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Blueprint

Solution
Manual

Page 1 of 43

s0010 CHAPTER 1
o0010

1-1
...
Toppling torque Ta 5 Fa × 1
...

1-2
...
1
...
E
...
2

f0010

L1

d1
q

q

L2

d2

p0040

The magnitudes of the two angles of the lever arm with respect to the
horizontal are equal therefore,
L1 ¼ d1 sin θ L2 ¼ d2 sin θ
L1 d 1
and
¼
L2 d2

Page 2 of 43

o0020

1-3
...
E
...
3, the sum of the two angles ω + 100o ¼ 180o
∴ ω ¼ 80o
x0 ¼ 30 × cos 80° ¼ 5:2 cm
y0 ¼ 30 sin 80° ¼ 29:4 cm
θ

¼

y0

tan —1

θ ¼ tan

x0 + 4
29:4
—1
5:2+4

¼ 72:6

°

f0015

100°
y
w

q
4 cm

x
o0025

1-4
...
1-13
W¼

Fm
10:5

¼ 335 N ¼ 75 lb

Page 3 of 43

o0030

1-5
...
E
...
5

f0020

y
160°
a


w

q


b

Fm
x



w

ω + 160° ¼ 180° ;ω ¼ 20°
a ¼ 30 sin 20 ¼ 10:3cm
b ¼ 30 cos 20 ¼ 28:2cm
a
10:3
θ ¼ tan—1
¼ tan—1
b +4
28:2+4
°
θ ¼ 17:7
p0065

The upper arm is at the same angle as in Fig
...
Using results from
Exercises 1
...
(1-10)
x component: Fm cos(θ + δ) ¼ Fr cos ϕ
y component: Fm sin(θ + δ) ¼ Fr sin ϕ + W
Torque is: 4 cm Fm sin (θ) ¼ 40 cm W × sin γ
From these we obtain 3 equations
1
...
7 ¼ Fr cos ϕ
2
...
7 ¼ Fr sin ϕ + 137 N
3
...
7 ¼ 10 × 137 × sin 30o
From 3
...
4o
Fr ¼ 2,386 N ¼ 536 lb

Page 4 of 43

o0050

1-6
...
1-12 (or 1-13) θ ¼ 72
...
1-12
4 cm × Fm sin θ ¼ 20cm × W
W ¼ 14 × 9:8 ¼ 137 N
Fm ¼ 720 N ¼ 162 lb

p0130

Following Eqs
...
(a) As in Eq
...
1-15
Fr cos ϕ ¼ 646 N
Fr sin ϕ ¼ 2, 060 — 2 × 137 ¼ 1, 790 N
F2r ¼ 3:61 × 106N2
Fr ¼ 1, 900 N
1, 790
°
tan ϕ ¼
¼ 2:77;ϕ ¼ 70:2
646

li7890
o0065
p0155
p0160
p0165

(b) yes
1-8
...

Referring to Problem 1-6
Added force Fm ¼ 720
¼ 103 N
7
Added force Fr ¼ 590
¼
84 N
7

Page 5 of 43

o0070

1-10
...
E
...
10 θ ¼ 72
...
3° ¼ 19
...

3:4

p0185

Speed of muscle contraction ¼ 4 cm=s
19:6 cm
Speed of weight displacement ¼
0:5s
¼ 38 cm=s:
Ratio of speeds is approximately inverse of mechanical advantage
...


Using data given in the text, torque regarding hip, torque about the
insertion point is:
Fm × 7cm ¼ W × 3+ ð7 — 5:56Þ0:185 W
;Fm ¼ 0:47 W

p0200

As seen from Fig
...
The y
components of forces set to 0 leads to:
Fm + W ¼ Fr + 0:185 W
1:47 W ¼ Fr + 0:185 W
Fr ¼ 1:28 W

Page 6 of 43

o0080

1-12
...

Angle of reaction force Fr at fifth lumbar with respect to y-axis is ϕ
x component of force ¼ 0
Fm sin 72° ¼ Fr sin ϕ
y comp of force ¼ 0
Fr cos ϕ ¼ 160 + 320 + Fm cos 72°
Fr sin ϕ ¼ 1, 902 N
Fr cos ϕ ¼ 1, 098 N
;Fr 2 ¼ 4:82 × 106 N2 and Fr ¼ 2, 200 N

o0085

(b) The added 20 kg mass is a force F ¼ 196 N
...
Torque conditions:
‘
2
‘ × 356 + 320 cos 30 ° ¼ Fm ‘sin 12°
3
2
Fm ¼ 3, 220 N

p0225

Following 1-12 (a)
Fr sin ϕ ¼ 3, 062
Fr cos ϕ ¼ 356 + 320 + Fm cos 72° ¼ 1, 671 N
Fr 2 ¼ 12:17 × 106 N2 ;Fr ¼ 3, 490 N

o0090
p0235
p0240
p0250
f5780

1-13
...

