Download Spring Practice Multiple Choice Answers 1 D Acceleration produces

Spring Practice Multiple Choice Answers
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D Acceleration produces a change in velocity and/or direction.
B Fair = bv = ma  aair = bv/m  overall a = g – bv/m (down)
F = -kx  -(0.50 kg)(10 m/s2) = -k(0.25 m)
C g
 k = 20 N/m
Because the surface is rough, some energy is lost, potential
E
energy after sliding is less and h' will be less.
C d = ½at2  3.0 m = ½a(1 s)2 a = 6 m/s2
F is proportional to mass on the Earth or Moon, but centripetal
C g
acceleration doesn't depend on mass (ac = v2/r).
Momentum is conserved, The total momentum after release equal
E momentum before, which is zero,  momentums must be equal
but opposite in direction.
Gravitational force between spheres is equal, but opposite in
E
direction.
At maximum displacement, the oscillating object is momentarily
A stopped while it changes direction, but the spring restoring force is
at a maximum (F = -kx).
Momentum is conserved (p = mv). Since the total mass is greater,
D
then the combined velocity must be less.
Total momentum increases, so an external force (impulse) must
B
have been applied, which could have come from a loaded spring.
Velocity is horizontal at that instant and acceleration is upward
B (centripetal acceleration causes the car to climb up the other side
of the hill).
Resistivity is temperature dependent. Some substances lose all
C
resistance at very low temperatures.
Loop rule states that the change in voltage around any complete
B
circuit equals zero, which is a conservation of energy concept.
C4 + 2 = 4 F + 2 F = 6 F
D 1/C4-2 + 3 = 1/6 F + 1/3 F = 1/2 F  C4-2 + 3 = 2 F
C4-2-3 + 5 = 2 F + 5 F = 7 F
V across the 5 F capacitor = 100 V (in parallel)
B
Q = CV = (5 F)(100 V) = 500 C
The electric field, E, is constant between the capacitor plates, 
B
force is constant (Fe = qE).
Right hand rule: Thumb pointed right, straight fingers point to the
E
top of the page, then curl 90o, tips are out of the page.
The magnetic fields are in the same direction between the wires,
E so the field can't be zero between the wires. Outside the wires, the
fields are opposite, but not equal,  nowhere.
D E = IR = (0.3 A)(4  + 16  + 20 ) = 12 V
C V = E – IR = 12 V – (0.3 A)(4 W) = 10.8 V
A P = I2R = (0.3 A)2(4 ) = 0.36 W
C U = Qin + Win = 275 J + 125 J – 50 J = 350 J
Qmetal = Qliquid  mmetalcmetalTmetal = mliquidcliquidTliquid
C (1.5 kg)(200 J/kg•Co)(100 -20)oC = (3 kg)(1,000 J/kg•Co)T
T = 8oC
B P = Q/t = (1.5)(200)(80)J/5 s = 4800 W
B Converging lenses are thicker in the center than the edges.
D v = f = (5 s-1)(2 m) = 10 m/s
B fn = nf1  5 s-1 = (2)f1  f1 = 2.5 s-1
Radio has the longest wavelength, lowest frequency (c = f) and
B
energy (E = hf) and gamma is opposite.
A Sound waves in air are longitudinal.
E Image position corresponds, but on the other side of the mirror.
B missing mass: 235 + 1 – (138 + 95) = 3 3 neutrons
Total mass is less because energy is released. Total nucleons is
A
conserved and how the nucleus breaks apart is random.
D It takes 3 ½-life to go from 1  1/8  3 x 5 yrs = 15 yrs.
4  3 = 1 eV, 4  2 = 3 eV, 4  1 = 6 eV, 3  2 = 2 eV,
D
3  1 = 5 eV, 2  1 = 3 eV (4 eV is impossible)
E Lowest energy transition produces the longest wavelength.
Photoelectric effect correlates each released electron with the
C
absorption of a quanta of energy (photon as particle).
C Interference is a wave phenomenon.
A E = vLB
C Fg = GMm/r2 and Fe = kQq/r2, but nuclear force decreases with r3.
Maximum positive acceleration occurs when the object is at the
A
lowest point in its motion (F = -kx = ma).
