# MCAT Physics Review

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 Author: Courtenay ID: 229539 Filename: MCAT Physics Review Updated: 2013-10-01 15:22:00 Tags: MCAT Physics Folders: Description: Conceptual premises, translational motion, forces, energy, work, power, momentum, fluids and solids, electrostatics, capacitors, electromagnetism, waves, oscillations, simple harmonic motion, lights and optics Show Answers:

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1. Power is measured in the SI unit of watts (W) and represents work/time. Work represents force x distance. Force represents mass x acceleration. Which of the following expresses the quantity of 1 watt in terms of the five basic SI units?

A.  kg*m2/s
B.  kg*m2/s3
C.  kg2/m3*s
D.  kg/m3*s3
• B is correct. Watts can be written as J/s. Joules can be broken down further to N*m, which can be broken down further to kg*m/s2.
• Thus:  W=J/s=
• (N*m)/s=
• (kg(m/s2)m)/s=
• kg*m2/s3
3. Give the sine of the following angles:
30° =
45° =
60° =
90° =
• sin 30° = 0.50
• sin 45° = 0.70
• sin 60° = 0.87
• sin 90° = 1
4. Give the cosine of the following angles:
30° =
45° =
60° =
90° =
• cos 30° = 0.87
• cos 45° = 0.70
• cos 60° = 0.50
• cos 90° = 0
5. Give the tangent of the following angle:
45° =
tan 45° = 1
6. The 30°, 60°,90° right triangle has sides that bear what ratio?
• The multiple represents the hypotenuse.
• The  multiple represents the side opposite the 60° angle.
•  represents the side opposite the 30° angle.
7. The 45°, 45°,90° right triangle has sides that bear what ratio?
• The  multiple is associated with the hypotenuse.
•  is the length of each leg.
8. The right triangle with sides  having a hypotenuse bearing a ratio of 5:4 to any other side will then have a ratio to the third side of what?
5:3 The right triangle would have the ratio of 3:4:5 with an angle approximately 53° between the 3 and 5 sides, and 37° between the 4 and 4 sides.
9. A right triangle has sides 4 centimeters, 8 centimeters, and  centimeters. What are the values of its angles?
• The triangle conforms to the right triangle with a ratio of 1:2 (sides 4 and 8).
• 4(1)=4     4(2)=8
• The triangle with sides in the ratio of always has angles of 30°, 60°, 90°
10. Find Ax and Ay:
• Given that A=8m and the cos30° =  approximately = 8.87, so Ax = 8*(0.87) = 7m.
• Similarly: sin30°=Ay/A and since sin30°=0.5, Ay = Asin30°= 8*(0.50)=4m
11. Average velocity is a vector, denoted by here by V. V is defined as:
•  (units: m/s)
• where  is the change in time
• If there is zero displacement, the average velocity will also be zero.
12. Speed is a scalar quantity. How does speed relate to velocity.
Speed is the magnitude of velocity.
13. Average acceleration is a vector quantity denoted here by a. How is a defined?
a=  = (units: m/s2)
14. An object moving to the right increases its speed from 0 m/s to 40 m/s in 5 s. What is the magnitude of its average acceleration?
15. If acceleration is constant then
16. Give the formula for uniformly accelerated motion in one dimension that relates displacement (d), initial velocity (vo), final velocity (vf), acceleration (a), and time (t) when missing a.
17. Give the formula for uniformly accelerated motion in one dimension that relates displacement (d), initial velocity (vo), final velocity (vf), acceleration (a), and time (t) when missing d.
18. Give the formula for uniformly accelerated motion in one dimension that relates displacement (d), initial velocity (vo), final velocity (vf), acceleration (a), and time (t) when missing vf.
19. Give the formula for uniformly accelerated motion in one dimension that relates displacement (d), initial velocity (vo), final velocity (vf), acceleration (a), and time (t) when missing vo.
20. Give the formula for uniformly accelerated motion in one dimension that relates displacement (d), initial velocity (vo), final velocity (vf), acceleration (a), and time (t) when missing t.
21. An object starts from rest and accelerates uniformly at 5 m/s2 and travels 40m. What is its final speed?
• vfo=02+2(5m/s2)(40m) =√400 =20m/s
22. An object accelerates uniformly from rest for T seconds and travels D meters. How far will it travel in total after 2T seconds?
d is proportional to t2 so since T doubles, T2 will quadruple and therefore so will D. The answer is 4D.

• vf is proportional to t
• d is proportional to t2
• vf2 is proportional to d
23. Give three important formulas for use when examining questions with gravity.
• Using the subscript "y" to distinguish from the x-direction...
24. If an object is dropped from a height of 80 meters above the earth, (a) how long does it take to reach the earth, and (b) what is the object's velocity at the instant prior to impact with the earth?
• 80m=0+1/2(10m/s2)t2
• t=4s
• vfy=0+(10m/s2)(4s)
• vfy=40m/s
25. An object is projected upward at 20 m/s and hits the ground 6s later. From what height was it thrown?
• dy=voyt+1/2(at2)
• dy=(20m/s)(6s)+1/2(-10m/s2)(6s)2
• dy=120m-180m = -60m
• dy is negative only because upward was chosen as positive. Therefore h=60m.
1. When an object moves upward, it slows down by approximately ________.
2. When an object moves downward, it speeds up by approximately ________.
3. An object projected upward from the ground will take __________ time to travel up ______.
4. An object projected upward from the ground will land _________.
• 1. When an object moves upward, it slows down by approximately 10 m/s every second.
• 2. When an object moves downward, it speeds up by approximately 10 m/s every second.
• 3. An object projected upward from the ground will take the same amount of time to travel up as it will to travel back to the ground.
• 4. An object projected upward from the ground will land with the same speed as it was initially projected.
27. An object is projected upward at 50 m/s from the ground. (a) How long will it take to reach the top of its trajectory, (b) how long will it travel in total before hitting the ground, and (c) with what speed will it land?
• a) Since an object slows down 10m/s every second on the way up, it will take 5s to slow down from 50m/s to 0. Thus, t=5s.
• b) Since time up equals time down, the total time is 10s.
• c) Since the object will land with the same speed as initially projected, the final speed = 50m/s.
28. An object is launched at 40 m/s at an angle of 60° above the horizontal. What will be its horizontal velocity 2 seconds later?
Since horizontal velocity never changes, it is equal to: =(40m/s)(0.5)=20m/s
29. For projectile motion, the x- and y- motions should be separated. Give the following equations for:
y- equations:
vfy=
dy=
vfy2=
x- equations:
dx=
• vfy=voy+at
• dy=voyt+(1/2)at2

dx=voxt
30. An object is projected horizontally off a cliff of height 20m at 15m/s. How far from the base of the cliff does the object land?
• First find t using the y- equations:
• 20m=0+(1/2)(10m/s2)t2
• t= 2s

• then plug into the x- equation:
• dx=(15m/s)(2s) = 30m
31. An object is projected from the ground a 100 m/s at an angle of 30° above the horizontal. Find (a) how long it takes to reach the top of its trajectory, (b) how high it goes, and (c) how far away it lands.
• (a) To find the time up, look at your y- equations: 0=50m/s +(-10m/s2)t
• t=5s
• (b) To find height, use y- equation for d: dy=(50m/s)(5s)+(1/2)(-10m/s2)5s)2 = 250m-125m = 125m
• (c)Now use your x- equation: dx=(87m/s)(10s)=870m
32. Centripetal acceleration is calculated with the formula: ac=
33. Centripetal velocity (speed) formula:
where v= speed and T=period or time required for one complete revolution
34. Assume that the objects travel in a circle at velocity 15m/s with radius of 30 meters. Determine (a) the magnitude of centripetal acceleration and (b) the period of its motion.
ac= (15m/s)2/30m = 7.5 m/s2

T=(2(3.14)(30m))/15m/s= approx. 12s
35. An object is moving in a circle with radius r and speed v. How many times greater must its acceleration be to travel with speed 2v?
Since ac=v2/r and r is a constant, we see that ac is proportional to v2, so it v doubles, then ac would need to quadruple to maintain circular motion.
36. Newton's first law:
Inertia: An object initially at rest of in motion with a constant velocity will remain in its initial state unless acted upon by a nonzero net external force.
37. Newton's second law:
F(net)=ma (net applied force equals the object's mass multiplied by the acceleration produced. It is in kilogram (kg) x m/s2 or newtons (N0.
38. What is the net force?
Sum of applied forces: Fnet= F1 + F2+...=ma
39. An object is subjected to a 40N force westward and a 15N force eastward. What net force does it experience?
Since the two forces are exactly opposite in direction, the answer is 40Nwest + -15Neast = 25N West (the 15 is negative because west was arbitrarily chosen as the positive direction)
40. A 5kg object is subjected to a 90N force to the north and a 120N force to the east. What is the magnitude of its acceleration?
Fnet = √((90N)2+(120N)2 = 150N (the Pythagorean Theorem used to obtain the net force). Since F=ma, 150N=(5kg)a. a=30m/s2.
41. 2Using the figure below, find ax and ay.
• First split F4 into components: F4x= F4cosθ=(20N)cos45°=14N and F4y=F4sinθ =(20N)sin45°=14N
• Fnetx= 14N-10N = 4N and since Fnetx=ma: 4N=(5kg)ax, ax=0.8m/s2
• Fnety=5N+14N-16N=3N and since Fnety=ma, 3N=(5kg)ay, ay=0.6m/s2.
42. In each case, what is the physical consequence:
Case 1: velocity and force are in the same direction.
Case 2: velocity and force are in opposite directions.
Case 3: velocity and force are perpendicular.
All other cases:
• Case 1: corresponds to speeding up
• Case 2: corresponds to slowing down
• Case 3: corresponds to a change in direction
• All other cases: correspond to both a change in direction and speeding up or slowing down.
43. Newton's third law
• When one body exerts a force on another, the second body will exert an equal and opposite force on the first.
• Mathematically, F2on1 = -F1on2.
44. A 1 kg block collides head-on with a 100 kg block. During the collision, which block feels the greater force?
Newton's Third Law states that they each feel the same magnitude force. However the effect of the force is different. m1a1=m2a2. The small block will experience 100 times the acceleration of the large block.
45. Give the formula for the force of gravity.
Fg=mg Where g is the acceleration due to gravity (=10m/s)
46. An object on earth has a weight of 50 newtons. What is its mass?
• Weight is equal to the product:
• weight (w) = Fg = mg
• so, 50N=m(10m/s2)
• m=5kg
47. An object situated on Planet X has mass of 80 kilograms and weight of 320 newtons on Planet X. The object is dropped from 18 meters above the surface of Planet X. Find the time before the object strikes the surface of the planet, and find the object's velocity at the instant before impact with the planet's surface.
• First determine the object's acceleration:
• Fnet=ma
• Choosing downward as the positive direction:
• 320N=80kg(a)
• a=4.0m/s2
• Knowing that the body accelerates at 4.0m/s2, calculate t:
• d=v0t+(1/2)at2
• 18m=0+(1/2)(4)(t2)
• 18m=2t2
• t2=9
• t=3s
• Knowing that the object travels for 3 seconds with an acceleration of 4.0m/s2, calculate its final velocity:
• vf=v0+at
• vf=0+(4.0m/s2)(3s)
• vf=12m/s
48. All objects with mass attract all other objects with mass. Newton's law of universal gravitation states that any pair of objects will produce a mutually attractive gravitational force which says that:
• Fg=force of gravity
• m1= mass of object 1
• m2= mass of object 2
• r= distance between the centers of the masses of objects 1 and 2
• G= universal gravitational constant=6.67x1011N*m2/kg2.
• The MCAT does not require you to know the value of the universal gravitational constant, but what is important is that gravitational force is:
• directly proportional to the mass of each object and
• inversely proportional to the square of the distance between the centers of the mass of the objects.
49. An object with mass 36kg and weight 360N rests on the surface of the earth. If it is transported to an altitude of twice the earth's radius, what are its new mass and weight?
A. m=36kg, w=90N
B. m=9kg, w=90N
C. m=36kg, w=40N
D. m=4kg, w=40N
• Since mass is an intrinsic property of the object, it will not change. This eliminates choices B and D.
• The only variable changing in the gravitational formula is r.
• Since Fg is inversely proportional to the square of the distance between the centers of mass of the objects, it is 1/32 = 1/9 so,
• 1/9(360N)=40N. C is the answer.
50. What is the normal force?
The force, Fn, that refers to the force in a direction that is perpendicular to the surface.
51. Find the normal force in all three cases:
• 1. Start by drawing all of the forces acting on the object. To find the normal force, we only need to look at Newton's Second Law in the y-direction.
• Fnet,y=may
• Calling upward positive, FN-Fg=0, or m*0, since there is no acceleration in the y-direction. Therefore:
• FN=Fg
• =mg
• =(100kg)(10m/s2)
• 2. First, split F into components. Again, we are only concerned with the y-direction.
• Fnet,y=may
• Calling upward positive, we see that:
• FN+Fsinθ-Fg=m*0  Therefore,
• FN=Fg-Fsinθ
• =mg-Fsinθ
• =(100kg)(10m/s2)-(100N)sin30°
• =1000N-50N
• =950N
• 3. Now there is an upward acceleration:
• Fnet,y=may
• FN-Fg=may
• FN=Fg+may
• =mg+may
• =(100kg)(10m/s2)+(100kg)(2m/s2)
• =1000N+200N
• =1200N
52. On the MCAT, all ropes, strings, and cables will be massless. If two different objects are connected by the same string, what will each object experience in regard to tension (FT)?
Since the ropes, strings, or cables are massless, the two objects will experience the same magnitude of tension, but in the opposite direction.