From Fig
...
5 W
x component of force:
FT sin ϕ ¼ FA sin 15° ¼ 0
...
42 W
;F2T ¼ 12:1 N2

f0025

p0290

and FT ¼ 3
...
(a) Ff ¼ μ Fn Fn ¼ m g
μ ¼ 0
...
33 meters/sec
Work done by frictional force bringing the skater to a halt is Ffd
1
Ff d ¼ mv2
2
1 2
μmgd
mv
¼
2
ð8:33Þ2
v2
d¼
¼
¼ 354 meters
2μg 2 × 0:01 × 9:8
(b) The distance of coasting is independent of mass
...
As in Chapter 1 Eq
...
5 ¼ W × 0
...
(a) Let the force at point be B ¼ Fb
Let the force at point be C ¼ Fc
Equilibrium torque conditions about point A ¼ 0
FB × 2:5 cm=cos 45° ¼ ðFC × 3 cmÞ=cos 45°
2:5
;FC ¼
FB
3
Sum of forces perpendicular to the fin are equal to zero:
F sin 25° + FC ¼ FB

p0370
p0375

p0380

With F ¼ 0
...
1 by Eq
...
01,
;45° — θ ¼ 44:4° and θ ¼ 0:6°

2-4
...
Volume of residual saliva in the mouth ¼ 0
...

p0410
Let radius of the tongue (assumed to be a sphere) be rt and radius of the
oral cavity be rc
...
17 cm3 and (rc)3 ¼ 9
...
93 cm and rc ¼ 2
...
8¼cm3
p0435
Let the thickness of the coating¼Δr, then the volume of the saliva coating on the tongue and the oral cavity respectively is 4 π (rt)2 x Δr and 4 π
(rc)2 x Δr respectively
...
2 Δr + 56
...
8 cm3 or Δr ¼ 0
...
0775 mm
This is approximately ~0
...


Page 10 of 43

s0020 CHAPTER 3
o0125
p0470
p0475
p0480
p0485

o0130

p0495

3-1
...
3-12
Work done on the body ¼ W(c + H)
W ¼ 70 kg × 9
...
2 meters
∴ work done ¼ 823 J
work
823 J
Power ¼
¼
¼ 4, 120 W:
time 0:2 sec
3-2
...
Assuming as in the text that Fm W
¼
From Eq
...
Square Eq
...

Fr2 ¼ 5 W2 — 4W2 sin θ

p0510

Obtain
sin θ ¼

p0515

p0520

5W2 — F2r
4W2

Substitute into 3-20 and obtain quadratic equation Fr2 + 1
...
41 ∴ θ ¼ 65
...
From the text vo ¼ 3
...
7 × sin 45° ¼ 2
...
3-1
vy ¼ vy o + at

p0560

At maximum height of jump vy ¼ 0
Therefore

p0565

—a vy o
vy o
¼
a
g
2:62
¼
¼ 0:267 sec
9:8
Duration of jump ¼ 2t ¼ 0
...
(a) Eq
...

Because weight on the moon is
Eq
...
3-19 and Eq
...
1 and obtain
Fr2 + 0:236 Fr W — 3:97 W2 ¼ 0

p0595
p0600
p0605
p0610
p0615
p0620

Solve for Fr taking positive answer
Fr ¼ 1
...
707 × 1
...
665
θ ¼ 48
...
4 × 0
...
1 m2/sec2
From Eq
...
3-15
2 θ
2v
ðvo2Þ cos
22:1
o
H¼
¼ 0¼
¼ 3:39 m
2g0
4g 4 × 1:63
(c) From Eq
...
Duration of jump ¼ 2 × t ¼ 4
...

