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5.4. Step 1: Assume room temperature (20.0°C) to obtain the Step 2:
velocity:
d _

v = ∴ d = vt
m _ t
)
0.600

s
7

f = 2.00 × 10 Hz v TP(m/s) = v 0 + ( _ T P ) ft _
(

°C = (1,127 ) (0.500 s)
s
λ = ?
ft _

v = ? m _ = 1,127 × 0.500 × s
s

0.600
m _ _ )
s

= 331 + ( (20.0°C) = 563.5 ft
s °C
m _ m _ = 564 ft

= 331 + 12.0
s s

Step 3: The distance to the building is one-half the distance
m _ the sound traveled, so
= 343
s
_
564

Step 2: = 282 ft
2
v _
v = λf ∴ λ = 5.7. Step 1:

f
d _
m _ t = 1.75 s v = ∴ d = vt

343

__ t
s

=
7 1 _
2.00 × 10 v = 1,530 m/s m _

s

d = ? = ( 1,530 ) (1.75 s)
s
_ m _ _ s
343
= × m _

7 s
2.00 × 10 1 = 1,530 × 1.75 × s

s
–5
= 1.715 × 10 m
= 2,677.5 m
–5
= 1.72 × 10 m

Step 2: The sonar signal traveled from the ship to the bottom,
5.5. Step 1: Assume room temperature (20.0°C) to obtain the then back to the ship, so the distance to the bottom is
velocity (yes, you can have room temperature outside one-half of the distance traveled:
a room):
2,677.5 m
_ =
m _ 2 1,338.75 m
0.600 )

s

(
d = 150 m v TP(m/s) = v 0 + ( _ T P ) = 1,340 m

°C
t = ? m _
)

0.600
m _
s

v = ? = 331 + ( _ (20.0°C) 5.8. f = 660 Hz v = λf

s °C λ = 9.0 m 1 _

= (9.0 m)(660 )
s
m _ m _ v = ?
= 331 + 12.0

s s 1 _
= 9.0 × 660 m ×

s
m _
= 343 m _

s = 5,940
s
m _

Step 2: = 5,900
s
d _ d _ 5.9. Step 1:

v = ∴ t =

t v m _
0.600 )

s

_ t = 2.5 s v TP(m/s) = v 0 + ( _ T P )
150.0 m
(

= m _ °C

343
s T = 20.0°C m _
0.600

_ s m _ _ )
150.0 m _ _
s
×
= d = ? = 331 + ( (20.0°C)

343 1 m s °C
= 0.4373177 s v = ? m _ m _

= 0.437 s = 331 + 12.0
s
s
m _
5.6. Step 1: Find the velocity of sound in ft/s at a temperature of = 343

20.0°C: s
ft _
)
2.00 Step 2:

s

t = 0.500 s v TP(ft/s) = v 0 + ( _ (T P ) d _

°C v = ∴ d = vt
T = 20.0°C ft _ t m _
)
2.00
ft _ _ = ( 343 ) (2.5 s)
s
s

= 1,087 + ( (20.0°C)

s °C m _

= 343 × 2.5 × s

ft _ ft _ s
d = ? = 1,087 + 40.0

s s = 857.5 m
ft _
v = ? = 1,127
s = 860 m
E-15 APPENDIX E Solutions for Group A Parallel Exercises 657