Morby, Grant – Homework 29 – Due: Apr 17 2006, noon – Inst: Drummond
1
This
printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Blood flows through a coronary artery that
is partially blocked by deposits along the
artery wall.
blood
f
lo
w
Through which part of the artery is the flux
(mass of blood per unit time) largest?
1.
the flux is the same in both parts
correct
2.
the narrow part
3.
the wide part
Explanation:
Because blood is incompressible, the flux
for a same flow is always the same.
002
(part 1 of 2) 10 points
A person rides up a lift to a mountain top,
but the person’s ears fail to ”pop”; that is,
the pressure of the inner ear does not equalize
with the outside atmosphere.
The radius of
each eardrum is 0.40 cm. The pressure of the
atmosphere drops from 1
.
008
×
10
5
Pa at the
bottom of the lift to 0
.
983
×
10
5
Pa at the top.
a) What is the net pressure on the inner ear
at the top of the mountain?
Correct answer: 2500 Pa.
Explanation:
Basic Concept:
P
net
=
P
b

P
t
Given:
P
b
= 1
.
008
×
10
5
Pa
P
t
= 0
.
983
×
10
5
Pa
Solution:
P
net
= 100800 Pa

98300 Pa
= 2500 Pa
003
(part 2 of 2) 10 points
b) What is the magnitude of the net force on
each eardrum?
Correct answer: 0
.
125664 N.
Explanation:
Basic Concepts:
P
=
F
A
A
=
πr
2
Given:
r
= 0
.
40 cm
Solution:
F
net
=
P
net
A
=
P
net
(
π r
2
)
= (2500 Pa)
π
(0
.
4 cm)
2
·
1 m
100 cm
¶
2
= 0
.
125664 N
004
(part 1 of 1) 10 points
In testing a new material for shielding space
craft, 138 small ball bearings, each moving at
a supersonic speed of 393.9 m/s, collide head
on and elastically with the material during a
0.689 min interval.
If the bearings each have a mass of 6.1 g
and the area of the tested material is 0.77 m
2
,
what pressure is exerted on the material?
Correct answer: 20
.
8336 Pa.
Explanation:
Basic Concepts:
P
=
F
A
Conservation of momentum:
Δ
p
=
F
Δ
t
Elastic collision:
v
f
=

v
i
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Morby, Grant – Homework 29 – Due: Apr 17 2006, noon – Inst: Drummond
2
Given:
Δ
t
= 0
.
689 min
N
= 138
m
= 6
.
1 g
v
= 393
.
9 m
/
s
A
= 0
.
77 m
2
Solution:
Δ
p
=
m v
f

m v
i
=
m
(

v
i
)

m v
i
=

2
m v
i
and
F
=
Δ
p
Δ
t
For one bearing,
P
=
fl
fl
fl
Δ
p
Δ
t
fl
fl
fl
A
=
2
m v
A
Δ
t
For all of the bearings,
P
=
N
2
m v
A
Δ
t
¶
=
2(138)(6
.
1 g)(393
.
9 m
/
s)
(0
.
77 m
2
)(0
.
689 min)
·
1 min
60 s
¶
1 kg
1000 g
¶
= 20
.
8336 Pa
005
(part 1 of 1) 10 points
The hypodermic syringe shown in the figure
contains a medicine with the same density as
water. The barrel of the syringe has a cross
sectional area of 4
.
71
×
10

5
m
2
. The cross
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 Spring '09
 KLEINMAN
 Physics, Force, Work, Correct Answer, Kilogram

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