E P = W/t = Fd/t = mgh/t
B F = ma  mg = (3 m + m)a  a = g/4
E The package slides because of inertia. There wasn't enough
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friction to generate the centripetal acceleration inward.
D K = Us  ½mv2 = ½kx2  (m)vo2 = kx2  x = vo(m/k)½
F = mg = GMm/r2  G = gr2/M  gErE2/ME = gMrM2/MM
D g
10 m/s2(1)2/(1) = gM(½)2/1/10  gM = 4 m/s2
D Kelectron = hfphoton –  frequency and nature of surface ().
B P = V2/R  power is greater when Rtot is lowest (parallel).
C IVt = (C/s)(J/C)s = J
E c = f  3 x 108 m/s = f(0.03 m)  f = 1 x 1010 s-1
Isotopes have the same atomic number (protons), but different
A
mass (p + n).
A e = W/Qin = (100 J – 60 J)/(100 J) = 40 %
Angular momentum deals with radius of orbit and velocity,
A
answer deals with orbital values.
When balanced counter clockwise torque = clockwise torgue m1r1
D
= m2r2  6x = 8(7 - x)  x = 4
C f' = f(vw ± vo)/(vw ± vs)
The left sides of p and Q combine to make an extra high crest and
A
the right sides of p and Q cancel each other.
I and II are both interference phenomena where  is similar to the
C bubble thickness and spacing between grating lines. Bending of
light is independent of  (n1sin1 = n2sin2).
D Fg = mg  m = 20 N/10 m/s2 = 2 kg
C Fn = mgcos37 = (20 N)(4/5) = 16 N
B W = F||d = mgh = (20 N)(3 m) = 60 J
v is constant, but vy decreases with elevation.
D x
v = (vx2 + vy2)
E Acceleration is due to gravity, which is downward.
A The only force acting on the ball is gravity,  downward.
Rate of change of magnetic field = B/t.
A
E = B/t = A x B/t = IR  B/t = IR/A = IR/a2
B E = kQ/r2  EA/EB = 4/1
d is more than 2f, let's say 3f.
D o
1/do + 1/di = 1/f  1/di = 1/f – 1/3f = 2/3f  di = 1.5f
Charge on the smaller sphere is more concentrated. When the two
E spheres are connected, charge flows from the smaller sphere to
the larger until same energy per charge.
B All objects accelerate at the same rate,  speeds are equal.
C P = W/t = mgh/t = (700 N)(8 m)/10 s = 560 W
Inelastic collision
E
mAvA + mBvB = (mA + mB)v'  v' = (mv + 0)/(m + M)
Momentum is conserved,  with an increase in mass there is a
C
corresponding decrease in speed.
C P = W/t = J/s = W (kWhr is a measure of energy).
E L = rp = rmv = (4 m)(2 kg)(3 m/s) = 24 kg•m2/s
A With no acceleration, the vector sum of the forces equal 0.
D K + U is a constant,  the sum of the two graphs = Vo.
A P = Ffv = mgv
X-ray energy is very high,  electrons must be close to the
A
nucleus to require so much energy to change energy level.
B Diffraction is a wave phenomena, the other options aren't.
A Charge is conserved (bottom number on nuclear symbol).
A Only charged objects are affected by an electric field.
E E = V/d: 2 x V  2E, 1/5 d  5E,  10E (20,000 N/C).
A Rs = 1 + 3 = 4 , 1/Rp = 1/4 + 1/2 = 3/4  R = 4/3 
E Q/t = I = V/R: I is greatest where R is least.
C E is directed away from +Q's, which is down and to the left.
D Vtot = kQ/d + kQ/d = 2kQ/d, Ue = qV = q(2kQ/d)
C B = B x A = (2 T)(0.05 m x 0.08 m) = 0.008 T•m2
E
= Pt = IVt/1000
D KWhr
EkWhr = (4.0 A)(120 V)(2 hr)/1000 = 0.96 kWhr x 10 cents =
Left hand: thumb toward bottom of the page, fingers out of the
A
page,  palm (force) is to the right.
A Work = area under the curve, which is greatest for 1.
A T  PV,  process 1.