53. From the diagram above, m1=10kg, m2=20kg, and F=60N. Find (a)the acceleration of the blocks and (b) the tension in the string between them.
• Examine the blocks separately. According to Newton's Second Law, the equation in the horizontal direction for block one is:
• FT=m1a
• For block two:
• F-FT=m2a
• Since the blocks move together, the acceleration is the same for each. Since they are connected by a single string, FT is the same for each. Plugging equation 1 into equation 2, we see:
• F-m1a=m2a
• F=m1a+m2a
• F=(m1+m2)a
• 60N=(10kg+20kg)a
• a=2m/s2
• To solve for FT, we can plug the value of a into either equation
• FT=m1a
• FT=(10kg)(2m/s2)
• =20N
54. What are the two types of friction?
Kinetic (sliding) and static (not sliding)
55. Kinetic friction always opposes the motion of each object. What formula is used to find the magnitude of Ff,k?
• Ff,kkFN
• where FN is the normal force and μk is referred to as the coefficient of kinetic friction and is experimentally determined.
56. An object of mass 100kg is moving along the floor. A force of 200N is applied in the direction of motion. If μk=0.3, what is the object's acceleration?
• In the horizontal direction:
• Fnet=ma
• F-Ff,x=ma
• F-μkFN=ma
• In the vertical direction:
• Fnet=ma
• FN-Fg=0
• FN=Fg
• FN=mg
• Plugging this into the horizontal eq.:
• F-μkmg=ma
• 200N-(0.3)(100kg)(10m/s2)=(100kg)(a)
• 200N-300N=(100kg)a
• a=-1m/s
• Note that the object is slowing down and that the friction force is greater than the applied force. A greater applied force was needed to start the object moving in the first place, but once the object was moving, the value of kinetic friction remains constant regardless of whether the applied force changes.
57. Static friction is defined with an inequality rather than an equation. What is that inequality?
• Ff,s≤μsFN
• where μs is the coefficient of static friction.
58. An object of mass 100kg is at rest on the floor. If μs=0.4, how much force is required to start the object moving?
• F>Ff,s,max
• F>μsFN
• F>μsmg
• F>(0.4)(100kg)(10m/s2)
• F>400N
• Warning: Static friction does not always equal to μsFN. In general, it is whatever it needs to be to make sure that the object does not move.
59. A 100 kg object rests on the floor. A horizontal force of 100 N is applied. If μs=0.4, what is the force of static friction?
A. 0N
B. 100N
C. 300N
D. 400N
• The "trap" answer would be μsFN, but remember this is the maximum static friction, not necessarily what static friction is in this particular case.
• Ff,s,maxsFN
• smg
• =(0.4)(100kg)(10m/s2)
• =400N
• This tells us that an applied force of more than 400N is required to start the object moving. Since the applied force is only 100N, the object is NOT MOVING.
• If the object is not moving, we can apply Newton's Second Law.
• Fnet=ma
• F-Ff,s=0
• Ff,s=F=100N
60. A 10kg object slides down a frictionless inclined plane raised 30 degrees above the horizontal. What is its acceleration?
• Fnet=ma
• mgsinθ=ma
• a=gsinθ
• =(10m/s2)sin30°
• =5m/s2
61. What is the smallest angle of incline that would cause a 10kg object, initially at rest, to start sliding down the inclined plane? Note that μs=0.4
A. 22°
B. 45°
C. 68°
D. The block will never slide.
• To make an object just start to move, the downward component of gravity must be slightly larger than the maximum static friction.
• mgsinθ>Ff,s,max
• mgsinθ>μsFN
• To find the normal, we look at the y-direction:
• Fnet=ma
• FN-mgcosθ=0
• FN=mgcosθ
• Plugging this into the above:
• mgsinθ>μsmgcosθ
• sinθ>μscosθ
• tanθ>μs
• θ>tan-1s)
• θ>tan-1(0.4)
• Since we know that tan-1(0)=0 and tan-1(1)=45°, tan-1(0.4) must be in between 0° and 45°. Choice A is the correct answer.
62. If two masses are connected by a string over a pulley, then what is the magnitude tension of each mass in relation to the other?
Each mass feels the same magnitude tension.
63. A block of mass 10kg rests on a frictionless table and is connected via string and pulley to a 5kg mass hanging off the edge of the table. What is the acceleration of the system?
• Since each block experiences the same magnitude of acceleration, it is useful to choose the direction of motion to be positive for each block.
• Using Newton's Second Law, we see that:
• FT=m1a and m2g-FT=m2a (if m1 is for the black on the edge and m2 is the block hanging off the edge)
• plugging the first equation into the second equation:
• m2g-m1a=m2a (or Fnet=ma, if treating the entire system as one object with mass=m1+m2)
• m2g=(m1+m2)a
• (5kg)(10m/s2)=a(5kg+10kg)
• a=3m/s2
64. In the diagram below, how much force is needed to lift the block?
• In this problem, the upper pulley is fixed in place while the lower pulley moves with the mass. Treating the movable pulley and mass as an object, FT is labeled on each side of the small pulley on the string.
• Even though it is one string lifting the pulley, the tension acts on each side of it, counting it as two forces. The minimum force required to lift the pulley-mass system is:
• 2FT-mg=0
• FT=1/2(mg)
• Looking at the original diagram, we can see that F is equal to FT (since it is all on one string). Therefore:
• F=FT
• =(1/2)mg
• =(1/2)(10kg)(10m/s2)
• =50N
65. The acceleration of an object moving in a circle would be...
for an object moving in a circle of radius r and speed v, this is the acceleration toward the center of the circle.
66. Centripetal force is found using what equation?
67. If a satellite of mass ms orbits the earth (mass me) with radius r, what is its speed?
• Solving for v, we get:
• Note that the mass of the satellite does not matter.
68. A ball of mass 5kg moves in a vertical circle with a constant speed of 10m/s while attached to a string of length 50cm. (a) What is the maximum tension in the string? (b)What is the minimum speed at which the ball must travel to keep it in a circle?
• (a) The maximum tension occurs when the ball is at the bottom of the circle, since the string needs to both support its weight and to provide necessary centripetal force. If we decide that toward the center is positive:
• Fnet=ma
• FT-mg=m(v2/r)
• FT=mg + mv2/r
• =(5kg)(10m/s2)+(5kg+(10m/s)2)/0.5m
• =1050N
• (b) To find the minimum speed needed to make it around the circle, we look at the top of the circle.
• FT+mg=m(v2/L)
• The minimum speed corresponds to the minimum tension (zero). Therefore:
• mg=m(vmin)2/L
• vmin=L/√(gL)
• =√((10m/s2)(0.5m))
• -2.2m/s
69. The center of mass is the point at which an object's mass is concentrated, or the balancing point. Give the equation to find the center of mass, sometimes called the center of gravity.
• For any object whose mass is uniformly distribute, the center of mass is at the geometric center.
• For a system of objects arranged along the x-axis:
• A useful method to choose a location of origin is to use the location referenced in the problem. If the question asks, "how far from the left is the center of mass?" choose the left as the origin.
70. A block of mass 2kg and a block of mass 6kg sit on the ends of a plank of length L. How far from the left end is the center of mass if (a) the plank is massless and (b) the plank has mass and is uniformly dense?
A. 1/4L
B. 1/3L
C. 2/3L
D. 3/4L
• (a) Conceptually, it can be seen that the center of mass will be closer to the heavier mass. This eliminates choices A and B. Choosing the left edge as the origin, we see that x=0 and x2=L. Therefore:
• xcm=((2kg)(0)+(6kg)(L))/(2kg+6kg)
• =3/4(L) Choice D
• (b) Even with the plank having mass, we can eliminate choices A and B. Since the mass of the plank is not given, we cannot use the formula. However, since the plank is uniformly dense, its center of mass is the actual center. This new mass would skew the answer to part (a) closer to the actual center.
• Therefore, 1/2<xcm<3/4L, choice C is the only answer that fits the criteria.
71. Give the equation for torque.
t=rFsinθ where r is the vector that points from the pivot to the location of the applied force and θ is the angle between r and F.
72. A pendulum with mass 10 kg and string length 50 cm is pulled 30° from the vertical. Find the torque due to (a) tension and (b) gravity.
• (a) Since the angle between r and FT is 180°,
• tT=L*FTsin180°θ=0
• (b) The angle between r and Fg is 30°, so
• tg=L*Fgsin30°=L*mgsin30°= (0.5)(10kg)(10m/s2)(0.5) =25N*m
73. Find the torque: t
• Using Ft,s≤usFN would be difficult since θ isn't given. t=lF where l is the distance from lever arm to the line of action (from the pivot point to the bottom).
• l=1/2(1m)=0.5m
• t=(0.5m)(20N)
• =10N*m
74. Explain the concepts of translational equilibrium and rotational equilibrium.
A body (or system of bodies) is in translational equilibrium if it experiences no net forces in any direction and is in rotational equilibrium if it experiences no net torque.
75. Ropes 1, 2, and 3 are knotted together, as shown in the figure below. Rope 2 is fastened to a vertical wall, and rope 3 is fastened to a horizontal ceiling. A mass with a weight of 50 newtons hangs on rope 1. The angle between rope 3 and the ceiling is 60°. Find T1, T2, and T3, the tension in each of the ropes. (Assume that the weight of the ropes is negligible.)
• Since the problem is in equilibrium, the net force on the mass is equal to zero.
• T1-Fg=0N
• T1=Fg=50N
• You can also examine the forces acting on the knot.
• Splitting T3 into horizontal and vertical components, we see:
• Horizontal: T2=T3cos 60°
• Vertical: T3sin60°=T1
• Starting with the vertical equation T3: T1/sin60°=50N/.87=60N (approx.)
• And therefore: T2=(60N)cos60°=30N (approx.)
76. If the plank is uniform with m=30kg, find m2, such that the system is balanced.
• While this problem can be solved using center of mass, using torque is easier. Treating the plank and blocks as one object, where masspl is the mass of the plank:
• Setting the clockwise torques equal to the counterclockwise torque:
• tcw=tccw
• (2m)(mpl)g+(6m)(m2)g=(2m)(m1)g
• (2m)(30kg)(10m/s2)+(6m)(m2)(10m/s2)=(2m)(60kg)(10m/s2)
• 60kg*m+(6m)m2=120kg*m
• m2=60kg*m/6m
• =10kg
77. The work done by a constant force, F, to move an object through a displacement, d, is given by what equation?
• W=Fdcosθ; where θ is the angle between F and d and is expressed in N*m, which is also called the joule (J) Torque is also expressed in N*m, but is NOT expressed in joules. This is because torque is a vector quantity. Work is scalar, and the joule is a scalar unit.
• **This is ONLY valid for constant forces!
78. If a 50kg object is pushed to the right with a 30N force in the same direction for a distance of 1 km along a frictionless surface, how much work is performed on the object?
• The mass of the object is irrelevant because the applied force is provided. Also, since the object is moving in the direction of the applied force, θ=0.
• W=Fdcosθ = (30N)(1,000m)(1)=30,000N*m
• =30,000J
79. A 10N force is applied as shown to a 10 kg object. If the object is displaced by 10 meters to the right, then how much work has been done by the applied force?
• The work equals only the force applied in the direction of displacement, which is 10N times the cosine of 60°.
• Thus: W=Fdcosθ
• =(10N)(10m)(.5)=50J
80. Work is scalar and can be positive, zero, or negative depending on θ. Describe this conditions of each case.
• Case 1: 0≤θ<90°
•      cosθ is positive -->W is positive
• Case 2: θ=90°
•      cosθ is zero --> W is zero
• Case 3: 90°<θ≤180°
•      cosθ is negative --> W is negative
81. Find the work done by F, Ff,k, Fg, and FN.
• F: since F is in the direction of d, θ=0°
•      WF=Fdcos0°
•      =(100N)(5m)(1)
•      =500J
• Ff,k: since Ff,k is opposite the direction of d, θ=180°
•      Wf,k=Ff,kdcos180°
•      =ukFNdcos180°
•      =ukmgdcos180°
•      =(0.3)(100kg)(10m/s2)(5m)(-1)
•      =-1500J
• Fg and FN: since these forces are perpendicular to d, θ=90°
•      =Wg=WN=0
82. _______ _______ always does negative work.
Kinetic friction
83. _______ _______ always does zero work (on MCAT problems).
Static friction, all centripetal forces (since the force is directed toward the center and the direction of motion is tangent to the circle); the normal force often (but not always) does zero work (since objects often move parallel to the surface and the normal is always perpendicular to the surface)
84. How do you find total work acting on an object?
• Since work is scalar, the total work acting on an object is merely the sum of the individual works.
• WTOTAL=WF+Wg+WN+...
85. If a graph of Force vs. Position is given, the work done can be calculate by finding the _________.
area under the curve
86. Gravity, along with Electrostatic Force and Spring Force, are called conservative forces. One feature of conservative forces is that the work done is __________.
path independent
87. Describe the work done in the following case:
• The object is moving in the direction of Fg.
• Therefore:  Wg=Fghcosθ
•                      =mgh
• Note that the work done by gravity can be calculated merely by looking at vertical displacement, rather than the particular path.
88. Describe the work done in the following case:
• The object is moving at an angle with respect to Fg. Knowing that mgsinθ is the component of gravity that points down the plane:
•      Wg=Fgd
•      =mgsinθ*L
•      =mg(Lsinθ)
•      =mgh
• Note: the work done by gravity can be calculated merely by looking at the overall vertical displacement, rather than the particular path.
89. Describe the work done in the following case:
In this case, we cannot use W=Fgd since the component of gravity down the plane changes. However, since gravity is a conservative force, we can say that Wg=mgh.
90. An object is moved along a path as shown:

Find the work done by gravity.
• Since there was no vertical displacement,
• Wg=0
91. Kinetic energy (KE) refers to the energy inherent in the movement of an object and can be determined by what formula?
KE=(1/2)mv2 with the units in joules (J)
92. A 40kg object falls from a height. When its velocity reaches 20 m/s, what kinetic energy does the object possess?
KE=(1/2)mv2=(0.5)(40kg)(20m/s)2=8,000J
93. Give the equation the connects work and energy called the Work-Kinetic Energy Theorem.
• WTOTAL=ΔKE=KEf-KE0
• The total work done on an object equals the object's change in kinetic energy.
• Conceptually, this means that if a force does positive work on an object, the object gains KE. If a force does negative work on an object, the object loses KE.
94. How much work is required to stop a 1,000kg car moving at 20m/s?
• We know that W=Fdcosθ, but no information was given about F, d, or θ. We do not need to know what is actually stopping the car (friction? a wall?), but we can calculate the work done based on how much kinetic energy is being lost.
• WTOTAL=KEf-KEo
• =(1/2)(1,000kg)(0)2-(1/2)(1,000kg)(20m/s)2
• =-200,000J
• We expect the answer to be negative since kinetic energy is being taken away from the car.
95. An object slides from rest down a frictionless ramp of height h (as shown below). Find an expression for its final speed.
• We cannot solve this with kinematics since the Big Five require uniform acceleration. Instead, we can use the Work-Kinetic Energy Theorem:
• WTOTAL=KEf-KEo
• Since the ramp is frictionless, the only forces acting on the object are gravity and the normal. The work done by the normal force is zero. Therefore:
• Wg=KEf-KEo
• =>mgh = (1/2)mvf2-(1/2)mvo2
• =>mgh=(1/2)mvf2-0
• =>vf=√(2gh)
• This problem can also be approached using Conservation of Mechanical Energy:
• KEo+PEo=KEf+PEf
• 0+mgh=(1/2)(mvf2)+0
• solving for vf=√(2gh)
96. Gravity has the potential to do work on an object positioned above the ground. Give the equation that represents this.
• PE=mgh
• A classic example of potential energy is a stretched or compressed spring; the spring has the potential to perform work.
• For an object falling from height h, we can say that the potential energy of the system was converted into kinetic energy.
• Note: the ground is an arbitrary point of reference - h can be measured from anywhere
97. A body with a mass of 75kg is situated at a height of 30m above the ground. What is its gravitational potential energy
(a) relative to the ground;
(b) relative to a point 30m above the ground; and
(c) relative to a point 60m above the ground?
• (a) PE=mgh, where h is 30m:
• =(75kg)(10m/s2)(30m)
• =22,500J
• (b) PE=mgh, where h is now 0:
• =(75kg)(10m/s2)(0)
• =0
• (c) PE=mgh, where h is now -30m:
• =(75kg)(10m/s2)(-30m)
• =-22,500J
• Note: the actual value of potential energy isn't important. How it changes it the important part. We can say that Wg=-ΔPE
98. The law of conservation of energy states:
The energy of an isolated system is _______.
• constant
• This means that the energy in the universe may change form, but it cannot be destroyed or created. MCAT questions will sometimes draw on the candidate's ability to apply this principle to situations and systems involving the conversion of energy from one form to another.
99. Give the equation for mechanical energy.
• E=KE+PE
• We know that WTOTAL=ΔKE
• If gravity is the only force doing the work, then Wg=ΔKE.
• We also know that Wg=-ΔPE
• Therefore: -ΔPE=ΔKE
• or: Δ(KE+PE)=0
• or: KEo+Eo=KEf+PEf

