Version PREVIEW – Practice 9 – carroll – (11108)
This print-out should have 8 questions.
Multiple-choice questions may continue on
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before answering.
Pressure by a Cylinder
001 10.0 points
A 77.5 kg solid cylinder, 1.95 m long and with
an end radius of 6.2 cm, stands on one end.
The acceleration of gravity is 9.8 m/s2 .
How much pressure does it exert?
Correct answer: 62.8919 kPa.
Explanation:
1
P = Patm + ρ g h
= 1.013 × 105 Pa
+ (1030 kg/m3 ) (9.8 m/s2 )
× 5872.85 m
= 5.93818 × 107 Pa .
003 (part 2 of 2) 10.0 points
If the inside of the vessel is maintained at
atmospheric pressure 1.013 × 105 Pa, what
is the net force on a porthole of diameter
15.1 cm?
Correct answer: 1.06159 × 106 N.
Explanation:
Let :
m = 77.5 kg and
r = 6.2 cm .
Let :
d = 15.1 cm = 0.151 m .
The gauge pressure at this depth is
F
mg
=
A
π r2
2
100 cm
(77.5 kg) (9.8 m/s2 )
·
=
π (6.2 cm)2
1m
1 kPa
×
1000 Pa
= 62.8919 kPa .
P =
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Deep in the Ocean
002 (part 1 of 2) 10.0 points
Engineers have developed a bathyscaph that
can reach ocean depths of 3.65 mi.
If the density of seawater is about
1030 kg/m3 , what is the absolute pressure
at that depth? The acceleration of gravity is
9.8 m/s2 .
Correct answer: 5.93818 × 107 Pa.
Explanation:
Let : g = 9.8 m/s2 ,
h = 3.65 mi = 5872.85 m ,
ρ = 1030 kg/m3 , and
Patm = 1.013 × 105 Pa .
Pg = P − Patm
= 5.93818 × 107 Pa − 1.013 × 105 Pa
= 5.92805 × 107 Pa ,
and the inward force on the port hole is
!
2
d
F = Pg A = Pg (π r 2 ) = Pg π
2
2
0.151 m
7
= (5.92805 × 10 Pa) π
2
= 1.06159 × 106 N .
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Cavendish Balance
004 10.0 points
In introductory physics laboratories, a typical
Cavendish balance for measuring the gravitational constant G uses lead spheres of masses
2.19 kg and 8.7 g whose centers are separated
by 5.85 cm.
Calculate the gravitational force between
these spheres, treating each as a point mass
located at the center of the sphere. The
value of the universal gravitational constant
Version PREVIEW – Practice 9 – carroll – (11108)
is 6.67259 × 10−11 N · m2 /kg2 .
Correct answer: 3.71489 × 10−10 N.
2
= (6.6726 × 10−11 N · m2 /kg2 )
8.7 × 1023 kg
×
(4.8 × 106 m)2
Explanation:
= 2.5196 m/s2 .
Let : m1 = 2.19 kg ,
m2 = 8.7 g = 0.0087 kg ,
r = 5.85 cm = 0.0585 m , and
G = 6.67259 × 10−11 N · m2 /kg2 .
The force of gravity is
m1 m2
r2
= (6.67259 × 10−11 N · m2 /kg2 )
(2.19 kg) (0.0087 kg)
×
(0.0585 m)2
F =G
= 3.71489 × 10
−10
N .
Clearly, this force is very tiny.
The
Cavendish balance is set up very delicately
to detect this force.
AP B 1993 MC 48
005 10.0 points
The planet Krypton has a mass of
8.7 × 1023 kg and radius of 4.8 × 106 m.
What is the acceleration of an object in free
fall near the surface of Krypton? The gravitational constant is 6.6726 × 10−11 N · m2 /kg2 .
Correct answer: 2.5196 m/s2 .
Explanation:
Let : M = 8.7 × 1023 kg ,
R = 4.8 × 106 m , and
G = 6.6726 × 10−11 N · m2 /kg2 .
Near the surface of Krypton, the gravitation force on an object of mass m is
F =G
Mm
,
R2
so the acceleration a of a free-fall object is
a = gKrypton =
=G
M
R2
F
m
Energy in Lifting
006 10.0 points
Energy is required to move a 1540 kg mass
from the Earth’s surface to an altitude 2.98
times the Earth’s radius RE .
The acceleration of gravity is 9.8 m/s2 .
What amount of energy is required to accomplish this move?
Correct answer: 7.19813 × 1010 J.
Explanation:
Using the formulae
U = −G
and
g=G
mME
R
ME
2 ,
RE
(where ME is the mass of the Earth and g is
the gravitational acceleration at the Earth’s
surface), we obtain that the change, ∆U , in
the potential energy of the mass-Earth system
is
∆U = (Ufinal − Uinitial )
1
1
−
= G m ME
Rinitial Rfinal
1
1
= G m ME
−
RE
RE (1 + a)
1
= m g RE 1 −
1+a
a
= m g RE
1+a
= 1540 kg × 9.8 m/s2 × 6.37 × 106 m
2.98
×
1 + 2.98
= 7.19813 × 1010 J .
Tipler PSE5 11 43
Version PREVIEW – Practice 9 – carroll – (11108)
007 10.0 points
An object is dropped from rest from a height
6.2 × 106 m above the surface of the earth.
The acceleration of gravity is 9.81 m/s2 .
If there is no air resistance, what is its speed
when it strikes the earth?
Correct answer: 7.85141 km/s.
3
Explanation:
Let : g = 9.8 m/s2 ,
Patm = 1.01 × 105 Pa ,
h = 12 m , and
ρw = 1000 kg/m3 .
Explanation:
6
Let : h = 6.2 × 10 m .
G ME
and the potential energy at a
2
RE
distance r from the surface of the earth is
The pressure at the bottom of the lake is
equal to the pressure at the surface plus the
pressure given by the mass of water above, so
g =
U (r) = −
G ME m
.
r
Using conservation of energy to relate the
initial potential energy of the system to its
energy as the object is about to strike the
earth (Ki = 0),
Kf + Uf − Ui = 0
K(RE ) + U (RE ) − U (RE + h) = 0
G ME m G ME m
1
m v2 −
= 0.
+
2
RE
RE + h
Solve for v:
s G ME
G ME
v= 2
−
RE
RE + h
s
h
= 2 g RE
RE + h
s
2 (9.81 m/s2 ) (6.37 × 106 m)
=
6.2 × 106 m + 6.37 × 106 m
p
1 km
× 6.2 × 106 m ·
1000 m
= 7.85141 km/s .
Lake Bottom Pressure
008 10.0 points
Determine the absolute pressure at the bottom of a lake that is 12 m deep. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 2.186 × 105 Pa.
P = Patm + ρw g h
= 1.01 × 105 Pa
+ (1000 kg/m3 ) (9.8 m/s2 ) (12 m)
= 2.186 × 105 Pa .
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