C U = Qin + Win = (400 J) + (-100 J) = 300 J
Q = mciT + mL + mcwT
B
Q = m[ci(273 – T1) + L + cw(T2 – 273)]
1/do + 1/di = 1/f  1/ + 1/di = 1/f  0 + 1/di = 1/f  di = f
B
f = ½R = ½(1.0 m) = 0.5 m
E nwater > nair, vn = c/n and n = /n
1/do + 1/di = 1/f
D 1/6 + 1/di = 1/9  1/di = 2/18 – 3/18 = -1/18  di = -18
M = -di/do = -(-18)/6 = -3 (virtual, upright, 3 x larger)
99
B
100 D
101 A
102 B
103 D
104 B
105 A
106 A
107 A
108 D
109 E
110 B
111 A
112 E
113 B
114 C
115 A
116 C
117 D
118 D
119 C
120 D
121 B
122 C
123 B
124 B
125 B
126 D
127 C
128 C
129 D
130 E
131 D
132 C
133 B
134 C
135 D
136 C
137 D
138 A
139 B
140 E
Transverse wave reflects off of a more dense surface undergoes
phase shift (v and A remain the same).
Real, inverted magnified image is formed for f < di < 2f.
n2 > n1 because light bend toward normal in medium 2.
n1 = n3 because rays are parallel.
4,000 cpm  2,000 cpm  1,000 cmp  500 cmp
12:00
12:30
1:00
1:30
214
0
214
82Pb  -1 +
83Bi
p = h/,  when p doubles,  is ½ as much.
Rutherford's experiment is with alpha particle, not photons.
At position A, the number of undecayed atoms is half the original
number, which is one ½-life.
p = h/ and c = f  p = hf/c,  when f doubles, p doubles.
4 kg will stretch the spring 16 cm to reach equilibrium position,
then it will stretch an additional 16 cm before turning around,  32
cm.
W = Fg = GMm/r2,  when r  4r, W  1/16W
Fc = Fg  mv2/r = GMm/r2  K = ½mv2 = ½GMm/R
Since there was no initial y momentum, then the total y momentum
= 0: (0.2 kg)(1 m/s) = (0.1 kg)v,  v = 2 m/s 
Refraction angle depends on the difference in nprism and nmedium.
Since the difference decreases, the angle of separation will
decrease, but the order remains the same.
Momentum increases when velocity increases, which produces a
curved d vs. t graph.
Fnet = 0 means no acceleration, where v = 0 or constant (graph is
straight).
No work is done because force and displacement are at right
angles from each other.
Fc = FB  mv2/r = qvB
 v = qBr/m = (1.6 x 10-19)(10-1)(10-1)/(1.6 x 10-27) = 106 m/s
Induced current occurs when B changes (a, e) or A changes (b, c).
In option (d) neither changes.
Right hand thumb points out of the page (don't flip), fingers wrap
around your thumb in a counterclockwise direction.
Frequency is greatest where the wavelength is shortest (v = f).
This occurs where the wave fronts are closest together.
sinc = n/d  d = n/sinc = (1)(0.12 m)/(3/5) = 0.20 m
Image formed is the "mirror" image of the object.
Pressure is doubled (PV = nRT), but the density remains the same
(d = m/V).
Atoms have diameters around 10-10 m (1 angstrom)
Area = ¼(10-3 m)2 = N¼(10-10 m)2
N = (10-3/10-10)2 = 1014
T  PV,  T4 < T1 = T3 < T2
Kelectron = hc/ – and number of electrons  light intensity, 
increase N = greater intensity and less K = longer .
From midpoint to one extreme position = ¼T,
 T = 4(0.1 s) = 0.4 s.
Impulse, J = FT = mv = (0.4 kg)(5.0 m/s) = 2 N•s
Counterclockwise torque = clockwise torque
Rm + ½RM = R(2M)  m + ½M = 2 M,  m = 3M/2
Collisions must be elastic, otherwise energy is lost and the
temperature would have to decrease over time.
d = ½at2  t = (2d/a)½ = (2h/g)½
K' = K + Ug = ½mvo2 + mgh
K = ½mv2 = 3/2RT  v = (3RT/MM)½
 v1/v2 = (T1/T2)½ = (600 K/300 K)½ = 2
If the battery remains connected, then V is constant. C  A/d: if d
is doubled, then C is halved, Q = CV,  ½C  ½Q.