¼1:63

Page 12 of 43

o0160

3-6
...
3-24
vt ¼

W
CA

1=2

ρ ¼ density

4
W ¼ πr3ρ A ¼ πr2
3
C ¼ 0:88 kg=m3ðsee text section 3:7Þ
r ¼ 0:5 cm ¼ 5 × 10—3m
ρ ¼ 1g=cm3 ¼ 103kg=m3
vt ¼
¼

4rgρ
3c

1=2

4 × 5 × 10—3 × 9:8 × 103

1=2

3 × 0:88
¼ 8:6 m= sec
o0165

m=sec

3-7
...
3-24
W
A¼ 2
vt C
W ¼ 70 × 9:8N
v2t ¼ 196 m2 C ¼ 0:88 kg=m3

p0665
p0670
o0170
o0175

Therefore A ¼ 3
...
13 m
3-8
...
3-24
W
vt ¼
(a)
CA

1=2

ρ ¼ density ¼ 0:92 × 103kg=m3
4
W ¼ πr3ρ A ¼ πr2
3
C ¼ 0:88 kg=m3ðsee textÞ
r ¼ 5 × 10—3mρ ¼ 0:92 × 103kg=m3
vt ¼
¼

4rgρ
3C

1=2

4 × 5 × 10—3 × 9:8 × 0:92 × 103
3 × 0:98

1=2

m=sec

¼ 8:3 m= sec

Page 13 of 43

o0180

p0690
o0185

(b) Note from Eq
...
3 m/s × (4)1/2 ¼ 16
...
Work done in each step ¼ mv2 (see text Section 3
...
Assume area of runner is 0
...
7)
Wind velocity ¼ 30 km/hr ¼ 8
...
33 + 4
...
8 m/sec
Fa ¼ C Av2 ¼ 0
...
2 × 165 ¼ 29
...
Centrifugal force Fc
Mass of the arm ¼ (0
...
06)70 kg ¼ 9
...

R location of center of mass from shoulder ¼90 cm ¼ 0
...
From Eq
...
(b) Assume as before that center of mass (m) of arm is at mid-length
of the arm
...
45 m
Let the force along the arm be Fa; this force is at 60° with respect to the
horizontal
...
Following Eq
...
The length of the step is 1m therefore, the runner does 100 steps in 10 sec
...
1 sec
Using the pendulum analogy,
T ¼ 2 × 0
...
2sec
A ¼ length of step (See Fig
...
4-11
amax ¼

o0215

1=2

1=2

2 0:9
¼ 2π 3 × 9:8

1=2

¼ 1:6 sec:

As discussed in text, time for 1 step is T2 ¼ 0:80 sec
Length of step is 0
...
The period of the arm swing is
T ¼ 2π

p0865
o0225
p0875

2

0:9
×

1=2

¼ 1:6 sec:

3 9:8
Same as that for the leg
4-8
...
of steps/sec 5 speed/length of step 5 vA(from Fig
...
of steps
...
4-12
T

Page 16 of 43

I ¼ m‘3 from Appendix A;
‘ ¼ length of leg
vmax
ωmax ¼
‘
Maximum energy/step 5 Er
2

p0880
p0885

p0890

1 2
Er ¼ Iωm see text
2
1 m‘2 π2v2 π2 2
¼
2 ¼ 6 mv
2 3 ‘2
¼ 1:64 mv
o0230

4-9
...
4
...
2 is
m‘2
I¼
‘ ¼ 2m
3
m‘2
T ¼ 2π
mg‘=2
¼ 2π

2‘
g

1=2

1=2

¼ 6:28 ×

4
9:8

1=2

¼ 4:0 sec:
p0905
o0235
p0915

Time of falling is T4 ¼ 1 sec
4-10
...
45 5 26
...
7° 5 0
...
)
The distance the center of mass is raised in the course of one step
is: 1m – 0
...
11m 5 11cm
...
From Eq
...
1)
E¼
SB ¼ 10 dyn=cm
2 y
14 × 100 × 1018
¼
2 14 × 1010
¼ 143 J

p0930

8

¼ 14:3 × 10 erg

Therefore the kinetic energy of runner that may cause fracture is
obtained from
1 2
mv ¼ 143 Jm ¼ 50 kg
2
v ¼ 2:39 m=sec ¼ 8:6 km=hr
¼ 5:3 m:p:h

o0245
p0940

5-2
...
However,
if the area of impact is smaller the maximum speed of fracture would correspondingly decrease
...
An impact force FB that will fracture the skull is
FB ¼ 1 cm2 × 109dyn=cm2 ¼ 109dyn
4

¼ 10 N
p0970

The impulsive force of the falling body is
F¼

mv
1=2
;v ¼ ð2ghÞ
Δt

Page 18 of 43

p0975
p0980
p0985

Therefore F ¼ (m(2gh)1/2)/Δt
Fracture will occur when F ¼ 104 N
or
F2Δt2

108 × 10—6

h¼

o0255

p0995

o0260

¼
¼ 5:1m
m22g 1 × 2 × 9:8
5-4
...
A1-4)
v v2
¼
t 2s
Therefore t ¼ 2sv
s ¼ 30 cm ¼ 0:3m
v ¼ 70 km=hr ¼ 19:4 m= sec ~ 20 m=sec
0:6
t¼
¼ 3 × 10—2 sec
20
5-5
...
1 rupture strength on stretching, (as in whiplash) is
83 × 107 dyn/cm2
FB ¼ 1 cm2 × 83 × 107 ¼ 83 × 102N
mv

p1005
p1010

Impulsive force is F ¼ Δt
Therefore the collision velocity that may fracture the neck is
v¼