100. Find the final speed.
• While the problem could be solved using kinematics, it would be very time-consuming. Conservation of Mechanical Energy is much easier:
• KEo+PEo=KEf+PEf
• Setting the ground as the reference point:
• (1/2)m(10m/s)2+m(10m/s2)(40m)=(1/2)mvf2+0
• canceling m from each term:
• 50m2/s2+400m2/s2=(1/2)vf2
• =>vf=√[2(50m2/s2+400m2/s2)]
• =30m/s
101. On the MCAT, the only conservative forces encountered will be gravity, the electrostatic force, and the spring force. Each of these forces has a potential energy associated with it and, if only these forces do work, what may you use for solving problems? If other forces do work (for example, friction and all applied forces) then what may be used?
Conservation of Energy; if F is the non-conservative force, you may use KEo+PEo+WF=KEf+PEf or WF=Δ(KE+PE)
102. What is the work required by F to push the block up a frictionless plane?

• The word "required" means that we are looking for the minimum work and, hence, the minimum force. The long way would be to solve for F using W-Fdcosθ. A shorter way is to look at work and energy. Since the force will not be accelerating the block, the kinetic energy will remain constant. Therefore, the work done by F is only giving the block potential energy.
• PEo+WF=PEf
• Setting the ground as the reference:
• WF=mgh
• =(10kg)(10m/s2)(5m)
• =500J
• Note: this is the same as if the force were applied directly up the plane or if the force were lifting the object straight upward.
103. What amount of force would be required to be applied to live an 80N object 12 meters from the ground along a plane inclined at 30 degrees (and find the work needed without mechanical advantage).
• To find the length of the inclined plane:
• sin30°=12m/d
• d(sin30°)=12m
• d(0.5)=12m
• d=24m
• If an 80N object is to be lifted a distance of 12 meters, the work required is equal to:
• Fd=W
• (80N)(12m)=960J
• To find the F needed to travel the inclined plane of 24m (as found previously) then:
• W=Fd
• 960J=F(24m)
• F=40N
• To lift the object straight up would take 80N and to have the mechanical advantage of the inclined plane would take a force of 40N
104. Give the formula used to find power needed:
• P=W/Δt (work over the time). It is expressed in J/s or watts (W)
• P=Fvcosθ since watts is equivalent to N*m/s which is the product: (force)(velocity)
105. How much power is needed to lift an object with a mass of 75kg straight upward to a height of 16m in 20s at a constant speed?
• Since the object is not accelerating, the upward force necessary to lift it is merely equal to the object's weight. The object's mass is 75kg, and its weight is:
• Fg=weight=75kg(10m/s2)=750N
• The lift will involve the work of:
• W=Fd=(750N)(16m)=12,000J
• If 12,000 joules of work are to be performed in 20 seconds, the associated power would be:
• P=W/t=12,000J/20s=600W
• Note: watt is equivalent to N*m/s, which also represents the product: (force)(velocity).
106. An engine that delivers power of 27,000 watts lifts an object straight upward against the force of earth's gravity with a constant velocity of 300 meters per second. Determine the mass of the object (and assume friction to be negligible).
• Since the force is acting in the same direction of the motion:
• P=Fvcosθ where cosθ=1
• 27,000W=F(300m/s)
• F=90N
• The force applied by the engine to keep the object moving upward at constant velocity=90N, which if we assume no air resistance, means the object's weight =90N. This means its mass is:
• F=ma where a = gravity
• 90N=m(10m/s2)
• m=9kg
107. What is the equation used to find linear momentum?
• p=mv  (mass of body)(velocity)
• and is expressed in units of (kg)(m/s) or kg*m/s
108. If an object with a mass of 100kg moves to the west with a velocity of 20m/s, what is its momentum?
• p=mv
• (100kg)(20m/s)=2,000kg*m/s
109. Give the formula for impulse (or the change in momentum.
• Impulse is denoted by J
• Fav*Δt=Δp  or J=Δp
• Impulse can be measured in N*s or kg*m/s
110. A batter strikes a thrown ball (m=0.15kg) that was moving horizontally at 40 m/s. The ball leaves the bat moving at 40m/s in the opposite direction. The bat was in contact with the ball for 15ms. Find:
(A) the impulse exerted by the batter;
(B) the magnitude of the average force exerted by the bat on the ball.
(C) the baseball's change in momentum;
(C) Momentum, like velocity, is a vector. If the direction of the initial velocity of the ball is considered positive:
po=mvo=(0.15kg)(40m/s)
=+6kg*m/s
and pf=mvf=(0.15kg)(-40m/s)
=-6kg*m/s
Therefore, the change in momentum is
Δp=pf-po=(-6kg*m/s)-(6kg*m/s)
=-12kg*m/s
Note: since direction matters, the answer is not zero.
(b) J=Δp=-12kg*m/s or -12N*s
(c) Since, by definition,
J=FavΔt
-12kg*m/s=Fav(15x10-3s)
Fav=-800N
The magnitude of the force is therefore 800N (the answer was positive because the bat is hitting the ball in the negative direction)
(this multiple choice question has been scrambled)
111. If the two objects involved in a collision are free of net external forces (i.e., the net force that each object feels is the force due to the other object), then the total momentum of the system is conserved. What does this mean mathematically?
• m1v1o+m2v2o=m1v1f+m2v2f
• This is known as the Law of Conservation of Momentum.
112. For the figure below, the surface is frictionless.

Find the final velocity of m2.
• Choosing the positive direction to the right:
• m1v1o+m2v2o=m1v1f+m2v2f
• (2kg)(6m/s)+(1kg)(0)=(2kg)(3m/s)+(1kg)v2f
• v2f=6 m/s
113. Express an elastic vs. and inelastic collision mathematically.
• For all collisions between objects free of net external forces, momentum is conserved.
• For collisions where kinetic energy is conserved, then
• Elastic collision: po=pf and KEo=KEf
• If kinetic energy is NOT conserved, then
• Inelastic collision: po=pf and KEo≠KEf
114. Determine if the collision in the example below is elastic or inelastic.

• Choosing the positive direction to the right:
• m1v1o+m2v2o=m1v1f+m2v2f
• (2kg)(6m/s)+(1kg)(0)=(2kg)(3m/s)+(1kg)v2f
• v2f=6 m/s
• KEo=(1/2)m1v1o2+(1/2)m2v2o2
• =(1/2)(2kg)(6m/s)2+(1/2)(1kg)(0)2
• =36J
• KEf=(1/2)m1v1f2+(1/2)m2v2f2
• KEf=(1/2)(2kg)(3m/s)2+(1/2)(1kg)(6m/s)2
• =9J+18J
• =27J
• Since KEo≠KEf the collision is inelastic.
115. The following collision is perfectly elastic:

Find the final velocity of the objects.
• Since the objects move together after the collision, we can write:
• m1v1o+m2v2o=(m1+v1)vf
• Choosing the positive direction to the right:
• (2kg)(6m/s)+(1kg)(-3m/s)=(2kg+1kg)vf
• vf=+3m/s (to the right)
116. A woman of mass 60kg stands at rest on a frictionless surface. If she throws a 2-kg ball horizontally at 10 m/s, with what speed will she move backward? Is this situation elastic?
First, a picture:

• Since the surface is frictionless, use Conservation of Momentum. Choosing the positive direction to the right:
• mpvpo+mbvbo=mpvpf+mbvbf
• 0+0=(60kg)vpf+2kg)(-10m/s)
• vpf=+0.3m/s
• To determine if it is elastic, look at initial and final kinetic energies:
• KEo=(1/2)mpvpo2+(1/2)mbvbo2
• =0+0=0
• KEf=(1/2)mpvpf2+(1/2)mbvbf2
• =(1/2)(60kg)(0.3m/s)2+(1/2)(2kg)(-10m/s)2
• =2.7J + 100J
• =102.7J
• Since KEo≠KEf, the situation is inelastic.
• Note: that KEf>KEo means that the system gained kinetic energy, as opposed to a perfectly inelastic collision, where kinetic energy was lost.
117. What is density?
• a measure of mass per volume and is expressed in the SI unit kilogram per cubic meter (kg/m3).
• Density=mass/volume
118. If a substance has a density of 9.3 x 104 kilograms per cubic meter, what volume is occupied by 88 kilograms of the substance?
• density = mass/volume
• 9.3x104=88kg/volume
• volume = 9.5x10-4m3
119. What is the specific gravity of a solid whose density is 2,000 kilograms per cubic meter?
• The density of water is 1,000 kilograms per cubic meter.
• Specific gravity represents the ratio of density of the solid (or substance) to density of water.
• So, this solid has a specific gravity of 2.
120. A fluid whose density is 600 kilograms per cubic meter has what specific gravity?
• The density of water is 1,000 kilograms per cubic meter.
• Specific gravity represents the ratio of density of the solid (or substance) to density of water.
• 600/1000=0.60
121. A container with total capacity of 1.0 x 10-3 cubic meters is half-filled with water so that the fluid level reaches the mark, 0.5 x 10-3 cubic meters. A solid object with mass of 0.2 kilograms is completely submerge in the water. The presence of the object causes the fluid level in the container to rise to 0.75 x 10-3 cubic meters. What is the density of the object? What is the specific gravity of the object?
• Density = mass/volume
• The object's mass is given as 0.2 kg.
• The object's volume is equal to the volume of the liquid it displaces:
• (0.75x10-3m3)-(0.50x10-3m3)=0.25x10-3m3
• Density: 0.2kg/0.25x10-3m3=
• (0.2x100kg)/(0.25x10-3m3)=
• 0.8x103kg/m3
• Specific gravity: 800/1000=0.8
122. What is tensile stress (Tss)?
• A force that (a) is applied equally at both ends of a solid object and (b) tends to stretch or compress it.
• Equals force/cross-sectional area
• Measured in pascals (Pa) where 1N/m2=1Pa
123. Consider a cylindrical rod with a radius of 3 centimeters. It is subjected to a 90-newton stretching force at each end. What is the amount of stress it experiences?
• Rod's cross-sectional area: =(3.14)(0.03)2=2.83x10-3m2
• Tensile stress experienced:
• force/cross-sectional area=
• 90N/2.83x10-3m2=
• 90x100N/2.83x10-3m2=
• 31.80x103Pa
124. A rectangular steel bar, the ends of which have two edges measuring 3 centimeters and 4 centimeters, respectively, is subjected at each end to a compressive force of 40 newtons. What tensile stress does the bar experience?
• Tss=F/Acs
• Cross-sectional area:
• (0.04m)(0.03m)=0.0012m2=1.2x10-3m2
• Stress:
• 40N/1.2x10-3m2=
• 40x100N/1.2x10-3m2=
• 33.33x103Pa
125. What is tensile strain?
• The degree to which tensile stress causes a solid object to undergo a change in length.
• (change in length)/(original length) = ΔL/Lo
• No units are used as strain is a dimensionless quantity.
126. If a solid object of length 2 meters undergoes tensile stress that stretches it and, as a result, actually lengthens it so that its new length is 2.01 meters, the object experiences what tensile strain (Tsn)?
• ΔL/Lo
• =(2.01m-2.0m)/2.0m
• =0.01m/2.0m=.005
127. A rectangular bar of length 34 meters is subjected to a compressive force at either end, and its length is reduced to 33.97 meters. What tensile strain has the object experienced?
• ΔL/Lo
• =(34m-33.97m)/34m
• =0.03m/34m=0.00088 or 8.8x10-4
128. What is Young's modulus?
• Every substance that obeys Hooke's law is associated with its own Young's modulus--a value that, for that particular substance, expresses the degree of strain that will be imposed by a given degree of stress.
• For any object that obeys Hooke's law (force, F, needed to extend or compress a spring by some distance, d, is proportional to that distance. That is, F=kd), it is the ratio:
• tensile stress/tensile strain
• It is expressed in pascals (Pa)
129. Consider an object composed of substance X, for which Young's modulus is 7.5x105 pascals. If it is subjected to a tensile stress of 50 pascals, what degree of strain would it undergo?
• 50 Pa/tensile strain=7.5x105 Pa
• tensile strain = 50 Pa/7.5x105 Pa
• =6.67x10-5
• The problem does not specify whether the imposed stress was stretching or compressive; the math is the same in either case.
130. A cylindrical rod measure 5 meters in length and 1.5 centimeters in radius. It obeys Hooke's law and has a Young's modulus of 5.2x106 pascals. What compressive force is necessary to reduce its length to 4.9981 meters?
• Calculate cross-sectional area:
• =(3.14)(0.015)2=0.00071m2=7.1x10-4m2
• Calculate strain:
• ΔL/Lo=(5m-4.9981m)/5m=0.0019m/5m
• =0.00038=3.8x10-4
• Calculate stress:
• stress/strain=5.2x106Pa
• stress=(5.2x106Pa)(3.8x10-4)
• =1.9x103Pa
• Calculate force:
• Tss=F/Acs
• F=(Tss)(Acs)=(1.9x103Pa)(7.1x10-4m2)
• F=1.4N
131. For any resting fluid that is not experiencing any external pressure, the fluid pressure at a specified depth is given by what formula? (In other words, what is gauge pressure?)
• Pressure=(density)(acceleration due to gravity)(depth below the surface of the fluid)
• P=ρ(g)(h)
•  where ρ=fluid density,
•  g=acceleration due to gravity
•  h=height of the fluid above the point in question
132. What is the difference between gauge pressure and absolute pressure?
• Gauge pressure: P=ρ(g)(h)
• Absolute pressure: P=ρ(g)(h)+atmospheric pressure. (or gauge pressure plus atm pressure)
• Absolute pressure is the actual pressure to which a submerged object is exposed.
133. What does Pascal's principle say about pressure applied to an enclosed fluid?
• Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
• It follows that fluid pressure at any given depth in a resting fluid is unrelated to the shape of the container in which the fluid is situated.
134. Fluid F has a density (ρ). At a given depth (h), within a sample of fluid F, fluid pressure is 3,000 pascals. What is the fluid pressure at depth 2h in a sample of fluid E, whose density is 0.4ρ?
• Because fluid pressure is directly proportional to fluid density and depth, the pressure at depth 2h in fluid E is calculated:
• P=(3000Pa)(0.4ρ)(2h)
• =0.8(3,000Pa)
• =2,400 Pa
135. Buoyancy refers to a fluid's tendency to propel submerged or partially submerged substances toward the surface of the fluid. The buoyancy force then is an upward force exerted by a fluid on another body or fluid. For any body immersed in fluid, the magnitude of buoyancy is calculated using Archimedes' principle:
Buoyancy (Fb)=?
• Buoyancy (Fb)=(V)(ρ)(g)
• =(volume of fluid displaced)(density of displaced fluid)(acceleration due to gravity)
• Buoyancy (Fb)= weight of the fluid displaced
136. An object with density of 11.3x103 kilograms per cubic meter and mass of 57 kilograms is immersed in water and subjected to no horizontal or rotational forces.
(1) What is the approximate net force experienced by the object?
(2) Will it accelerate? If so, with what magnitude?
• (1) The only forces to which the object is subject are those of its own weight, and the upward for of buoyancy.
•      Object's weight:
•       w=57kgx10m/s2=570N
•      For buoyancy calculate volume:
•       volume = mass/density
•       volume = 57kg/11.3x103kg/m3
•           =5.0x10-3m3
• Knowing that the solid displaces 5.0x10-3 cubic meters of water and using the known density of water (1,000 kilograms per cubic meter)...
•      Buoyancy=(V)(ρ)(g):
•       =(5.0x10-3m3)(1x103kg/m3)(10m/s2)
•       = 50N
• So, the object experiences a downward force of 570 newtons and an upward force of 50 newtons. It experiences, therefore, a net force of 520 newtons downward because:
•      Fnet=50N-570N=-520N
• (2)The object will accelerate downward. Since its downward net force has a magnitude of 520N:
•       F=ma
•      520N=57kg(a)
•      a=approx. 9.1m/s2
137. Flow (Q) refers to the quantity of fluid that passes a given point in a given period of time. Give this concept as a mathematical formula for any fluid flowing through a vessel (pipe or tube), and give several phenomena associated with ideal flow.
• Flow=(area of pipe or tube)(velocity of fluid)
• Ideal flow:
•    is constant at all points,
•    has no viscosity,
•    has no turbulence,
•    has no drag (fluid friction), and
•    =(area)(velocity)
138. To what does each refer:
Laminar flow
Turbulent flow
• Laminar flow refers to flow in which whirlpools and eddies are absent.
• Turbulent flow is flow in which whirlpools and eddies are present.
139. How is velocity related to cross-sectional area for fluid flowing within a vessel system of variable caliber?
Velocity is inversely proportional to cross-sectional area
140. What is the relationship between pressure and the vessel caliber for a fluid flowing within a vessel system of variable caliber?
An increase in one corresponds to an increase in the other, but the relationship is not proportional.
141. Fluid flows through a system of vessels of various calibers, as shown below:

(1) In which section of the system is the fluid under the greatest pressure?
(2) Through which section of the system does the fluid travel at the greatest velocity?
(3) If the fluid's velocity is 0.5 meters per second through section B, (a) what is its velocity through section D? (b) what is its flow at section D? and (c) what is the flow through section A?
• (1) Where caliber is greatest, pressure is greatest; thus pressure is greatest at section A.
• (2) Where caliber is lease, velocity is greatest; thus velocity is greatest at section C.
• (3)(a) Area and velocity are inversely proportional. Ascertain the ratio. area of section B : area of section D
• Looking to the formula for the area of a circle, the area of the circle is proportional to the square of the radius. The radius of section B is 3 meters, and the radius of section D is 1 meter. The two radii bear the ratio 3:1, which means that the ratio of their areas is:
•      32:12 = 9:1
• Since velocity is inversely proportional to area, and the velocity in section B is 0.5 m/s, the velocity in section D is:
•      (0.5m/s)(9)=4.5m/s
• (b) Since flow equals the product of area and velocity; flow of section D equals
•      x(4.5m/s)=(3.14)(1m)2x(4.5m/s)
•       =14.13m3/s
• (c) Flow is equal at all points. Having calculated flow in section D, section A will have the same flow of 14.13m3/s
142. Describe what it means to be a poor conductor of electricity.
• A poor conductor has relatively poor electron mobility.
• A poor conductor will neither produce nor experience contact charge.
143. A positively charged body is placed in contact with an uncharged body. Which of the following choices is (are) true?

I. If either of the bodies is a poor conductor, each body will tend to maintain its initial charge status.
II. If both bodies are good conductors, each body will tend to maintain its initial charge status.
III. If both bodies are good conductors, protons will move from the uncharged body to the negatively charged body.

A. I only
B. II only
C. I and III only
D. I, II, and III
A is correct. Contact charge between a charged and uncharged body involves the movement of electrons (NOT PROTONS) from one body to another and proceeds only if both bodies experience relatively good electron mobility; in other words, both bodies must be good conductors.
144. A negatively charged body, X, and a positively charged body, Y, come into contact. Each body is composed of a metallic substance that features a high degree of electron mobility. Which of the following choices is (are) true?

I. Body Y will undergo a contact charge.
II. Body X will undergo a contact charge.
III. Electrons will move from body X to body Y.

A. I only
B. II only
C. I and II only
D. I, II, and III
D is correct. The question concerns contact charge as applied to a negatively and positively charged body. Both bodies are good conductors. Body X will impart electrons to body Y, which makes choice III correct. The negative charge on body X will be reduced in magnitude, as will the positive charge in body Y. Choices I and II are also accurate. Note that simply knowing the direction of electron flow allows you to choose III, and by extension D.
145. A negatively charged object, A, is brought near the end of a conductive metal object, B, which has no charge. At its opposite end, object B is connected directly to another metal object, C. Object C is removed first, and then object A is removed. Which of the following statements is true at the end of this experiment?