Momentum is conserved,  their momenta must be equal in
magnitude and opposite in direction.
R = L/A: If the wire loop has twice the radius, then it is 4 x longer,
 R = 4R.
Since A is the enclosed area of the magnetic field, B, and it doesn't
change than B is unchanged by increasing b to 2b and E is the
same as before.
px: 3mvx = mV,  vx = V/3, py: 3mvy = mV,  vy = V/3
v = (vx2 + vy2)½ = ((V/3)2 + (V/3)2)½ = 2V/3
torque,  = r x F = r x Fsin = rsin x F
Pauli exclusion principle states that no two electrons in an atom
can have the same set of quantum numbers.
Uc = ½QV and Q = CV,
 Uc = ½CV2 = ½(4 x 10-6 F)(102 V)2 = 2 x 10-2 J
141 E
142 B
143 D
144 E
145 B
146 A
147 E
148 E
149 C
150 B
151 D
152 C
153 A
154 C
155 A
156 A
157 B
158 E
159 D
160 D
161 E
162 D
163 C
164 B
165 B
166 A
167 C
168 E
169 A
170 C
171 E
172 B
173 B
174 A
175 D
176 C
177 E
178 A
179 E
180 B
181 A
182 B
183 E
184 D
185 D
186 D
187 C
Only non-accelerated motion is constant straight line motion, but
straight line can also be accelerated.
ma = F – Fg = 50 N – 30 N = 20 N  a = 20 N/3 kg = 6.7 m/s2
Velocity is vector sum of compressed spring () and the rotation
()  
Velocity is zero, but acceleration is downward  K = 0.
F + F b = Fg  F b = F g – F = 4 N – 3 N = 1 N
Maximum K occurs at the equilibrium position and maximum U
occurs at the release point.
Ts = Tp  2(m/k)½ = 2(L/g)½  m1/k = L/g  k = m1g/L
T = 2(m/k)½  period T depends on mass m and force constant k.
The order will not change the final velocity, since it only depends
on the total mass at the end.
Fg = GMm/r2, where M and r are doubled  Fg = ½ as much (250
N).
vav = d/t = (8 m – 2 m)/1 s = 6 m/s
Thermal equilibrium means that the temperatures are the same.
Change in internal energy U depends only on the initial and final
temperatures.
FB = kI1I2L/r  F'/F = I1'I2'/I1I2 = (2)(2)  F' = 4F
The electric field inside a conductor is zero at all locations.
Circuit elements in parallel have the same potential difference
(voltage) and Q = CV.
R = L/r2  R'/R = L'r2/Lr2' = ½/½2  R' = 2R
e = Plift/Pcurrent = mgv/IV
e = (9 kg)(10 m/s2)(0.5 m/s)/(0.5 A)(120 V) = 0.75
The electric field is away from +Q () and toward –Q ()  
E+Q = E-Q = kQ/d2 = [(kQ/d2)2 + (kQ/d2)2]½ = 2kQ/d2
F = qE, since q is the same for the electron and proton, and they
are in the same field E, then Fe = Fp
The hole expands in the same proportion as the steel plate 
1.01/1.00 = x/0.100  x = 0.101 m
Increased speed means T increases. PV = nRT  P also
increases. Volume and density are unchanged..
Center of curvature is twice as far from the mirror as the focal
point.
fn = nf1: f2 = 2f1  f1 = f2/2 = f/2
When the object is placed inside the focal length, then the image is
virtual, upright and larger.
c = f  3 x 108 m/s = (100 x 106 s-1)   = 3 m
All waves show interference. Sound is longitudinal and light is
transverse. Light is 106 x faster.
slope = y/x = 20 N/4 s = 5 N/s
Ft = p. Ft = area = ½(4)(20) + (4)(5) = 60 N•s
The air mattress lengthens the stopping time, which reduces the
decelerating force on the stunt person.
