionvd AB t ABd BC t BCd km d km 70 k m hd km d km 90 km h78.75 km h . The average speed is 78.75 km/h.16. The sounds will not occur at equal time intervals because the longer any particular nut f
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editiondirection. The object stops moving at t = 90 s and stays at rest until about t = 108 s. Then the object begins to move to the right again, at first with a large acceleration, and then a lesser ac
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimensiona ve s p e e dd total ttotal2 5 7 .5 k m 3.33 h77 km h8.The speed of sound is intimated in the problem as 1 mile per 5 seconds. The speed is calculated by: d i s t an c e 1 mi 1 6 1 0 m s p ee
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionALTERNATE SOLUTION: The speed of the car relative to the truck is 88 km h 75 km h 13 km h . In the reference frame of the truck, the car must travel 110 m to catch it. 0.11 km 3600 s t 30.46 s 13
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension17. (a) The average acceleration of the sprinter is a (b ) av t1 0 .0 m s 0 .0 m s 1.35 s7 .4 1 m s 2 .7 .4 1 m s21 km 10 00 m3600 s 1h29 .6 0 1 0 4 k m h 218. The time can be found from t
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionGraph of the velocity30 25 v (m/s) 20 15 10 5 0 0 1 2 3 t (s ) 4 5 6a ( m/s 2) 6 5 4 3 2 1 0Graph of the acceleration0123 t (s )45621. By definition, the acceleration is av v02 5 m s
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension27. The final velocity of the driver is zero. The acceleration is found from Eq. 2-11c with v solving for a .0 andav2 2xv02 x0085 km h1m s 3 .6 k m h22 0 .8 0 m3 .4 8 4 1 0 2 m s 2 9 .8 m
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionSet the two expressions for40 m tpass 2 5 m s t pa ss 80s 2 8. 9 4 sxpassing equal to each other in order to find the time required to pass.car2 5 m s t p as s1 21.0 m s 2 t p2ass40 m1 2
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension33. Choose downward to be the positive direction, and take y 0 velocity is v0 0 , and the acceleration is a (2-11b), with x replaced by y.20 at the top of the cliff. The initial9.80 m s . The dis
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThe height can be calculated from Eq. 2-11c, with a final velocity of vv22 v00 at the top of the path.22a yy0yy0v2v 2a2 000 21 4 .7 m s 9 .8 m s 211 m38. Choose downward to be th
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One DimensionWe could do this in general.yn dn11 2 1 2gn 2yn 1 yn1 2 1 2gn 1 gn11 22yn 1 g n221 2gn 21 2gn12n22n 1 n 2g 2n 1The value of 2 n 1 is always odd, in the sequence 1, 3, 5, 7, 41. C
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThe 3rd picture after the t 0 picture (the first one that is not overlapping with another image) has the stem 16.5 mm from the origin of coordinates, at a time of t 3T . The actual position would
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimensionto give t1TH vs, where T3.2 s is the total time elapsed from dropping the rock to hearing thesound. Insert this expression for t1 into the equation for H, and solve for H.H1 2gTH vs2g 2v
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) At the top of its path, the velocity will be 0, and so we can use the initial velocity as found above, along with Eq. 2-11c.v2v2 02a yy0yy0v22 v02a00 227 m s 9.8 m s 2237 m(c
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimensionabout t 27 s . (c) The indication of a 0 velocity on a position-time graph is a slope of 0, which occurs at about from t (d )38 s .The object moves in both directions. When the slope is positive,
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editiontwo segments, from 0 to 50 seconds and then from 50 to 60 seconds. v 3 8 m s 14 m s (a) t = 0 to 50: a1 0 .4 8 m s 2 t 50 s 0 sd1 a2 d2 d1vo1t1 v t vo 2 t 2 d21 2a1t1214 m s 50 s1 20 .4 8
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One Dimension57. (a) For the free-falling part of the motion, choose downward to be the positive direction, and y 0 0 to be the height from which the person jumped. The initial velocity is v0 0 , acceleration is
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionSo he should jump when the truck is about 1.5 poles away from the bridge. 61. (a) Choose downward to be the positive direction, and y 00 to be the level from which thecar was dropped. The initi
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One DimensionFor the motion in the water, again choose down to be positive, but redefine y 0 surface of the water. For this motion, v0 8.85 m s , v acceleration from Eq. 2-11c, with x replaced by y.v22 v00 to
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) If the stations are spaced 3.0 km =3000 m apart, then there is a total of3 inter3000 m station segments. A train making the entire trip would thus have a total of 3 inter-station segments an
Arcadia University - CHEM - 101
Chapter 2Describing Motion: Kinematics in One DimensionBecause 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
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