I. Object A is negatively charged.
II. Object A has no charge.
III. Object B is negatively charged.

A. I only
B. II only
C. I and III only
D. I, II, and III
A is correct. The question concerns charge by induction. Object A neither gains nor loses electrons throughout the process and so maintains its original negative charge (it was brought near but was no touching object B). Some of the electrons in object B are repelled by the negative charge of object A and passed to object C. When object C is removed, object B is left positively charged
146. A particle with a charge of -4 C lies 3 centimeters from a particle with a charge of -3 C. The particles affect each other with a repulsive force, Fr1. Assume the distance between the particles is decreased to 0.75 centimeters, thus modifying the repulsive force, so that it acquires a new value, Fr2. Write an equation that expresses Fr2 in terms of Fr1.
• The problem draws on the proportionalities associated with Coulomb's law.
• Coulomb's law:
•  where q1=charge on object 1
•            q2=charge on object 2
•            r=distance between the centers of the charge
•            the first fraction is a constant
• The repulsive (or attractive) force created by two point charges is inversely proportional to the square of the distance between them.
• Distance is reduced by a factor of:
• 3/0.75=4
• repulsive force increased:
• 42=16
• So the equation that expresses Fr2:
• Fr2=16(Fr1)
147. Two charged particles, A and B, separated by a distance, D, carry charges of +X coulombs and -Y coulombs, respectively. Another pair of charged particles, G and H, separated by a distance, 0.5D, carry charges of +3X and +6Y, respectively. Write an equation that expresses the magnitude of the attractive force Fa(ab) created by particles A and B in terms of the repulsive force Fr(gh) created by particles G and H.
• As noted, A and B experience an attractive force, since their charges are of opposite sign, and particles G and H experience a repulsive force, since their charges are of like sign.
• The question concerns the relationship between the magnitude of these forces, and the answer lies in the proportionalities associated with Coulomb's law:
• F=k[(q1q2)/r2]
• In relation to the first pair of particles, the second pair of particles generates a numerator that is:
• (3)x(6)=18 times as large.
• However, the second pair of particles is separated by one half the distance that separates the first pair. Since the applicable force is inversely proportional to the square of the distance between the particles:
• Fa(ab)=[1/22](Fr(gh))=[1/4](Fr(gh))
• Combine the effects of the numerator and denominator:
• Fa(ab)=(1/4)(1/18)(Fr(gh))
• Fa(ab)=Fr(gh)/72
148. Two charged particles experience a repulsive force of 33 newtons. If the charge on one of the particles is tripled, and the distance between the particles is also tripled, what will be the magnitude of the repulsive force experienced by the two particles?
• According to Coulomb's law (F=k[q1q2/r2]), tripling the magnitude of charge on one of the  particles will triple the repulsive force between the particles. Tripling the distance between the particles reduces the force by a factor of:
• (3)2=9.
• Multiply the new force by 3/9 or 1/3:
• 33N*(1/3)=33/3=11N
149. A particle, A, carries a charge of magnitude x coulombs. At a point, P1, located y meters from particle A, particle A creates an electrostatic field strength of 0.008 per coulomb.(a) What electrostatic field strength does particle A exert at point P2, located 6y meters from particle A?(b) Among the following choices, the equation that properly expresses x in terms of y, π, and ε0 is:
A. x=[y(0008)]/(πε0)
B. x=[y2(0.008)]/(π2ε02)
C. x=y2(0.008)(4πε0)
D. x=y(0.004)(2πε02)
• The problem requires the use of the mathematical relationships inherent in Coulomb's law:
• F=[1/(4πε0)][q1q2/r2]
• (a) The force between two charged particles is inversely proportional to the square of the distance between them. The ratio 6y:y is equivalent to the ratio 6:1. Since (6)2=36, the electrostatic field strength at P2=x/36 = 0.008/36 N/C = 2.2x10-4 N/C.
• (b) C is correct. Using Coulomb's law:
• 0.008N=(1/4πε0)[(x*1)/y2]
• Solving for x:
• 0.008÷[1/4πε0]=(x*1)/y2
• (0.008)(4πε0)=(x)/y2
• x=y2(0.008)(4πε0)
150. What is the force experienced by any charged particle located at a position of known electrostatic field strength equal to?
• Forcees equals (electrostatic field strength)(charge on particle) or:
• Fes=Efs(q)
151. A positively charged particle situated at point P establishes an electrostatic field with strength = 6 newtons per coulomb at point P1. What is the magnitude of force experienced by a particle with a charge of 0.0022 coulombs located at point P1?
• Fes=(electrostatic field strength)(charge on particle)
• Fes=Efs(q)
• Fes=(6 N/C)(0.0022 C)
• Fes=0.0132 N
152. How do you calculate the electric field strength?
• Charge on one plate
•  (ε0) (area of plate)
• or
• E=Q/[(ε0)(A)]
153. Two plates, A and B, each measuring 1.5 centimeters x 2 centimeters, are closely spaced, one carrying a charge of (+)1x10-5 coulombs, and the other, a charge of (-)1x10-5 coulombs. A separate pair of plates, C and D, each measuring 1.8 centimeters x 2 centimeters, are closely spaced, one carrying a charge of (+)1.2x10-5 coulombs, and the other, a charge of (-)1.2x10-5 coulombs. Write an expression that describes the electrostatic field strength associated with the first pair of plates (E1) in terms of that associated with the second (E2).
• The charge on each plate in the first pair in relation to the charge on each plate in the second pair gives rise to the ratio 1:1.2 = 5/6.
• The area of each plate in the first pair in relation to the area of each plate in the second pair gives rise to the ratio 3:3.6 = 5/6.
• This second ratio is the ratio of the areas, which is inversely proportional to the electric field and, thus, should be inverted. The electric field strength associated with the first pair of plates in relation to that associated with the second creates the ration (5/6):(5/6)=1.
• The equation for the electrostatic field, E1, in terms of E2 is simply E1=E2.
154. How is the electrostatic potential calculated?
• Electrostatic potential = (Electrostatic field strength) (distance between plates)
• V=Ed
• calculated in volts
• 1V=1J/C
155. Consider a capacitor and imagine a massless charge -1coulomb situated on the positively charged plate. The hypothetical particle is attracted to the positively charged plate and held there with a force equal to the capacitor's electrostatic field strength. Suppose that the plates of a particular capacitor have such area and charge as to produce an electrostatic field strength of 800N/C. Suppose further that the plates are separated by a distance of 1.5 centimeters. A massless particle with charge of -1coulomb situated on the positively charged plate will be held to the plate with a force of 800 newtons. To move the -1coulomb particle from the positive plate to the negative plate-thereby opposing the force imposed by the electrostatic field-one would have to perform work on the particle. Calculate the work that must be performed.
800Nx(.015m)=12N*m=12J
156. A capacitor is composed of two plates separated by a distance, d, each plate having an area A and carrying a charge of magnitude Q1. The capacitor's electrostatic potential is V1. If:
1. the capacitor is replaced with another for which the area of each plate is doubled,
2. the distance between the plates is tripled, and
3. the charge between the plates is maintained at Q1,
what is the electrical potential, V2, of the new capacitor expressed in terms of V1?
• Recall that:
• 1. electrostatic field strength (E) is directly proportional to the charge on either plate (Q) and inversely proportional to the area of either plate (A), and
• 2. electrical potential (V) is directly proportional to electrostatic field strength (E) and also directly proportional to the distance between the plates (d).
• V is directly proportional to the fraction:
• (Q)(d)
•     A
• The doubling of (A) tends to reduce V by a factor of 2, and the tripling of d tends to increase V by a factor of 3. The alterations, therefore, multiply V by the fraction 3/2 = 1.5, which means that:
• V2=(1.5)(V1)
157. What is capacitance?
• The term capacitance refers to the degree to which a capacitor stores an electric charge in relation to its electrostatic potential. For any capacitor, capacitance (C) is proportional to the quotient:
• (charge)/(electrostatic potential) or
• Q/V
• expressed in C/v (Coulombs/volt) or a farad
158. What is the effect of capacitance on plate size?
• 1. the area of the plates is inversely proportional to electric field strength (E).
• 2. The electric field strength is directly proportional to electrostatic potential (V).
• 3. The electrostatic potential is inversely proportional to capacitance (C); increased size of the capacitor's plates is directly proportional to capacitance.
• In terms of algebra:
• C is proportional to Q/V
• V is proportional to (E)(d)
• E is proportional to Q/(ε0)(A)
• C is plainly proportional to:
• (d)(Q) = (Q)(ε0)(A) = 0)(A)
•      (ε0)(A)       (d)(Q)          d
• which is proportional to A
• So, if you have the dielectric constant...
• C= [(K)(ε0)(A)]/d
159. If a particle with charge of (+)3x10-7 coulombs is first situated on the negative plate of a 100-volt capacitor and then moved to the positive plate, the amount of work performed on it will be:
A. 3.0x10-5 joules.
B. 9.0x10-5 joules.
C. 1.5x10-4 joules.
D. 1.8x10-4 joules.
• A. is correct. The designation "100-volt capacitor" signifies a capacitor in which a hypothetical particle with charge of magnitude 1 coulomb would acquire 100 joules of potential energy when moved from the plate with a charge of opposite sign to the plate with a charge of like sign.
• 100V= 100J/C
• 100J/Cx3x10-7C=300x10-7J=3.0x10-5J
160. Assuming that the value of the permittivity constant, ε0 = 8.85x10-12C2/N*m2, consider two parallel plates of equal area and equal but opposite charges. Each plate measures 10 centimeters x 9 centimeters, and the magnitude of charge on each is 9x10-4 coulomb. The electrostatic field strength produced by such a pair is most nearly equal to:
A. 0.1 newtons per coulomb.
B. 1.13x1010 newtons per coulomb.
C. 71.65x105 newtons per coulomb.
D. 88.5x1012 newtons per coulomb.
• Knowing that the electrostatic field strength is equal to the fraction:
• (Q)/[(ε0)(A)]
• Calculate:
• 9x10-4C/[8.85x10-12C2/N*m2)(9x10-3m2)=
• 90x10-5C/79.65x10-15C2/N=
• 1.13x1010N/C
161. Within an electrostatic field, the movement of a charged body from point A to point B requires work of 40 joules. The movement of the charged body back to point A from point B produces 38 joules of work. Which of the following best explains the finding?
A. Two joules of energy were lost as heat.
B. The charged body followed two different pathways in traveling between points A and B.
C. The charged body acquired 78 joules of potential energy when it moved from point A to point B.
D. The sign of the body's charge is positive.
A. is correct. If the movement of a charged body between two points in an electrostatic field requires work, then the body is acquiring potential energy equal to the amount of work required minus any energy that's converted to other forms, such as heat. In no event can the body acquire potential energy greater than the work performed on it. When the charged body is allowed to move back to its original position, its potential energy might be harnessed to perform work. The maximum amount of work that might be performed is equal to the potential energy, which, in turn, is equal, at maximum, to the amount of work originally performed to move the object from its original position. If the work done by the object during its "return" trip is less than that done on the object during its initial movement, the explanation might be that the potential energy it acquired has been converted, in part, to heat, and not directed to the production of work.
(this multiple choice question has been scrambled)
162. An investigator makes appropriate electrical connections between a capacitor and an electrical lifting device so that energy stored within the capacitor may be used to lift a 5-kilogram object. She hypothesizes that if the capacitor causes the machine to lift the object a total distance of 2 meters upward, it will, in the process, lose more than 100 joules of potential energy. Is the hypothesis plausible?
A. Yes, because the first law of thermodynamics is inapplicable to charged plates so long as they are spaced relatively close together.
B. Yes, because the operation of the machine itself will occasion the performance of work and the generation of heat.
C. No, because the gravitational potential energy of the object would then be greater than the potential energy expended.
D. No, because the gravitational potential energy cannot be related mathematically to electrical potential energy.
• B. is correct. The investigator intends to apply the energy stored within a capacitor (the potential energy represented by its electrical potential) toward the performance of work. The lifting of a 5 kilogram body 2 meters upward imparts gravitational potential energy of:
• (m)(g)(h)=(5kg)(10m/s2)(2m)=100 joules
• which corresponds to 100 joules of potential energy. Since the operation of the machine itself will require the performance of work and hence the conversion of potential energy, it is plausible that the actual amount of potential energy lost by the capacitor will be greater than that acquired by the lifted object.
163. The flow of a current is promoted by the potential difference between positive and negative poles; increased potential difference between positive and negative poles increases current flow. The flow of current is impaired by resistance, and increased resistance decreases current flow. These relationships constitute the essence of Ohm's law which give what equation?
• I=V/R
•   where
• I= current (amperes-A)
• V= voltage (volts-v), and
• R= resistance (ohms-Ω)
• Current is directly proportional to voltage (electrical potential difference) and inversely proportional to resistance.
164. A cathode of a 12-volt battery is connected by a conductive wire to an electric motor, which is then connected by another wire to the battery's anode. The total resistance of the circuit-through the wire, motor, and battery-is 2 ohms. What is the magnitude of current traveling through the circuit?
• I=V/R
• 12v/2Ω
• =6A
165. A piece of equipment requires for its operation a minimum current of 4,800 amperes. It draws electricity from an 1,800-volt source. What is the maximum resistance of the system under which the equipment will remain operational?
• I=V/R
• 4800A=1800v/R
• R= 1800v/4800A
• =0.375Ω
166. Consider a power source of 50 volts that drives equipment with a current of 40 amperes. What is the total resistance in the circuit through which the equipment operates?
• I=V/R
• 40A=50v/R
• R=1.25Ω
167. An electrical device is driven by a power source of voltage V. The circuit through which it operates has resistance R and carries a current of I. Assuming that the conductor in the circuit is modified so that the total resistance of the circuit is 3R, and the power source is modified so that its voltage is 1/2V, write an equation that expresses the resulting modified current, I1, in terms of the original current, I.
• The problem requires an appreciation of the proportionalities of Ohm's law. Current is directly proportional to voltage and inversely proportional to resistance. Since resistance is multiplied by a factor of 3, and voltage by a factor of 1/2, the new current, I1, comes from the original current, I, divided by a factor of 3 and multiplied by a factor of 1/2:
• I1= (I)x[(1/2)/3]
• I1= I/6
168. Power represents the quantity of work per unit time, and is expressed in the SI unit watt (W). One must be able to derive and manipulate power, P, using the electrical variables current, voltage, and resistance. Give the following equations:
1. How the power of electricity flowing through a circuit is related to current and voltage.
2. If the resistance and current of the system is known, derive another expression for electrical power.
3. Give another expression for power in terms of resistance of the system and voltage.
• 1. P=IV
• 2. since V=IR; P=I(IR) or P=I2R
• 3. since I=V/R; P=(V/R)(V) or P=V2/R
169. A 200-watt device is operated within a circuit that draws 1.60 amperes. What is the voltage of the power source?
• P=IV
• 200W=(1.60A)(V)
• V=125-volts
170. Power does NOT represent one concept in the context of electricity and another in the context of force and motion. Express power in both electrical and mechanical terms.
•      work   force x dist   (mass x accl)x dist
• P= time =     time      =       time
• and
• P=IV
171. An electric lifting device operates from a 3,000-volt power source, and the circuit through which it operates has total resistance of 1,200 ohms. If the lift is attached to a mass of 200,000 kilograms, during what time period will it raise the mass to a height of 8 meters? (Assume that no energy is lost to heart and that g=10m/s2.)
• P=work/t
• P=V2/R
• Power=(3000V)2/1200Ω
•   =7500W
• which by definition means
•   =7500 J/s
• Knowing W=(force)(distance)
• =[(200,000kg)(10m/s2)][(8m]
•   =1.6x107J
• Then solve for time...
• 7500J/s=(1.6x107J)/time
• time=2133s
•   approx. = 36 minutes
172. How do you ascertain the total resistance in a circuit (ignoring that of the conducting wire itself and the battery)?
• Simply sum the resistances of each of the resistors:
• Rt=R1+R2+R3+...
173. How do you obtain the amount of resistance when two or more resistors are connected in parallel?
• Total resistance is equal to the reciprocal of the sum of reciprocals of individual resistances.
• 1/Rt=(1/R1)+(1/R2)+(1/R3)+...
174. The electric circuit depicted below presents a 220-volt power source that operates an article of equipment having 3 ohms of resistance. The circuit further features a 3-ohm resistor, a 2-ohm resistor, and the set of resistors connect in parallel, as shown. Within the set of resistors connected in parallel are resistors connected in series. Ignoring any resistance offered by the conducting wire itself or the power source, please find (a) the total resistance in the circuit, (b) the total current drawn by the device, and (c) the total power associated with the circuit.
• (a) The total resistance within the circuit is the sum of the four resistors connected in series, including, as one resistor, the equipment itself, and as another resistor, the set of resistors connected in parallel. The resistors other than those connected in parallel have a total resistance of:
• 3Ω+2Ω+3Ω=8Ω
• The set connected in parallel features three resisting entities, each composed of two resistors in series.
• The series on the right has a resistance of 6Ω+18Ω=24Ω;
• the middle series, 5Ω+3Ω=8Ω
• and on the left, 5Ω+7Ω=12Ω
• The entire set, therefore, is equivalent to three parallel resistors of 24Ω, 8Ω, and 12 Ω, respectively.
• To determine the total resistance of the parallel resistors:
• 1/Rt=1/24Ω+1/8Ω+1/12Ω
•   =1/24Ω+3/24Ω+2/24Ω
•   =6/24Ω  = 1/4Ω
• The reciprocal of 1/4 is 4, so the total resistance of the set is equal to 4Ω.
• The total resistance of the circuit is 4Ω+8Ω=12Ω
• (b) Knowing now that the circuit's total resistance is 12Ω, the current may be calculated using Ohm's law:
• V=IR;  220V=I(12Ω)
• I=18.33A
• ((c) knowing that the circuit's resistance is 112Ω, and its current is 18.33 amperes, power can be calculated using  P=I2 or P=IV
• P=IV; P=(18.33A)(220v)  =4,032 watts
175. The electric circuit depicted below presents a 450-volt power source that operates an article of equipment having 7 ohms of resistance. The circuit further features a 3-ohm resistor, and two sets of resistors connected in parallel, as shown. Within the second set of parallel resistors are two resistors connected in series. Ignoring any resistance offered by the conducting wire itself or the power source, please find (a) the total resistance in the circuit, (b) the total current drawn by the device, and (c) the total power associated with the circuit.
• (a) The circuit features two sets of parallel resistors. The total resistance of the circuit is equal to the sum of all resistors connected in series, which means the sum of (1) the first set of parallel resistors, (2) the equipment with resistance of 7 ohms, (3) the 3-ohm resistor, and (4) the second set of parallel resistors. The total resistance of the first set of parallel resistors is:
• 1/Rt=1/40Ω+1/40Ω
• 1/Rt=2/40Ω
•   Rt=20Ω
• Total resistance of the second set of parallel resistors is determined first by summing two resistors connected in series:
• R=22Ω+23Ω = 45Ω
• The parallel set is thus equivalent to a set consisting of a 45-ohm resistor, a 90-ohm resistor, and a 30-ohm resistor. Total resistance of the set is:
• 1/Rt=1/45Ω+1/90Ω+1/30Ω
•   =2/90Ω+1/90Ω+3/90Ω
•   =6/90Ω
•   Rt=90Ω/6  =15Ω
• Total resistance of the circuit is equal to
•   20Ω+7Ω+3Ω+15Ω = 45Ω
• (b) Knowing that the circuit's total resistance is 45Ω, its current is calculated by reference to Ohm's law:
• V=IR => 450V=I(45Ω)
•  I=10A
• (c) Knowing that the circuit's current is 10 amperes, the associated power is calculated according to the equation, P=I2R, or P=IV
• P=IV => P=(10A)(450V) = 4500W
176. Consider the circuit given below. Calculate the current. Give the voltage drop between points A and B.
• The total resistance = 20Ω+7Ω+3Ω+15Ω=45Ω
• The current: V=IR; 450V=I(45Ω); I=10A
• The total resistance between point A and B is 20Ω.
• The voltage drop: V=IR =(10A)(20Ω)=200V
177. Consider the circuit shown below. Calculate the current and total resistance of the parallel resistors. What is the voltage drop between points A and B, points A and C, points B and C, points C and D and points A and D?
• Parallel resistors total: 1/Rt=1/24Ω+1/8Ω+1/12Ω
•    =1/24Ω+3/24Ω+2/24Ω
•    =6/24Ω =1/4Ω
• Rt=4Ω
• Current: V=IR; 220V=I(12Ω); I=18.33A
• A and B: V=(18.33A)(3Ω) = 55V
• A and C: V=(18.33A)(5Ω) = approx. 92V
• B and C: V=(18.33A)(2Ω) = approx. 37V
• C and D: V=(18.33A)(4Ω) = approx. 73V
• A and D: 55V+37V+73 = 165V
•      or V=(18.33A)(9Ω) = 165V
• Note: if the 165-voltage drop between A and D is added to the voltage drop between the negative terminal itself, and point D, where a 3Ω resistor is interposed, the result is
• 165V + [(18.33A)(3Ω)=165V+55V=220V
178. Magnetic field strength is given in units called tesla (T). What is the equivalent of T in relation to newtons, amperes and meters?
T= N/A*m
179. The magnetic field generated by the current in a wire is continuous at all distances from the wire, so that it can be thought of as an infinite series of circles from the wire outward. There are two possible directions. What is an easy way to identify the direction of the direction of the magnetic field surrounding the wire?
• The right-hand rule:
• Imagine a current-a stream of positive charge-flowing through a conducting wire...
• ___________________________
• (-)  <-------------- I(current)  (+)
• position the right hand so that the thumb points in the direction of the current in the wire. If the hand grasps the wire in this way, the bend of the fingers around the wire will indicate the direction of the induced field surrounding the wire.
180. Consider a current moving through the conducting wire as shown below and determine which of the illustrations that follows it correctly depicts the direction of the resulting magnetic field.
____________________
I  ----------------->

A. _______------->___
________________
------->
B. _______<------____
_________________
<------
C. _______________
_________________