TL + TR = Fg  TL + 250 N = 125 N + 500 N  TL = 375 N
 =   (250 N)(4 m) = (500 N)x + (125 N)(2 m)  x = 1.5 m
K = hf – , which is a graph of a straight line with a negative yintercept.
Number of photons (intensity) is directly proportional to number of
electrons (current).
Light elements have a 1:1 ratio of protons:neutrons. Heavier
elements have a 1:1.5 ratio.
P = F/A = gh  F = ghA, which is the same for the four beakers.
The constricted air speeds up, which reduces pressure  the liquid
rises in the tube.
s.g. = mair/(mair – mfluid) = 0.45 kg/(0.45 kg – 0.36 kg) = 5
object = (s.g.)(fluid) = 5(1000 kg/m3) = 5000 kg/m3
0 = mAvA + mBvB  mA/mB = -vB/vA = 2/5
Fnet = ma  10 N – 30cos60 N = (10 kg)a
a = -5 N/10 kg = -0.5 m/s2
ac = v2/r = 42r2/T2r = 42R/T2
Tp = 2(L/g)½, g = GM/r2: with 2 x m, g is 2 x and T is 1/2 shorter;
Ts = 2(m/k)½, which is unchanged.
In order for the ball to stay in circular motion, there must be a net
force Fc toward the center of the circle.
E = V/d, which is constant. 8 V/0.04 m = X V/0.01 m
 X = 2 V. With the lower plate at 2 V, point P = 4 V.
E = V/d = 8 V/0.04 m = 200 V/m
Loop 1: B is downward and B is decreasing (don't flip)  I is
clockwise. Loop 2: B is constant  No I.
188 D
189 A
190 E
191 A
192 A
193 E
194 D
195 D
196 D
197 A
198 E
199 E
200 C
201 D
202 E
203 B
204 D
205 D
206 D
207 C
208 E
209 C
R|| = 1/(1/2000 + 1/6000) = 1500 
R = 2500 + 1500 = 4000 I = V/R = 12/4000 = 0.003 A
I1 is total current, I2 and I3 share the total current, but I2 > I3
because I = V/R and V is the same for ||.
Left interface, light bends toward normal (D-E).
Right interface, light bends away from normal (E).
Left interface, light bends away from normal (A-B). Right interface,
light bends toward normal (A).
f/f  v/vw.
If frequency from X > frequency from Y, then vx > vy.
nf is extreme with bright reflection  T = ¼
1/do + 1/di = 1/f  1/8 cm + 1/di = 1/2cm  di = 8/3 cm
Less light reaches the screen  the image is dimmer.
Qin/t  A(TH – TL)/L. Decreasing A and T2 – T1 or increasing L
would decrease Qin/t.
U = -Wout – Qout = -32 J – 16 J = -48 J
All kinetic energy and energy equivalent of the decrease in mass
becomes neutron kinetic energy.
214
0
214
82Pb  -1e +
83X
P = NEphoton/t = Nhc/t = N(6 x 10-34)(3 x 108)/(4)t = 5 x 104
 N/t = (104)(100)(1034)(10-8) = 1030
Mv + 2M(-v) = 3Mv'  v' = -v/3
K = ½[(3M)(-v/3)2 – (3M)v2] = ½[-8/3Mv2] = -4/3Mv2
The first force is , the second force is   the velocity is .
Each force causes velocity to increase by the square of time
(v2 = vo + at)  parabolic shape increases.
Current reverses twice per revolution (each time it is vertical to the
plane of the page)  6 times.
K = QV  K' = 2QV  K' = 2K
The direction of the electric force on a negative charge is opposite
the direction of the field.
