johnson (mjj622) – HW 17 – Coker – (56625)
1
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17
questions.
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before answering.
This is the last chapter of material to be
covered on Quiz 4. Remember that no Quiz 4
material will appear on the final to be given
on December 11.
001 (part 1 of 2) 10.0 points
A 0
.
8 kg physics book with dimensions of
28
.
4 cm by 19
.
7 cm is on a table.
What force does the book apply to the
table? The acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 7
.
84 N.
Explanation:
Let :
m
= 0
.
8 kg
and
g
= 9
.
8 m
/
s
2
.
The force is
F
=
m g
= (0
.
8 kg) (9
.
8 m
/
s
2
)
=
7
.
84 N
.
002 (part 2 of 2) 10.0 points
What pressure does the book apply?
Correct answer: 140
.
13 Pa.
Explanation:
Let :
ℓ
= 28
.
4 cm
and
w
= 19
.
7 cm
.
The pressure is
P
=
F
A
=
F
ℓ w
=
7
.
84 N
(28
.
4 cm) (19
.
7 cm)
·
parenleftbigg
100 cm
1 m
parenrightbigg
2
=
140
.
13 Pa
.
003
10.0 points
The air pressure above the liquid in figure is
1
.
21 atm
.
The depth of the air bubble in the
liquid is 41
.
6 cm and the liquid’s density is
832 kg
/
m
3
.
41
.
6 cm
air
air
Determine the air pressure in the bubble
suspended in the liquid. The acceleration of
gravity is 9
.
8 m
/
s
2
.
Correct answer: 1
.
25965
×
10
5
Pa.
Explanation:
Let :
ρ
= 832 kg
/
m
3
,
P
0
= 1
.
21 atm
,
g
= 9
.
8 m
/
s
2
,
and
h
= 41
.
6 cm = 0
.
416 m
.
P
=
P
0
+
ρ g h
= (1
.
21 atm)
·
1
.
013
×
10
5
Pa
atm
+ (832 kg
/
m
3
) (9
.
8 m
/
s
2
) (0
.
416 m)
=
1
.
25965
×
10
5
Pa
.
004
10.0 points
A heavy liquid with a density 13 g
/
cm
3
is
poured into a U-tube as shown in the left-
hand figure below. The left-hand arm of the
tube has a cross-sectional area of 13 cm
2
, and
the right-hand arm has a cross-sectional area
of 3
.
95 cm
2
.
A quantity of 107 g of a light
liquid with a density 1
.
7 g
/
cm
3
is then poured
into the right-hand arm as shown in the right-
hand figure below.
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johnson (mjj622) – HW 17 – Coker – (56625)
2
h
1
h
2
13 cm
2
3
.
95 cm
2
heavy liquid
13 g
/
cm
3
L
13 cm
2
3
.
95 cm
2
light liquid
1
.
7 g
/
cm
3
If
the
density
of
the
heavy
liquid
is
13 g
/
cm
3
, by what height
h
1
does the heavy
liquid rise in the left arm?
Correct answer: 0
.
485591 cm.
Explanation:
Let :
m
ℓ
= 107 g
,
A
1
= 13 cm
2
,
A
2
= 3
.
95 cm
2
,
ρ
ℓ
= 1
.
7 g
/
cm
3
,
and
ρ
h
= 13 g
/
cm
3
.
Using the definition of density
ρ
ℓ
=
m
ℓ
V
ℓ
=
m
ℓ
A
2
L
L
=
m
ℓ
A
2
ρ
ℓ
=
107 g
(3
.
95 cm
2
) (1
.
7 g
/
cm
3
)
= 15
.
9345 cm
.
After the light liquid has been added to the
right side of the tube, a volume
A
2
h
2
of heavy
liquid is displaced to the left side, raising the
heavy liquid on the left side by a height of
h
1
with a displaced volume of
A
1
h
1
.
Since
the volume of heavy liquid is not changed, we
have
A
1
h
1
=
A
2
h
2
.
At the level of the heavy-light liquid interface
in the right side, the absolute pressure is
P
=
P
atm
+
ρ
ℓ
g L ,
and at the same level in the left tube,
P
=
P
atm
+
ρ
h
g
(
h
1
+
h
2
)
.
Equating these two values, we obtain
ρ
ℓ
g L
=
ρ
h
g
(
h
1
+
h
2
)
=
ρ
h
g
parenleftbigg
h
1
+
A
1
A
2
h
1
parenrightbigg
=
ρ
h
g h
1
parenleftbigg
1 +
A
1
A
2
parenrightbigg
.
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- Fall '08
- Turner
- Coker
-
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