D. ________________
_________________
C. is correct. Using the right-hand rule, the fingers curl upward, then toward the reader over the top of the wire, then downward to curl underneath, with the thumb pointing toward the right
181. With what equation can the magnitude of the force the magnetic field exerts upon a particle? In what way can the direction of the force determined?
• F=qvB
• Force=(charge of particle)(velocity)(magnetic field strength)
• Another right-hand rule: with the thumb pointed in the direction of the current, the direction the palm is facing is the direction of the force.
182. Consider an electron that moves upward, as shown below. Assume that the particle passes through a magnetic field causing the electron to be deflected away from the reader--into the page. Characterize by illustration the direction of the magnetic field that the particle encounters.
/
|
|
* e- path
• Since you know the direction of deflection, you can place your right hand with your palm facing the direction indicated for the force. Next point your thumb in the direction of the current (opposite electron flow). Your fingers should be pointing to the left. This is the direction of the magnetic field.
•         /
• <-----|------
• <-----|------
•          * e- path
183. If, at a very small distance r from the conducting wire, the associated magnetic field has strength of B tesla, what is the strength of the magnetic field at a distance 2.5r from the conducting wire?
A. (2.5)B tesla
B. (2.5π)B2 tesla
C. I2B2π tesla
D. (0.4)B tesla
D. is correct. Strength is inversely proportional to r, the distance from the conducting wire. If at distance r, field strength equals B, then at a distance of 2.5r, the field strength will equal 1/2.5(B) or (0.4)B.
(this multiple choice question has been scrambled)
184. The transverse wave's cycle is expressed in terms of what? What is its reciprocal?
• Period: time/cycle seconds (a wave that has a period of 4 seconds is one that undergoes one cycle in 4 seconds)
• Frequency: cycle/second (if a wave undergoes 1 cycle in 4 seconds, then it  undergoes 0.25 cycles in one second. It is expressed in hertz (Hz) which is the same as "cycles per second".
185. A transverse wave travels through a visible medium. An observer directs her attention to one point on the medium, and with the appearance of a given crest, which she numbers as the first, she marks the time as zero. Including the first crest, she notes that a total of 5 crests pass her view in a period of 8 seconds. Find the wave's (a)period and (b)frequency.
• (a) The appearance of 5 crests in a period of 8 seconds indicates that the wave has undergone 4 (not 5) cycles in 8 seconds. The first crest does not represent a cycle. Rather, the appearance of 2 crests represents a cycle, and the appearance of 5 crests represents 4 cycles.
• The wave undergoes 4 cycles in 8 seconds. The fraction (8 seconds)/(4 cycles) is equivalent to the faction (2 seconds)/1, and so
• Period: the period of the wave is 2 seconds.
• (b) The wave's period is 2 seconds, and so it undergoes 1 cycle in 2 seconds. The frequency is how many cycles it undergoes in 1 second, which is the same as the reciprocal of the period.
• Frequency: 1 cycle/2 seconds=0.5 cycles/second = 0.5 Hz.
186. A transverse wave travels through a visible medium. An observer directs her attention to one point on the medium, and with the appearance of a given trough, which she numbers as the first trough, she marks the time as zero. Including the first trough, she notes that a total of 13 troughs pass her view in a period of 60 seconds. Find the wave's (a) period and (b) frequency.
• The second of the 13 troughs that pass the observer's view represents the completion of 1 cycle, and the 13 troughs in total represent the completion of 12 cycles. The wave, therefore, undergoes 12 cycles in 60 seconds.
• (a) The wave's period is derived from the fraction 60 seconds/12 cycles = 5 seconds/cycle, which is conventionally expressed as 5 seconds.
• (b) The wave's frequency is the reciprocal of its period. The reciprocal of 5/1 is 1/5=0.2, and the frequency, therefore, is 0.2 Hz.
187. What is wavelength for any wave and how can it be used to calculate the speed of any transverse wave?
• Wavelength: the distance from one crest to the next (or from one trough to the next) and is symbolized by the Greek letter lambda (λ).
• Speed: (frequency)(wavelength) = fλ m/s
188. A given transvers wave has a period of 4x10-5 seconds and a speed of 680 m/s. Find (a) its frequency and (b) its wavelength.
• a) If period = 4x10-5s, frequency = 1/(4x10-5)=0.25x105 or 2.5x104Hz
• which means it undergoes 25,000 cycles per second
• b) v=fλ; 680m/s=λ(25,000Hz)
•    =2.72x10-2m
189. With respect to any transverse wave, what does amplitude represent?
Amplitude represents the wave height; the distance from the central portion of one wave unit to the crest or the distance between the central portion of the wave unit and the trough.
190. What is a longitudinal compression wave?
a wave that travels parallel to the plane in which its own medium oscillates and gives rise cyclically to areas of increased and decreased density
191. What is rarefaction?
Since the overall mass of the bar through which a compression will travel longitudinally does not change, each region of increased density must leave behind an area of reduced density. The area of reduced density is called an area of rarefaction, and for a longitudinal compression wave, the analogs to crest and trough are compression and rarefaction, respectively.
192. If a particular sound wave has a period of approximately 0.0286 seconds and speed of 1,190 meters per second, find (a) its frequency and (b) its wavelength.
• a) Knowing the wave has a period of 0.0286 seconds (meaning it undergoes one cycle in 0.0286 seconds), identify its frequency by taking the reciprocal of that number:
• 1/0.0286 = approx. 3.5x101Hz or 35Hz
• b) Knowing that the wave's frequency is 35 Hz (meaning that it undergoes 35 cycles in one second) and that its speed is 1,190 meters per second, its wavelength (which is in reality meters/cycle) is calculated by the equation:
• speed=(frequency)(wavelength)
• 1190m/s=(35Hz)λ
• λ= 34m
193. If sound travels through two media that are equal in their resistance to compression, then it will travel:
A. more quickly or less quickly through the medium of higher density, depending on the mass of each medium.
B. more quickly through the medium of higher density.
C. more quickly through the medium of lower density.
D. at equal speed through the two media.
C. is correct. Since the two media are equally resistant to compression, density is the only factor affecting speed. The higher the density of the medium, the slower the conduction.
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194. If sound travels through two media, one more resistant to compression than the other, then it will travel:
A. more quickly through the medium that is less resistant to compression.
B. more quickly through the medium that is more resistant to compression.
C. more quickly or less quickly through the medium that is more resistant to compression, depending on the densities of the two media.
D. at equal speed through the two media.
C. is correct. The speed of sound is dependent not only on the medium's resistance to compression but also on its density.
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195. Sound has both loudness and intensity. What is the unit of each? What is the conventional equation relating loudness to intensity?
• Loudness (β) = decibel (dB)
• Intensity (I) = watts per square meter (W/m2)
• β=10 log (I/I0)
•    where β is loudness in decibels, I is intensity in watts per square meter, and I0 is the threshold of human hearing = 10-12 watts per square meter.
• NOTE: If any given intensity, I1, corresponds to a particular loudness, β1, then for any x>1, a new intensity I2, which is equal to (I1)(10x), corresponds to a new loudness, B2, which is equal to (β1+10x).
• So,
• Intensity (Z W/m2)                Loudness
• 100(Z) W/m2  corresponds to  Y+20dB (x=2)
• 1,000(Z)W/m2 corresponds to Y+30dB (x=3)
• 10000(Z)W/m2 corresponds to Y+40dB(x=4)
196. A whistle has a loudness of 88 decibels. If its loudness is increased to 108 decibels, its intensity will be:
A. multiplied by a factor of 200.
B. increased by 200.
C. multiplied by a factor of 100.
D. increased by 100.
C. is correct. For any x>1, a new intensity I2, which is equal to (I2)(10x) corresponds to a new loudness, β2, which is equal to β1 + 10x. In this instance, loudness has be increased by 20, which is equal to 10(2). The x value, therefore, is 2. Intensity, therefore, is multiplied by 10x, which in this instance, is 102=100.
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197. An engine's sound has intensity of 8x10-12 W/m2. If the intensity is increased to 8x10-9W/m2, the associated loudness, in decibels, will be:
A. increased by 3,000.
B. multiplied by a factor of 300.
C. increased by 30.
D. divided by a factor of 3,000.
C. is correct. The x value as used in this context is 3. When intensity is multiplied by 1,000, loudness is increased by 10x, or 30.
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198. The Doppler effect arises when a source of sound moves in relation of an observer (listener). From the observer's perspective, as the source of sound approaches, the wavelength shortens and the frequency increases causing the observer to hear an elevated pitch. Give the equation that expresses the Doppler effect.
• fo = frequency perceived by the observer,
• fs = frequency as actually emitted,
• v = the speed of the wave in the given medium (usually air),
• vo = velocity at which the observer moves, which is (+) if the observer moves toward the source and (-) if the observer moves away from it, and
• vs = velocity at which the source moves, which is (-) if the source moves toward the observer and (+) if it moves away from the observer.
199. A sound source moves toward an observer at a rate of 80 m/s. Assuming that the speed of sound in air is 340 m/s and that the true frequency of the source's sound is 400 Hz, what frequency would the observer perceive?
• The observer is stationary, and the sound source is moving toward the observer. The numerator of the fraction is v=340m/s, and the denominator is (340m/s-80m/s). The subtraction of 80 from 340 increases the size of the fraction, wince the sound source is approaching the observer. The equation becomes:
• fo=(400Hz)x(340m/s)/(340m/s - 80m/s)
• fo=(400Hz)(340m/s/260m/s)
•   =400Hz(1.3)= approx. 520 Hz
200. A sound source emits a sound with frequency of 280 Hz. It moves away from a listener at a rate of 30 m/s, but the listener moves toward the source at a rate of 70 m/s. Assuming that the speed of sound in air is 340 m/s, what frequency would the observer perceive?
• Both sound source and observer are in motion. The sound source moves away from the observer, which means the sign within the denominator will be (+), reducing the size of the fraction. The observer moves toward the sound source, which means the sign in the numerator will also be positive (+), increasing the size of the fraction:
• fo=(280Hz)x[(340m/s+70m/s)/(340m/s +30m/s)]
•    =(280Hz)(410m/s/370m/s)
•    =280Hz(1.1)= approx. 310 Hz
201. Simple harmonic motion is associated with force. In the case of a pendulum, the associated force is the pendulum's weight. In the case of the object affixed to a spring, the associated force is the tension within the spring. Give the equation from which force is derived.
• Hooke's law:
• F=-k(x)
• where
•    F=force
•    k=spring constant (expressed N/m), and
•    x=displacement
• Note: In the case of a pendulum, the spring constant is approximately equal to the fraction (weight of pendulum)/(length of pendulum)
202. A Hooke's law spring has a spring constant of 2,500 newtons per meter. It is stretched to a position located 0.06 meters from its relaxation point. What is the force that acts to restore the spring to its relaxation point?
• The problem calls for the application of Hooke's law
• F=-k(x)
• k is given as 2.5x103 N/m
• x is given as 0.06 meters
• F=(-2500 N/m)(0.06m)= -150N
• The (-) sign in the Hooke's law formula indicates that force is opposite the direction of displacement. The force at issue is restorative; it tends to restore the moving object to the point at which it has no potential energy. The force of a swinging pendulum is downward as the pendulum moves upward. The force of a spring is toward the relaxation point as the spring moves away from it.
203. Imagine that a pendulum has a mass of 25 kilograms and that its height at the upward end of its motion is 2 meters. What is its gravitational potential energy (relative to the bottommost position of its path)? What is its potential energy when it reaches the bottommost position of its path?
• Upward end:
• PE=mgh=(25kg)(10m/s)(2m)=500 joules (J)
• Bottommost position:
• When the pendulum reaches the bottommost position of its path, all of its potential energy has been converted to kinetic energy, potential energy is zero, and kinetic energy is at a maximum
• NOTE: Since the total sum (kinetic energy)+(potential energy) is unchanged at all points in the path, the sum of (mgh)+1/2(mv2) is constant throughout the path of simple harmonic motion.
204. A pendulum swings in simple harmonic motion and has a mass of 240 kilograms. Each time it passes the bottommost point in its path, it has speed of 8 m/s. Ignoring all forces of friction, determine the maximum height it reaches on each side of its path.
• In the case of a swinging pendulum, the potential energy to be considered is gravitational potential energy = mgh. Kinetic energy = 1/2mv2. The sum: (mgh)+(1/2)(mv2) is constant throughout the path of simple harmonic motion. At the bottommost position, potential energy is zero, and the entire sum of potential and kinetic energy is attributable to kinetic energy alone:
• 1/2(mv2)=1/2 x (240kg) x (82) = 7,680J
• At the uppermost portion of the pendulum's path, kinetic energy is zero and potential energy, therefore, must be equal to 7,680J. Knowing that (mgh)=7,680 joules, you can then solve for h,
• mgh=7680J; (249kg)(10m/s2)h=7,680J
•   2,400h=7,680J
•   h=3.2 meters
205. One musical instrument sounds a frequency of 440 cycles per second and another simultaneously sounds a frequency of 600 cycles per second. Both instruments are located on a vehicle that travels toward a listener at a rate of 90 meters per second. Assuming the speed of sound through air is approximately 340 meters per second, which of the following numbers best approximates the factor by which each frequency is multiplied in order to yield the apparent frequency perceived by the listener?
A. 0.2
B. 0.8
C. 9.0
D. 1.36
• D. is correct. Each sound source moves toward the observer at a rate of 90 m/s. The perceived frequencies will be greater than the true frequencies emitted by the instruments. The factor by which each frequency is multiplied to yield the apparent frequency is equal to the fractional component of the Doppler formula:
• f0=fs [(v±v0)/(v±vs)
• Since the observer does not move, the sum (v±vo)=340m/s. Since the sound source is moving toward the observer, frequency will increase, and the denominator must be reduced to reflect an increase in the overall size of the fraction. The sign employed in the denominator, therefore, must be (-), and the fraction becomes:
• 340/(340-90)=340/250=1.36
206. Two longitudinal compression waves, 1 and 2, travel through the same space simultaneously. Wave 1 has a period of P1 and a speed of S1. Wave 2 has a period of P2 and a speed of S2. If the wavelengths of waves 1 and 2 are represented by WL1 and WL2, respectively, which of the following represents the ratio WL2:WL1?
A. S1S2:P1P2
B. S2P2:S1P1
C. S2/P1:(P2P2)
D. P2/P1:(S1P1)
• B. is correct. The question tests your mastery of (a) algebraic manipulation and (b) the relationships among period, speed, frequency, and wavelength.
• Recalling that period is the reciprocal of frequency, and that
•      (speed)=(wavelength)(frequency)
•      (speed)=(wavelength)(1/period)
•      (speed)(period)=(wavelength)
• Wavelength is directly proportional to speed and frequency, so the ratio of WL2 to WL1 is simply S2P2:S1P1.
207. An experimenter designs a pendulum according to a carefully devised set of specifications. In a variety of experimental trials, she demonstrates that when the pendulum swings and is raised above the lowest position of its path, the force tending to draw it downward is NOT proportional to its displacement. The investigator concludes that the pendulum she has designed does not conform to Hooke's law. Is her conclusion justified?
A. No, because the conclusion cannot be drawn unless the investigator has calculated the pendulum's spring constant.
B. Yes, because according to Hooke's law, restorative force and displacement are proportional.
C. Yes, because the absence of the proportionality indicates the absence of a spring constant.
D. No, because Hooke's law applies to simple harmonic motion involving a spring.
B. is correct. For any oscillatory motion that conforms to Hooke's law, restorative force and displacement are proportional. The question describes a swinging pendulum for which restorative force and displacement are not proportional. It is a simple logical deduction, therefore, that the pendulum does not conform to Hooke's law--for the reason as stated in the indicated correct choice, that "according to Hooke's law, restorative force and displacement are proportional."
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208. Given that Hooke's law represents a linear equation: A laboratory worker experiments with a spring and mass that exhibit classic simple harmonic motion. For several positions relating to the path of motion, he records the tension in the spring and the corresponding displacement from the spring's relaxation point. On a Cartesian plane, he then plots his results, with force recorded on the y-axis and displacement on the x-axis. Which of the following descriptions would most likely characterize the resulting graph?
A. The graph would be linear, with a slope of 1.
B. The graph would be linear, with a slope of magnitude equal to the spring constant.
C. The graph would be linear, with an x-intercept equal to the spring constant.
D. The graph would be nonlinear.
• B. is correct. A linear equation as one that takes the form y=mx+b. Hooke's law represents a linear equation. Since Hooke's law, F=-k(x), represents such an equation, where m=-k and x=displacement, the graph drawn by the laboratory worker will be linear.
• For any linear equation, the slope has a magnitude equal to m, which in connection with Hooke's law, is -k, the spring constant. The graph drawn by this laboratory worker will be linear and will have a slope equal to the magnitude of the spring constant.
209. What is the speed of light?
3x108 meters per second (assigned the symbol c) and is the approximate speed at which electromagnetic radiation moves when traveling through space (a vacuum).
210. As light moves through some physical medium, its speed is reduced according to the medium's refractive index. What is this?
• The index of refraction is defined in terms of the degree to which it slows the travel of light. For any medium it is equal to the fraction
• (speed of light in vacuum)/(speed of light in medium)
• or n=c/v
• where n = refractive index of medium,
• c = speed of light in vacuum (3x108 m/s)
• v = speed of light in medium
• Index of refraction is dimensionless (without units)
211. What do the terms visible light, white light, infrared, and ultraviolet light mean?
• visible light: electromagnetic radiation whose wavelength falls between 390x10-9 meters and 700x10-9 meters
• white light: a mixture of all wavelengths within the spectrum of visible light
• infrared light: electromagnetic radiation with wavelength somewhat above 700nm
• ultraviolet light: electromagnetic radiation falling somewhat below 390 nm
212. The speed of light through water is approximately 2.26x108 meters per second. Find the refractive index of water.
• The question requires algebraic application of the equation for refractive index..
• n=c/v = (3x108m/s)/(2.26x108m/s) = approx. 1.33
213. The refractive index of fused quartz is approximately 1.46. At what speed does light travel through the substance?
• The problems relies on the equation that defines refractive index.
• n=c/v
• 1.46=(3x108m/s)/v
• v=approx. 2.5x108m/s
214. With respect to reflection and refraction, what information is known about the angle of incidence?
• Reflection: the angles of incidence and reflection are equal.
• Refraction: the angles of incidence and refraction are related by Snell's law, which provides what for each such angle, there is equality between the product --
•    sin θ x index of refraction
• Thus, n1sin θ1=n2sin θ2
• where n1 is the index of refraction for medium 1,
• n2 is the index of refraction for medium 2,
• θ1 is the angle of incidence, and
• θ2 is the angle of refraction
• NOTE: The angles of incidence, refraction, and reflection are all measured by reference to an imaginary line oriented perpendicular to the medium interface.
215. A light ray travels through a first medium and, on encountering a second, is reflected so that it makes an angle of 75 degrees with the interface between the two media. What is the associated angle of incidence?
• The 75 degree angle to which the problem refers is made between the reflected ray and the interface between the two media. It is NOT the angle of reflection. Rather the angle of reflection falls between the reflected ray and a line that runs perpendicular to the interface. The angle of reflection is complementary to the 75 degree angle mentioned in the problem:
• 90 degrees - 75 degrees = angle of reflection
•  = 15 degrees
• Since the angle of incidence and the angle of reflection are equal, the angle of incidence in this case is also 15 degrees.
216. A light ray traveling through a medium encounters a second medium, forming an angle of 60° with the interface between the two media. If the first medium has a refractive index of R1 and the second, a refractive index of R2, which of the following expressions best characterizes θ2, the angle of refraction (sin 30°=0.5; sin 60°=0.867)?
A. arccos(0.867)xR1/R2
B. arccos(R1)(R2)(0.867)
C. arcsin (90-60)(R2)
D. arcsinR1/2R2
• D. is correct. The 60° angle to which the problem refers does not constitute the angle of incidence, as it is formed by the media interface and the incident light ray. The angle of incidence is equal to the difference between that angle and 90°.
• The angle of incidence is, then, 90°-60°=30°. The sin of 30° is 0.5. Using Snell's law:
• n1sinθ1=n2sinθ2  R1(0.5)=R2sinθ2
• [(0.5)(R1)]/R2=sinθ2
• 1  x  R1  =  sinθ2      R1   =  sinθ2
• 2      R2                 2(R2
•      θ2=arcsin(R1)/2(R2)
217. What is the critical angle (θcr) and what modification of Snell's law is used to calculate it?
• Total internal reflection refers to a ray of light that travels through a medium and encounters a second medium with a lower index of refraction than the first. If the angle between the incident ray and the line perpendicular to the interface exceeds the critical angle (θcr), all of the incident light will be reflected back into the first medium. The quantitative aspects of total internal reflection pertain to the evaluation of the critical angle for two interfaces, and are calculated with a modification of Snell's law:
• sin θcr=nb/na
• where θcr = the critical angle that applies to the two media in question,
• nb = the index of refraction for the second medium, and
• na = the index of refraction for the first medium
218. All spherical mirrors feature a focal length, at the end of which sits a focal point. Give the equation used to determine focal length. Where is the focal point for a convex and a concave mirror?
• f=(1/2)Rc
• where
• f= focal length, and
• Rc = mirror's radius of curvature
• Convex: the focal length extends into the mirror, and the focal point is said to be located behind it. (By convention, when the focal point is located behind a mirror, its associated focal length is given a negative sign.)
• Concave: the focal length is directed away from the mirror, and the focal point is said to be located in front of it. (By convention, when the focal point is located in front of a mirror, its associated focal length is assigned a positive sign.)
219. A concave mirror has a radius of curvature equal to 40 centimeters. Identify the location of its focal point in terms of distance and sign.
• The mirror is concave, and its focal point is located in front of the mirror, which gives it a positive sign.
• f=(1/2)Rc; Rc = 40cm
• f=(1/2)40cm = 20cm
• The focal length is +20cm.
220. A convex mirror has a radius of curvature equal to 162 centimeters. Identify the location of its focal point in terms of distance and sign.
• The focal length is directed into the mirror, and the focal point therefore carries a (-) sign. The absolute value of the focal length is equal to:
• f=(1/2)Rc; Rc = 162cm
• f=(1/2)162cm = 81cm
• The focal point is located -81 centimeters form the mirror.
221. The location of a mirror's focal point is not necessarily the location of the image the mirror produces. Rather, the position of a mirror image produced by some object is determined by the thin lens equation, which, despite its name, applies to mirrors as well as lenses. Give this equation.
• (1/f)=(1/o)+(1/i)
• where
•   f = the mirror's focal length,
•   o = the distance of the object from the mirror, and
•   i = the distance of the image from the mirror.
• Although a concave mirror has its focal point located in front of the mirror, it may produce images behind the mirror.
222. Explain the distinction between real and virtual images. Also describe the difference between upright and inverted images.
• Real: A mirror image that is located in front of the mirror -- which means it cannot be seen in the mirror but rather on a screen held in front of the mirror -- and it is designated with a positive (+) sign.
• Inverted: Any real image.
• Virtual: A mirror image that is located behind a mirror -- which means that it is visible in the mirror -- and it is designated with a negative (-) sign.
• Upright: Any virtual mirror image.