Fg = Fc  GMm/r2 = mv2/r  r = r = GM/v2
P = Fv = (900 N)(4 m/s) = 3600 W
FB = Fe  qvB = qE  v = E/B = (6 N/C)/(2 T) = 3 m/s
e
a
b
4
c
d
a
5
b
c
The horizontal width of the wire approaches the width of the
magnet such that the ends of the wire are exposed to a weaker
magnetic field, which reduces the total magnetic flux experienced
by the wire loop.
vy2 = vyo2 + 2ady
0 = (vosin50)2 + 2(10 m/s2)(0.150 m)  vo = 2.26 m/s
V/t = A•L/t = Av = (4.0 x 10-3 m)2(2.26 m/s)
V/t = 1.14 x 10-4 m3/s
v1A1 = v2A2  v1 = v2A2/A1
v1 = (2.26 m/s)(4.00 x 10-3 m)2/(7.00 x 10-3 m)2
v1 = 0.738 m/s
P1 + gy1 + ½v12 = P2 + gy2 + ½v22
P1 + 0 + ½(1000 kg/m3)(0.738 m/s)2 =
0 + (103 kg/m3)(10 m/s2)(3 m) + ½(103 kg/m3)(2.26 m/s)2
P1 = 3.23 x 104 Pa (which is gauge pressure when P2,
atmospheric pressure, is set at zero)
Process
W
Q
U
0
+
+
AB
–
+
0
BC
+
–
–
CA
Win = 0 because V = 0.
U > 0 because U = 3/2nRT and T > 0.
Q > 0 because Q = U – Win and U > 0, while Win = 0.
PBVB = PCVC (TB = TC)
 VC = PBVB/PC = (5 atm)(0.001 m3)/(1 atm) = 0.005 m3
Spring Practice Free Response Answers
a = v/t  t = v/a = (5.0 m/s – 2.0 m/s)/(1.5 m/s2) = 2 s
vav = d/t  d = (vav)t = (3.5 m/s)(2 s) = 7 m
a
vav = d/t  t = d/vav = (15 m – 7 m)/5.0 m/s = 1.6 s
ttotal = 2 s + 1.6 s = 3.6 s
mAvA + mBvB = mAvA' + mBvB'
b(1) (250 kg)(5 m/s) = (250 kg)vA' + (200 kg)(4.8 m/s)
 vA' = 1.2 m/s
1
b(2) to the right because vA' > 0 (same direction as vA)
KA = ½mAvA2 = ½(250 kg)(5 m/s)2 = 3125 J
KA' = ½mAvA'2 = ½(250 kg)(1.2 m/s)2 = 180 J
c KB' = ½mBvB'2 = ½(200 kg)(4.8 m/s)2 = 2304 J
3125 J  (2304 + 180 J)
 kinetic energy is lost  the collision is not perfectly elastic.
F = ma  4.0 N = (10 kg)a  a = 0.4 m/s2
a
F = ma = (2.0 kg)(0.4 m/s2) = 0.8 N
b F = kx  x = F/k = 0.8 N/80 N/m = 0.01 m
c The same because the same force is pulling the same mass.
2
Greater because a greater force is used to stretch the spring
d F = ma = (8.0 kg)(0.4 m/s2) = 3.2 N and F = kx
 x = F/k = 3.2 N/80 N/m = 0.04 m
KB = Us  ½mv2 = ½kx2
e
 x = (mv2/k)½ = [(8.0 kg)(0.50 m/s)2/80 N/m)½ = 0.16 m
V = IR  R = V/I = 16 V/4.0 A = 4 
4  = Rwire + Rinternal  Rwire = 4  – 0.50  = 3.5 
a
R = L/A  L = RA/
3
L = (3.5 W )(3.5 x 10-9 m2)/(1.7 x 10-8 •m) = 0.72 m
Downward Using hand rules, thumb (I) into the page, fingers (B) to
b
the right, palm (FB) down.
c FB = ILB  B = FB/IL = (0.060 N)/(4.0 A)(0.020 m) = 0.75 T
a
6
The image is real because the rays converge on the same side of
the mirror as the object.
1/do + 1/di = 1/f
c
1/di = 1/f – 1/do = 1/6.0 cm – 1/8.0 cm  di = 24 cm
The image is smaller because diverging mirrors can only make
smaller images.
