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Physics 110 Quiz #7, November 11, 2016
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worth 10 points total.
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1. A tsunami is a wave pulse consisting of several crests (high points of a wave) and
troughs (low points of a wave) that become drastically large as they enter shallow
water at the shore. Tsunamis are large amplitude ocean waves and as the wave
climbs up the shoreline the height of the approaching wall of water gets very tall on
land as shown in the diagram below. Assume that the tsunami has a wavelength of
λ = 235km and a velocity of v = 153 ms that was caused by an earthquake out in the
Pacific Ocean. As the tsunami approaches Hawaii, people are drawn to the shoreline
to observe the drastic withdrawal of water as the tsunami approaches.
Approximately how much time do
they have to run to safety before
the wave hits the shoreline? (In the
absence of knowledge and warning
systems people have died during
tsunamis, some of them attracted to
the shore to see stranded fish and
boats.)
From the crest of the wave to the ground represents one half of a wavelength. Thus,
Δx =
Δx
Δx λ 235 × 10 2 m
1min
λ
→ Δt =
=
=
= 768s ×
= 12.8 min .
and v =
m
Δt
v 2v
2 × 153 s
60s
2
2. A string of length L = 3m and mass 0.5kg has a weight of 48kg × 9.8 sm2 = 470N
suspended from one end. The end of string with the attached weight passes over a
massive pulley ( m p = 3kg ) and the other end is oscillated at a frequency f = 475Hz .
How many standing loops will be produced if the distance between the point of
oscillation and the pulley is l = 8m ?
v=
FT
= fλ =
µ
µ
⎛ 2l ⎞
f ⎜ ⎟ → n = 2lf
=
⎝ n⎠
FT
0.5kg
n = 2 × 8m × 475s
n = 143
−1
3m
470N
3. A stone is dropped from the top of a cliff. The splash it makes is heard 2.7s later.
How high is the cliff?
2
The time for the stone to fall is y f = yi + viyt + 12 ayt 2 → h = 12 gt down
. When the stone
hits the water the sound of the splash travels up the same distance the stone fell and
h
h
this time is given by vsound =
. The total time from dropping the stone
→ tup =
tup
vsound
to hearing the sound is T = tup + t down . Thus t down = T − tup = T −
h
vsound
and
2
h = gt
1
2
2
down
⎛
h ⎞
. Expanding this and solving for the height we have:
= g⎜T −
vsound ⎟⎠
⎝
1
2
2
⎛
⎛ 2
h ⎞
h
h2 ⎞
1
h = g⎜T −
=
g
T
−
2T
+
2 ⎜
2
⎟⎠
vsound ⎟⎠
vsound vsound
⎝
⎝
1
2
0=
⎛ g ⎞
gT 2 ⎛
gT ⎞
− ⎜ 1+
h + ⎜ 2 ⎟ h2
⎟
2
⎝ vsound ⎠
⎝ 2vsound ⎠
−5
0 = 35.7 − 1.08h + 4.2 × 10 h
.
2
⎧⎪
33m
h=⎨
4
⎩⎪ 2.6 × 10 m
Since the total time of travel is only 2.7s , we choose the smaller distance. Thus the
height of the cliff is 33m .
4. Two sounds A and B differ in intensity level by 20dB . What is the ratio of the
M
intensity of sound A to that of sound B? Hint: log M − log N = log .
N
β A − β B = 20dB = 10 log
IA
I
I
− 10 log B = 10 log A
I th
I th
IB
β A −β B
20
IA
= 10 10 = 10 10 = 10 2 = 100
IB
Useful formulas:
Motion in the r = x, y or z-directions
rf = r0 + v 0r t + ar t
1
2
Uniform Circular Motion
v2
ar =
r
v2
Fr = mar = m
r
2πr
v=
T
mm
FG = G 1 2 2
r
2
v fr = v 0r + ar t
v fr = v 0r + 2arΔr
2
2
Vectors
Geometry /Algebra
Circles
Triangles
Spheres
C = 2πr
A = 12 bh
A = 4 πr 2
A = πr 2
Quadratic equation : ax 2 + bx + c = 0,
whose solutions are given by : x =
g = 9.8 m s2
⎛v ⎞
direction of avector → φ = tan−1⎜ y ⎟
⎝ vx ⎠
N A = 6.02×10
2
Linear Momentum/Forces
→
p = mv
→
→
→
F = ma
→
→
G = 6.67×10−11 Nm
23 atoms
σ = 5.67×10
−8 W
K r = 12 Iω
Heat
TC = 59 [TF − 32]
2
TF = 95 TC + 32
Lnew = Lold (1+ αΔT )
U g = mgh
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
Anew = Aold (1+ 2αΔT )
Vnew = Vold (1+ βΔT ) : β = 3α
PV = Nk B T
W R = τθ = ΔE R
W net = W R + WT = ΔE R + ΔE T
3
2
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
Fluids
θ f = θ i + ωi t + 12 αt 2
M
ρ=
V
F
P=
A
Pd = P0 + ρgd
ω f = ωi + α t
ω 2 f = ω 2 i + 2αΔθ
τ = Iα = rF
L = Iω
L f = Li + τΔt
Δs = rΔθ : v = rω : at = rα
a r = rω
2
Sound
v = fλ = (331+ 0.6T) ms
I
; Io = 1×10−12 mW2
I0
v
v
f n = nf1 = n ; f n = nf1 = n
2L
4L
β = 10log
Simple Harmonic Motion/Waves
2π
ω = 2πf =
T
TS = 2π
m
k
TP = 2π
l
g
FB = ρgV
A1v1 = A2 v2
ρ1 A1v1 = ρ 2 A2v2
P1 + 12 ρv 21 + ρgh1 = P2 + 12 ρv 2 2 + ρgh2
kB T = 12 mv 2
ΔQ = mcΔT
ΔQ kA
PC =
= ΔT
Δt L
ΔQ
PR =
= εσAΔT 4
ΔT
ΔU = ΔQ − ΔW
ΔE R + ΔE T + ΔU g + ΔU S = 0
Rotational Motion
kg 2
vsound = 343 m s
m 2K 4
2
Fs = −k x
2
kB = 1.38×10−23 J K
mole
Work/Energy
K t = 12 mv
→
p f = p i + F Δt
→
−b ± b 2 − 4ac
2a
Useful Constants
magnitude of avector = v x + v 2 y
→
V = 43 πr 3
1
v =±
k ⎛ x 2 ⎞2
A⎜1− ⎟
m ⎝ A2 ⎠
x ( t ) = Asin( 2Tπt )
k
cos( 2Tπt )
m
k
a( t ) = −A sin( 2Tπt )
m
FT
v = fλ =
µ
v(t) = A
v
2L
I = 2π 2 f 2 ρvA 2
f n = nf1 = n