• Mirror Image Local ||  R/V   || Orientn  || Sign
• behind mirror        || virtual || upright  || (-)
• in front of mirror    ||  real   || inverted || (+)
223. A convex mirror has a radius of curvature equal to 80 centimeters, and an object is situated 20 centimeters in front of it. Identify the location of the resulting image in terms of distance and sign.
• Find focal length
• f=(1/2)Rc; Rc = 80cm
• f=1/2(80cm) = 40cm
• Because the mirror is convex, f is negative.
• Focal length: =-40cm
• Find image location
• 1/f=1/o+1/i
• 1/-40cm = 1/20cm + 1/i
• 1/-40cm - 1/20cm = 1/i
• -1/40cm - 2/40cm = 1/i
• -3/40cm = 1/i
• Image location: =-(40cm/3) = appro.-13.33cm
• The negative sign indicates that the image is behind the mirror and thus virtual and upright.
224. Spherical mirrors may produce magnification, and the size of an image may therefore vary from size of the object that generates. Give the fraction equal to magnification.
• m=-i/o
• where
• m= magnification
• i= distance of image from mirror, and
• o= distance of object from mirror
225. A convex mirror has a radius of curvature equal to 60 centimeters, and a 20-centimeter object is positioned 15 centimeters in front of it. Determine the size of the image that results.
• First, calculate the distance of the image from the mirror.
• Find focal length:
• f=(1/2)Rc; Rc = 60cm
• f=1/2(60cm) = 30cm
• Focal length: -30 cm (negative because it is a convex mirror)
• Find image distance
• 1/f = 1/o + 1/i
• 1/-30cm = 1/15cm + 1/i
• -1/30cm - 2/30cm = 1/i
• Image distance: =-30cm/3 = -10cm
• Find magnification;
• m=-i/o
• m=-(-10cm)/15cm = 10cm/15cm = 0.67
• Magnification: 0.67
• Since the object's true size is 20 cm,
• the mirror image will have a size equal to:
•   (0.67)(20cm) = 13.4 cm
• Note: A negative magnification would indicate an inverted image.
226. Lenses may be converging or diverging. Each term refers to the effect of the given lens on light rays that pass through it. Describe the difference.
• Converging lenses produce the convergence of rays toward the lens's central axis; the focal point is that point at which parallel rays traversing the lens will come together though refraction. Diverging lenses produce the divergence of rays away from the central axis; the focal point represents that point at which parallel rays that have traversed the lens, and so have been subjected to divergence, will meet by hypothetical linear extensions.
•     Converging           Diverging
227. Through the lensmaker's equation, the focal length of a lens is determined through (a) the refractive index that applies to the material of which the lens is made and (b) the radius of curvature associated with each side of the lens. Give the lensmaker's equation.
• where
•   f = focal length
•   n = refractive index of the lens material
•   R1 = radius of curvature for side 1, and
•   R2 = radius of curvature for side 2
228. The left side of a convergent glass lens has a 20-centimeter radius of curvature, and the right side has a 5-centimeter radius of curvature. Assuming that the refractive index of glass is 1.5, find the focal length of the lens.
• Imagine that light is incident from the right side, then side 1 is on the right, and R1 = 5cm. Side 2 is on the left, and R2 = 20cm. Since side 1 is on the right (and is convex), its center of curvature is on the left - the outgoing side - and it takes a positive sign, R1 = +5cm. R2, then, with center of curvature on the incoming side, takes a negative (-) sign and is equal to -20 centimeters. Application of the lensmaker's equation yields:
• 1/f=(n-1)[(1/R1)-(1/R2)]
•    =(1.5-1)[(1/5)-(1/-20)]
•    =(0.5)[(4/20)+(1/20)]
•    =(0.5)(5/20)
•    =(1/2)(1/4) = 1/8
• f=8.0cm
229. How are the images formed with lenses unlike mirrors?
• With lenses, an image formed behind a lens is real and inverted, and one formed in front of a lens is virtual and upright.
•           ||   Type    ||    Image    || Focal point
•    For   || concave || real, invrt  || in front (+)
• Mirrors||  convex  ||virtual, uprt|| behind (-)
•    For   || divrgng  ||virtual, uprt|| in front (-)
• Lenses ||  cvrgng  ||real, invrtd || behind (+)
230. A divergent glass lens has a focal length of 80-centimeters. A 40-centimeter object is held in front of it at a distance of 20 centimeters. Ascertain the location and size of the image, and determine whether it will be real, virtual, upright, or inverted.
• The absolute value of the lens's focal length is 80cm; since the lens is divergent, the sign is negative (-). The focal length, therefore, is -80cm. To locate the image, apply the equation:
• 1/f=1/o+1/i
• -1/80cm=1/20cm+1/i
• -1/80cm-1/20cm=1/i
• -1/80cm-4/80cm=1/i
• -5/80cm=1/i
• i=-80/5cm = -16cm
• Image location: -16cm from the lens, which means it is 16 cm in front of the lens. Any image located in front of the lens is virtual and upright.
• Image size is = object size x magnification
• m=-i/o = -(-16cm)/20cm = 0.8
• Image size: (40cm)(0.8) = 32cm
231. What is focal power?
• Focal power expresses the degree to which a lens imposes a convergence or divergence on the light rays that traverse it. The converging lens of relatively greater power produces greater convergence than does one of relatively lesser power; the diverging lens of relatively greater power produces greater divergence than does one of relatively lesser power.
• It is expressed in the unit diopter.
• It is equal to the reciprocal of the focal length:
•   Focal Power (Pf)=1/f
• Power of converging lens carries positive sign.
• Power of diverging lens carries negative sign.
232. A lens bring parallel light rays to a focus, as shown below. Determine its focal power.
• The illustration reveals a convergent lens with a focal length of +22cm. Expressed in meters, the focal length is +0.22meters, and the focal power is:
• 1/0.22 diopters
• = approx. 4.5 diopters
233. A patient complains of blurred vision when viewing objects located in the distance, but reports visual clarity when viewing objects located close to his eye. A physician hypothesizes that the difficulty is attributable to a condition in which the right retina is located at an abnormally large distance from the anterior surface of the right eye. Which of the following findings would best confirm her hypothesis?
A. The patient reports improvement when a convex lens is placed in front of his right eye.
B. The patient reports improvement when a concave lens is placed in front of his right eye.
C. The patient reports improvement when he opens his eyes more widely.
D. The patient reports that he does not experience the symptom when either eye is closed and the other remains open.
• B. is correct. The patient is unable to focus objects in the distance, suggesting that as parallel light rays traverse the refractive components of his eye, they experience such convergence as produces a focal point anterior to the retina. Either one or both of his eyes has excessive focal power or his retina is located too far back.
• That the patient clearly sees objects located near to him tends to confirm the physician's hypothesis, since, as an object comes closer to a lens, the rays that emanate from it show increased divergence in relation to one another. For such rays, the focal point would be moved backward and might focus on the retina as they should.
• If the physician's hypothesis is correct--that in relation to the retina's location, parallel rays of light are focused too far forward--some improvement should accompany the placement of a concave lens in front of the eye. The lens would diverge the incoming light rays before they reached the eye, and would present the eye with rays that, in their divergence, resemble rays that come from objects situated close to it. If the placement of a concave lens brings improvement, the hypothesis is supported.
234. Give the diagram below, what must the approximate value of θ2 be such that it his the chest as shown?