d
1/di = 1/f – 1/do = 1/-6.0 cm – 1/8.0 cm  di = -3.4 cm
M = -di/do = -(-3.40)/8 = 0.43  image is smaller
 = h/p
a
p = h/ = (6.63 x 10-34 J•s)/(0.038 x 10-9 m) = 1.74 x 10-23 kg•m/s
p2 = m2v2  K = p2/2m
b
K = (1.74 x 10-23 kg•m/s)2/2(9.11 x 10-31 kg) = 1.66 x 10-16 J
7
c qV = E  V = E/q =(1.66 x 10-16 J)/(1.60 x 10-19 C) = 1040 V
Kmax = hf –  = 0
d  f = /h = (4.5 eV x 1.6 x 10-19 J/eV)/(6.63 x 10-34 J•s)
f = 1.09 x 1015 s-1
a Us = Ug  ½kx2 = mgh h = (kx2/2g)(1/m)
b(1) 1/m and h
1/m (kg-1)
m (kg)
h (m)
50
0.020
0.49
8
33
0.030
0.34
b(2)
25
0.040
0.28
20
0.050
0.19
17
0.060
0.18
b
h (m)
0.5
d
c
0.4
0.3
0.2
e
0.1
0
d
e
a
b
10
30
40
50
1/m ((kg-1)
slope = kx2/2g = (0.5 – 0.1)/(45 – 10) = 0.011
k = (0.011)(2)(9.8)/(0.020)2 = 540 N/m
Measure the time that the toy is in the air. Since it takes as much
time to rise as it does to fall then starting at the highest point it
takes t/2 s to fall h distance  -h = ½g(t/2)2  h = gt2/8.

 
 

V = kQ1/r1 + kQ2/r2
V = kQ/Lsin + kQ/Lsin = 2kQ/Lsin
Ft
20
a
b
16
9
Fe
c
E2 < E1, The kinetic energy is transformed into non-mechanical
energy during the inelastic collision.
FB
FB = Fg  oVog = (mC + mS)g  Vo = (mC + mS)/o
Vo (x 10-6 m3)
80
60
c

Fg = (mC + mS)g
40
20
d
a
b
10
c
d
e
a
b
11
c
d
a
b
12
c
d
a
b
13
c
d
a
14
b
c
15
a
b
Fg
Tx = Fe  Tsin = kQ2/(2Lsin)2  T = kQ2/4L2sin3
Ty = Fg  Tcos = mg  T = mg/cos
I = E/R = BLv/R = (0.80 T)(0.52 m)(1.8 m/s)/3.0 
I = 0.25 A
F = Ff + FB = mg + ILB
F = (0.20)(0.22 kg)(10 m/s2) + (0.25 A)(0.52 m)(0.80 T)
F = 0.44 N + 0.10 N = 0.54 N
P = E/t = I2R
E = I2Rt = (0.25 A)2(3.0 )(2.0 s) = 0.38 J
W = Pt = Fvt = (0.54 N)(1.8 m/s)(2.0 s) = 1.9 J
Unequal because the work needed to pull the rod includes the heat
loss due to friction, which is not included in the answer for part (c).
PV = nRT, where (n = 2.2 kg/0.018 kg/mol = 122 mol)
(3 x 105 Pa)(2 m3) = (122 mol)(8.31 J/mol•K)T
 T = 590 K
P1V1/T1 = P2V2/T2 
(3 x 105)(2.0)/590 = (4 x 105)(2.5)/T2  T2 = 980K
Increase. Internal energy is proportional to temperature, which
increases from A  C.
WA  B = 0 (no change in volume)
WB  C = -PV = -(4.0 x 105 Pa)(0.5 m3) = -2.0 x 105 J
Wtot = 0 – 2.0 x 105 J = -2.0 x 105 J
No. Ft = Fg – Fb. Fg is the same, but Fb depends on volume (Fb =
Vog), which might be different.
Fb = Fg – Ft = (1300 kg/m3)(1 x 10-5 m3)(10 m/s2) – .0098 N
Fb = 0.12 N
Fb = fluidVog  0.12 N = fluid(1.0 x 10-5 m3)(10 m/s2)
 fluid = 1200 kg/m3
Increase because buoyant force is less since less liquid is
displaced by partially submerged objects.
f = c/ = 3.0 x 108 m/s/550 x 10-9 m = 5.5 x 1014 s-1
x/L  (m + ½)/d – (m – ½)/d = /d
x/2.2 m = (550 x 10-9 m)/(1.8 x 10-5 m)  x = 0.067 m
5.5 x 1014 s-1 (frequency is the same in all mediums)
Wavelength decreases (n = /n)  distance between fringes
decrease (x  mL/d).
