66. B if the wave pulses are moving toward each other, and they have the same amplitude (height), then they will sum together to give zero amplitude.

67. Yes, the velocity of a wave depends on the wavelength and the frequency. If it is on a string then it will be a transverse wave. The wave velocity will be moving along the direction of the wave, the particle will move up and down vertically and perpendicularly to the direction of the wave.

68. a) V = λf  λ = V/f = (335m/s)/(410Hz) = 0.817 m

b) if the bottom number above is doubled, then the whole fraction is half as big. (0.409 m)

c) amplitude

69. A V = λf  f = V/λ =( 10m/s)/20m = 0.5Hz

70. A There are 2.5 waves between point X & Y, so we can find the frequency by understanding that the frequency is a certain amount of things happening (waves) in a certain amount of time. So the frequency is = 2.5 waves / 0.02s = 125Hz

Now we can find the period by the equation T = 1/f  T = 1/125Hz = 0.008s

71. We assume that this is light, so we can use the equation n = c/V = (3 X 108 m/s)/(2.1 X 108 m/s) = 1.4

The frequency is extra information that you don’t need.

72. D The frequency of light doesn’t change, but the speed and wavelength will change. Light slows down when it moves through a medium that is more dense like air into water.

73.

Destructive interference Constructive interference

The above waves will sum together to give a resultant wave. The destructive interference example on the left sum together to give a zero amplitude, and the constructive interference example on the right sum together to give an amplitude of A + B.

74. C When light moves from one medium into another the speed depends on the density, or the index of refraction. The smaller the index the faster it goes. So the light slows down from air (n=1.00) to glass (n=1.52 or n=1.61) and then from glass into water (n=1.33) it will speed up because from glass to water the index decreases.

75. A Use the equation n1sinθ1 = n2sinθ2 where n1 = the index of refraction of air n = 1.00, and n2 = the index of refraction of water n = 1.33. θ1 = 30.6 degrees.

(1)sin(30.6) = 1.33sinθ  divide both sides by 1.33  0.3827379 = sinθ  take inverse sin of both sides

Θ = 22.5 degrees.

Notice how the light bends towards the normal (dotted line) when going from a medium of low index to one with a high index.

If it were reversed where the light was going from water to air then the light would bend away from the normal.

76. B Sound needs to be in a medium, and in this case the medium is air. When the pump is turned on the sound’s intensity will decrease until it stops completely.

77. a) Light waves are produced by the sun, whereas sound is produced by the compression of a material like air, water or metal.

b) The speed of light is 3X108m/s in a vacuum and decreases when it gets into a denser medium whereas the speed of sound in air is about 330m/s and increases as it gets into a denser medium.

c) (Reference answer #73) When light goes through destructive interference the light gets dimmer than the sources and gets more intense than the sources when undergoing constructive interference. Sound can do the same, but we say that it can get quieter during destructive interference and louder during constructive interference.

78. V = IR In a voltage vs. resistance graph you would see a straight diagonal line showing a direct relationship. The slope would be the current. However, in a circuit with constant voltage (same battery) if the resistance decreases then the current must increase.

79. C A will have only one lit bulb, B will have all bulbs lit, and D is a completely open circuit where no bulbs are lit.

80. A R2 is directly under the battery. If you take out R2 then it will be a series circuit. The resistors add in a series circuit, so the total resistance would be 5.1Ω + 5.1Ω = 10.2Ω. To find the current you need to use V = IR and solve for I.

I = V/R = 12V/10.2Ω = 1.18A or 1.2A

81. C The bigger the current the more electrons can give energy to the bulb. So the circuit with the biggest current will have the brightest bulbs. Use V = IR and solve for I for each circuit. The one with the biggest current will be the one with the brightest bulbs. When a circuit has low resistance and high voltage the current will be at a maximum.

82. C The potential difference is like an electric field making the charges move.

83. A Use V = IR and solve for current. I = 6v/6Ω = 1A

84. B The longer the wire the more the resistance, the thinner the wire the more the resistance and the more resistivity the wire has the more resistance. The plastic on the wire won’t affect the resistance of the wire.

85. B The electrons in insulators are not free to move as easily as in conductors. The electrons are held tightly to the nucleus.

86. C Understand protons don’t move. So A & B are eliminated. In the metal sphere the electrons can move to the rod.

87. D Alternating current alternates (reverses) direction several times per second. Remember that we use AC to send energy to buildings because it is easier to step up or down voltage using a transformer.

88. a) V = IR  I = V/R = 12V/6Ω = 2A

b) P = IV = 2A(12V) = 24W

89. A P = E/t The rate of anything is that anything divided by time. So the rate that energy is used is energy

90. D Find the total resistance using 1/12 + 1/6. The total resistance is 12Ω/3 = 4Ω. Now use V = IR = 4A(4Ω) = 16V

91. C The total resistance is 10.5Ω . You get this by adding the parallel branch to the lone resistor. First you have to get the total resistance of the parallel branch by 1/5Ω + 1/5Ω = 2/5Ω . Flip both sides RT = 5Ω/2 = 2.5Ω. The total resistance is equal to the lone resistor plus the parallel branch, or 10.5Ω = RL +2.5Ω  RL = 8Ω

92. C To find the equivalent resistance, first add the 25Ω and the 31Ω resistor in series to give 56Ω. Now the 56Ω and the 41Ω are in parallel, add those to get 23.67Ω. Now the 23.67Ω and the 15Ω are in series, add together to get 38.67Ω.

93. C 1/4Ω+1/5Ω+1/6Ω find the common denominator, flip both sides and get 1.6Ω.

94. D If the resistors are the same then the voltage across R2 will be the same as R1, so we attach the voltmeter in parallel and then the ammeter should be connected in series, right after (or before) the resistor.

95. BThe net charge is zero, but the comb is more negative and the hair is more positive.

96. The negatively charged comb pushes the electrons in the paper to the other side of the atoms in the paper. The resultant positive side is attracted to the negative comb.

97. Like #95, before the comb moves through the hair, the comb and hair are neutrally charged. When the comb moves through the hair and takes electrons off the hair, the comb is negative and the hair is positive, but the total charge of that system is the same as before = conservation of charge.

98. D Outward from positive charges and inward into negative charges.

99. The closer you get the stronger the field. The bigger the charge creating the field the stronger the field.

100. D F = (kqq)/r2 one q is positive and one is negative. Since they are a proton and an electron the force will be negative, and the magnitude of the charge of a proton and electron are the same. On the front of the reference table look for elementary charge 1.6X10-19C