A. 30.4o
B. 63.4o
C. 15.2o
D. 45.6o
A. is correct. Using the rules for right-angled triangles, tan θ2 = 1000/1700 = 1/1.7. From knowledge, tan 30o = 1/(√3), which is approximately equal to 1/1.7. Therefore, θ2 is approximately 30o.
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235. How does the refractive index in water for violet light compare with that of red light given that violet light travels more slowly in water than red light?
A. nviolet = nred
B. nviolet < nred
C. nviolet > nred
D. This depends on the relative speeds of the different colors in a vacuum.
• C. is correct.  n = c/v, n is inversely proportional to v
• c=vn, or c=v1n1 and c=v2n2,
• ∴ v1n1 = v2n2 = constant
• Thus vrednred = vvioletnviolet = constant
• ∴ given ↓vviolet then ↑nviolet
• and since the value of vviolet is lower than the value of vred, then the increase for nviolet will be greater than the value of nred.
236. Snell's law allows one to follow the behavior of light in terms of its path when moving from a material of one refractive index to another with the same, or different refractive index. It is given by: n1sinθ1=n2sinθ2, where "1" refers to the first medium through which the ray passes, "2" refers to the second medium, and the angles refer to the angle of incidence in the first medium (θ1) and the angle of refraction in the second (θ2).
Total internal reflection first occurs when a beam of light travels from one medium to another medium which has a smaller refractive index at such an angle of incidence that the angle of refraction is 90o. This angle of incidence is called the critical angle. What is the value of the sine of this angle when the ray moves from water towards air?
A. 0.50
B. 0.75
C. 0
D. 2π
B. is correct. Using the derivation of Snell's law which refers to the critical angle, sin θc = n2/n1= nair/nwater (see question text), where "θc" is the unknown critical angle. Thus sin θc = 1.00/1.33 ≈ 1/(4/3) = 3/4 = 0.75.
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237. In the case where it is stated that total internal reflection first occurs when a beam of light travels from one medium to another medium which has a smaller refractive index at such an angle of incidence that the angle of refraction is 90o. This angle of incidence is called the critical angle. Given that the value of the sine of this angle when the ray moves from water towards air is 0.75, what would happen to the critical angle if the beam of light was travelling from water to a substance with a greater refractive index than air, but a lower refractive index than water?
A. It would increase.
B. It would decrease.
C. It would remain the same.
D. Total internal reflection would not be possible.
• B. is correct. First: sin θc; = n2/n1 = nair/nwater
• Now: sin θc2 = (nair + x)/nwater
• Thus we increased the numerator by some value "x" which means sin θc2 > sin θc, which also means θc2 > θc.
238. The refractive index of a transparent material is related to a number of the physical properties of light. In terms of velocity, the refractive index represents the ratio of velocity of light in a vacuum to its velocity in the material. From this ratio, it can be seen that light is retarded when it passes through most types of matter. It is worth noting that prisms break up white light into seven "colors of the rainbow" because each color has a slightly different velocity in the medium.
Which of the following would you expect to remain constant when light travels from one medium to another and the media differ in their refractive indices?
A. Velocity
B. Frequency
C. Intensity
D. Wavelength
B. is correct. Both velocity and wavelength (different colors) change (recall velocity and wavelength are directly proportional). One should also know that the intensity (= the rate of energy propagation through space) of a light ray decreases when it passes through a medium other than air (its velocity decreases in the medium and this leads to an overall loss of energy).
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239. The following system includes a frictionless pulley and a cord of negligible mass. Since the system is at rest, what can be said about the force of friction between the platform and the large weight?

A. It is 190 N.
B. It is 10N.
C. in this case, the force of friction is not necessarily present.
D. It is 200N.
B. is correct. Since the 200 N block is not in motion, the value of the frictional force is equal and opposite in direction to the applied force, that is, 10 N.
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240. Two charged particles are a distance x apart. If the charges on the two particles remains the same while the distance between them is doubled, how does the force between the particles change?
A. It decreases by a factor of 4.
B. It decreases by a factor of 2.
C. It stays the same.
D. It decreases by a factor of √2.
A. is correct. From the equation F = kq1q2/r2, if the distance r is doubled then (2r)2 = 4r2, which means 4 x r2 in the denominator. Thus the original force is quartered (=decreased by a factor of 4).
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241. The intensity level of sound X is 1000 times that of sound Y. What is the difference in the intensity levels of X and Y in terms of decibels?
A. 1000
B. 3
C. 100
D. 30
• D. is correct. Let the number of decibels of sound Y be dBY = 10 log (I/Io). If the intensity 'I' of sound X increases by a factor of 1000 then:
• dBX = 10 log (1000 I/Io) = 10 log (10)3 + 10 log (I/Io) = 30 + dBY.
242. A mass of 100 kg is placed on a uniform bar at a point 0.5 m to the left of a fulcrum. Where must a 75 kg mass be placed relative to the fulcrum in order to establish a state of equilibrium given that the bar was in equilibrium before any weights were applied?
A. 0.66 m to the right of the fulcrum.
B. 0.66 m to the left of the fulcrum.
C. 0.38 m to the right of the fulcrum.
D. 0.38 m to the left of the fulcrum.
• A. is correct. A simple torque force problem: let the fulcrum (= the center of gravity of the bar) be the pivot point (draw a vector diagram):
• ∑L; = CCW - CW
•  =(100 kg)(g)(0.5 m) - (75 kg)(g)(x) = 0
• Thus (100 kg)(g)(0.5 m) = (75 kg)(g)(x)
• g cancels, manipulate to get: x = 50/75 = 2/3 m = 0.67 m (CW is to the right of the fulcrum).
243. A collision is an isolated event in which two or more moving or colliding bodies exert forces on each other for a relatively short time.
A 1200kg car is travelling at 7.5m/s in a northerly direction on an icy road. It crashes into a 8000kg truck in the same direction as the car with a velocity of 3.0m/s before the collision. The speed of the car after the collision is 3.0m/s in its original direction.
Which of the following is true regarding the relationship between energy and momentum in the above text?
A. The collision is not perfectly elastic, both momentum and energy are not conserved.
B. The collision is inelastic, kinetic energy is conserved but momentum is not.
C. The collision is not perfectly elastic, momentum is conserved but total energy is not.
D. The collision is not perfectly elastic, momentum is conserved but kinetic energy is not.
D. is correct. The external force acting on the cars is tiny compared to the large forces of the colliding vehicles (i.e. friction is negligible), so momentum is conserved (PHY 4.3). The total kinetic energy before collision is not equal to the total kinetic energy after the collision, thus it is not conserved and the collision is not completely elastic. It must be remembered that during a collision energy may change form (i.e. kinetic, heat, sound, etc.), but total energy must be conserved.
244. A collision is an isolated event in which two or more moving or colliding bodies exert forces on each other for a relatively short time.
A 1200kg car is travelling at 7.5m/s in a northerly direction on an icy road. It crashes into a 8000kg truck in the same direction as the car with a velocity of 3.0m/s before the collision. The speed of the car after the collision is 3.0m/s in its original direction.
What is the velocity of the truck after the collision?
A. 7.5 m/s
B. 3.7 m/s
C. 3.0 m/s
D. 1.1 m/s
• B. is correct. Using the principle of Conservation of Momentum:
• Momentum before collision = Momentum after collision
• (1200 kg)(7.5 m/s) + (8000 kg)(3.0 m/s) = (1200 kg)(3.0 m/s) + (8000 kg)(x m/s)
• Divide through by 400:
• (3)(7.5) + (20)(3) = (3)(3) + 20x
• Isolate x: (7.5 + 20 - 3)3/20 = x
• Thus x = (24.5)(3/20) = 73.5/20 = 3 13.5/20 ≈ 3 7/10 = 3.7 m/s.
• {Recall: momentum is a vector; in this problem the sign for the velocities are always positive because we are told that the vehicles are always moving in the same (i.e. northerly) direction}
245. A collision is an isolated event in which two or more moving or colliding bodies exert forces on each other for a relatively short time.
A 1200kg car is travelling at 7.5m/s in a northerly direction on an icy road. It crashes into a 8000kg truck in the same direction as the car with a velocity of 3.0m/s before the collision. The speed of the car after the collision is 3.0m/s in its original direction.
The car then proceeds to a garage. To get there, the driver turns off onto a smooth road with a coefficient of friction = 1/3. He then stops for a snack and then tries to drive off. What is the value of frictional force when the force the car exerts is 300 N?
A. 0 N
B. 100 N
C. 300 N
D. 4000 N
• C. is correct. F = f < fmax; KW: tries
• This question is fun! We begin with the fact that the maximum frictional force that can be exerted occurs when the car is just about to move, and is equal to the product of the coefficient of friction (static) and the normal force on the car. Since the car is not moving perpendicular to the surface, using Newton's Second Law (F = ma), since the acceleration 'a' when the car is about to move is zero, the normal force is equal to the weight of the car (i.e. ∑F = N - Mg = ma = 0; thus N = Mg).
• Therefore, the maximum frictional force = (1/3)(1200 kg)(10 m/s2)= 4000 N, approximately since we estimated 'g'. Since the force the car exerts (300 N) is less than the maximum frictional force on the car (4000 N), the car is not in motion. For an object not in motion, the sum of forces must be equal to zero (Newton # II !). Because the exerted force is 300 N, the actual frictional force must be 300 N in the opposite direction.
246. After leaving a garage, the driver of the car follows the road and eventually has to go up a hill. How does the frictional force on the car now compare to the value when the car was driving on level ground?
A. It increased.
B. No change
C. The direction of change depends on the angle of elevation.
D. It decreased.
D. is correct. The normal force on an object always acts perpendicular to the surface, in this case, the road. Friction depends on the normal force. Since only a component of the weight of the car acts perpendicular to the hill, the value of the normal force decreases. Since the coefficient of friction remains the same, the value of the maximum frictional force decreases. {Once an object is on an incline: N < weight of the object as determined by the cosine of the angle of the incline}
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247. If the 12000N car is moving up a 30o hill at 5 m/s and the car is 40m up the hill, how much potential energy does the car possess at that point? (g=9.8m/s2)
A. 2.40 x 105 J
B. 2.40 x 104 J
C. 4.95 x 105 J
D. 4.95 x 104 J
• A. is correct. The potential energy (P.E.) = mass (m) x gravitational acceleration (g) x perpendicular height (h).
• "h" is given by 40sin30o = (40)(1/2) = 20 m.
• The weight "m x g" is given already as
• 12000 N. Thus (20)(12000) = 240000 J.
248. A solar collector measure 20m by 20m is exposed to the sun. The sun radiates energy at a rate of about 3.9x1026W and the solar energy incident on our planet is received at a rate of approximately 1600W/m2.
Approximately 4 L of water is passed through the solar collector per second. The specific heat capacity of water is 4.2J/goC and the combined heat capacity of the collector walls and copper pipes is negligible.
Assuming that the thermal conductivity of the copper is large enough to allow almost instantaneous energy transfer, what is the temperature difference between the water in the supplying pipes and the water in the draining pipes using the data in the passage? (Assume that the water covers the whole area of the collector) Density of water = 1 g/cm3.
A. 38,100.0 oC
B. 38.1 oC
C. 76.2 oC
D. 9.5 oC
• B. is correct. Dimensional analysis is cool because it means you don't need to memorize an equation (!), you just have to define the terms and follow the units.
• Area of collector = (20 m)(20 m) = 400 m2
• Incident solar energy = 1600 W/m2 = 1600 J/sm2
• Therefore, energy received per second by collector = 1600 J/sm2 x 400 m2 = 640000 J/s
• Rate of flow of water = 4 L/s = 4000 cm3/s
• Mass of water going through collector in one second = (4000 cm3)(1 g/cm3) = 4000 g
• From Q = mcΔt: 640000 J = (4000 g)(4.2 J/goC)(Δt)
• Thus Δt = 38.1 oC
• {The answer could be quickly estimated by approximating 'c' as 4 and dividing through by 16000; the answer obtained would be 40 oC which is sufficiently accurate for this problem}
249. The amount of electricity that can be generated from the fluid coming from the draining pipe of the collector is dependent upon its temperature. Therefore, it might be more efficient to use a fluid with which of the following properties?
A. A lower specific heat capacity
B. A lower thermal conductivity.
C. A higher specific heat capacity.
D. A higher thermal conductivity.
A. is correct. The specific heat capacity is the energy required to raise the temperature of a unit of mass of a substance by one degree. The smaller the specific heat capacity, the less energy is required to raise the temperature of that substance by a certain amount.
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250. A 0.2 kg ball accelerates from rest at 5 m/s2 for five seconds. It then collides with a 0.5 kg ball which is initially at rest. The smaller ball stops moving while the larger ball begins its motion. After the collision, how fast is the larger ball initially moving?
A. 2.0 m/s
B. 10 m/s
C. 2.5 m/s
D. 5.0 m/s
• B. is correct. There are two parts to this problem. We begin by calculating the impact velocity of the 0.2 kg ball using the equation for speed V at any time t:
• V = Vo + at, where Vo is the original speed. Since the ball starts at rest we get:
• V = 0 m/s + (5 m/s2)(5 s) = 25 m/s
• Using the principle of Conservation of Momentum and momentum = mass x velocity, we get:
• Total momentum before collision = total momentum after collision
• (0.2 kg)(25 m/s) + (0.5 kg)(0 m/s)
• = (0.2 kg)(0 m/s) + (0.5 kg)(x m/s)
• Isolate x: x = 10.0 m/s
251. A vertically oriented spring is stretched by 0.50 m when a mass of 1 kg is suspended from it. What is the work done on the spring?
A. 5.0 J
B. 2.5 J
C. 0.5 J
D. 20.0 J
• B. is correct. F = mg = kx; k leading to W
• At equilibrium the force of the weight (W = mg) will be equal to the spring force (Fs = -kx), thus: kx = mg, k = mg/x
• = (1 kg x 10 m/s2)/(0.5 m) = 20 kg/s2
• Work done (spring) = 1/2 kx2
• = 1/2 (20)(0.5)2
• = 1/2 (20)(1/4)
• = (20)/(8) = 5/2 = 2.5 J.
• {Note F = mg, g is estimated as 10 m/s2}

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