p = h/ = 6.63 x 10-34 J•s/.85 x 10-9 m = 7.8 x 10-25 kg•m/s
K = p2/2m = (7.8 E-25kg•m/s)2/2(9.1 E-31kg) = 3.3 x 10-19 J
E = hc/ = (2 x 10-25 J•m)/250 x 10-9 m = 8.0 x 10-19 J
K = E – 3.3 x 10-19J = 8 x 10-19J –    = 4.7 x 10-19 J
d, The transition had to go to a lower energy state to emit a photon
and have a greater gap (E250 > E400).
dy = vyot + ½at2
0.80 m = 0 + ½(-10 m/s2)t2  t = 0.4 s
vx = dx/t = 1.2 m/0.4 s = 3 m/s
Us = K  ½kx = ½mv
(650 N/m)x2 = (4.0 kg)(3 m/s)2  x = 0.24 m
mAvA + mBvB = (mA + mB)v'
(4.0 kg)(3 m/s) + 0 = (4 kg + 4 kg)v' v' = 1.5 m/s
dx = vxt = (1.5 m/s)(0.4 s) = 0.6 m
2
c
d
2
0
d
e
a
0.010
0.030
0.050
0.070
mS (kg)
slope = (y2 – y1)/(x2 – x1) = 55 x 10-6/0.05
slope = 1.1 x 10-3 m3/kg
Vo = (o-1)mS + mC/o  o = 1/slope = 900 kg/m3
y-intercept = mC/o = volume of oil displaced by the mass of the
cup.
q1 is negative, q2 is positive
F2
b
17
c
d
F1
F1 = kq1q3/r2 = (9 x 109)(4 x 10-6)(1 x 10-6) = 2.25 x 10-3 N
F2 = kq1q3/r2 = (9 x 109)(1.7 x 10-6)(1 x 10-6) = 1.7 x 10-3 N
F = F2cos53 + F1cos37
F = 1.7 x 10-3(3/5) + 2.25 x 10-3(4/5) = 2.8 x 10-3 N
F = qE
 E = F/q = 2.8 x 10-3 N/1.0 x 10-6 C = 2.8 x 103 N/C
X
e
a
b
18
c
d
a
19
b
c
A positive charge directly in front of the force would cancel it.
0.12 = W/Qin = 4.5 x 106 W  Qin = 3.8 x 107 W
P = Fv
4.5 x 106 W = F(7.0 m/s)  F = 6.4 x 105 N
(1) The area represents the work done per cycle.
(2) P = W/t = 4(2 x 105 N/m2)(0.40 m3)/1 s = 3.2 x 105 W
AB and BC
n1sin1 = n2sin2
(1)sin40 = 1.65sin2  2 = 23o
For 4 > 90o, 3 would have to increase. For 3 to be larger, 2
would have to decrease. For 2 to be smaller, 1 would have to
decrease.
(1) n = /n = 6.65 x 10-7 m/1.38 = 4.82 x 10-7 m
(2) nfilm is middle value with destructive interference
T= ¼film = (4.82 x 10-7 m)4 = 1.21 x 10-7 m
a
20
B is increasing into the page, which generates a current with an
outward B  current is counterclockwise.
b(1) I = vLB/R = (3 m/s)(0.10 m)(2 T)/4  = 0.15 A
B(2) FB = ILB = (0.15 A)(0.10 m)(2 T) = 0.03 N
c
The net force is zero because the induced current at the two ends
go in the opposite direction resulting in no current and no net
force.
a
21
b
c
d
c = f
 f = c/ = (3 x 10 m/s)/(400 x 10-9 m) = 7.5 x 1014 s-1
Kelectron = Ephoton – 
1.1 x 10-19 J= (6.63 x 10-34 J•s)(7.5 x 1014 s-1) – 
K = qV  V = K/q = 1.1 x 10-19 J/1.6 x 10-19 C = 0.69 V
K = p2/2m  p = (2mK)½
p = [(2)(9.11 x 10-31 kg)(1.1 x 10-19 J)]½ = 4.5 x 10-25 kg•m/s