FB
83 × 102 × 10—2
Δt ¼
5
m
¼ 17m=sec
¼ 60 km=hr
¼ 37 m:p:h

o0265

5-6
...
A 1-8
v2
v ¼ 62:5 m= sec s ¼ 1m
2s
2
a ¼ ð62:5Þ ¼ 1:95× 103m=sec 2
2
The decelerating force is
a¼

p1020

F ¼ ma ¼ 137 × 103N
¼ 137 × 10 8 dyn
p1025
p1030
p1035

Area of impact
5 0
...
3 8 × 104 cm2
Force/cm2 on the body 5 137×10 ¼ 4:6 × 106dyn=cm2
0:3×104

This is below the rupture limit
...
Let m ¼ mass of hand
M ¼ mass of bag
V ¼ rebound velocity
mv ¼ MV
m
5
V ¼ v ¼ × 7 ¼ 0:7m=sec
M
50
1

mv2
2
¼ 122:5J
1 2
Final K:E:
mv
¼2

Initial K:E:

¼

¼ 12:25 J
p1060

Kinetic energy is not conserved
...


Page 20 of 43

s0035 CHAPTER 6
o0275

p1075

6-1
...
The upward force must restore the insect to the original height during the same time interval
...

During the downward wing stroke (which raises the insect) a force FW is
applied to the insect by the wings
...
To raise the insect in the same time interval Δt
FW — W ¼ W ;FW ¼ 2W

o0280

6-2
...
03

l

70°

p1085
o0285

Therefore ‘ ¼ 0
...
Energy stored in resilin is
E¼

1 YAΔ‘2

ðEq: 6 — 11Þ
2
‘
As given for a flea A ¼ 10—4cm2
‘ ¼ 2 × 10—2cm
Δ‘ ¼ 10—2cm
Y ¼ 1:8 × 107 dyn=cm2

Page 21 of 43

p1095

Energy stored per leg
E ¼

p1100
p1105

1 1:8 × 107 × 10—4 × 10—4

¼ 4:5 erg
2
2 × 10—2
Total stored in 2 legs ¼ 9
...

E ¼ mgh m ¼ 0:5 × 10—3 g
g ¼ 980 cm=sec 2
9:0
h¼
0:5 × 10—3 × 980

o0290

¼ 18:4 cm

6-4
...
As the limbs of area A move down a distance ‘, they sweep out a volume
A × ‘
...

Therefore volume swept out per unit time is
Aב
¼ Av
t

p1145
p1150

p1155
p1160

The mass swept out per unit time is Avρw
...
7-11
...
7-11 and 7-13
P ¼

1
2

Aρωv

3

1
gvðρ — f ρ !3=2
ωÞ
¼ Aρω
1=2
2
ðAρω Þ
3=2
1 ½W ð1 — f ρω =ρÞ]
¼
2
ðAρωÞ1=2
p1165
o0300
p1175
p1180

which is Eq
...
A ¼ 600 cm2 ¼ 6 × 10–2 m2
ρω ¼ 103 kg/m3
From 7-14
P ¼

o0305

1 ½50 × 9:8ð1 — 0:95Þ]
3

3=2

¼ 7:8 W

1=2

6 × 10—2

2

× 10

7-3
...
7-13 substitute
ðρw — ρÞ for ðρ — f ρwÞ then
gVðρω — ρÞ
Aρ ω

v¼

P ¼
o0310
p1195
p1200

1
2

W

1=2

ρω
—1
ρ

3=2

ðAρω Þ1=2

7-4
...
H
...
7-15
7-5
...

Weight of the column of unit area ¼ Pressure
150 × 102 cm × 980 cm/sec2 × 1
...
1 × 106 dyn/cm2
1 atm ¼ 1
...
The density of the air bladder is negligible
...
7-15 becomes
1:026 ¼

p1240
p1245
o0325

and
x ¼ 3
...
W1 ¼ ρ2V1 V1 ¼

ð100 — xÞ1:067
100

W1
ρ2

W2 ¼ ðρ2V1 — ρ1V1Þ ¼ V1ðρ2 — ρ1Þ
p1255

Substitute for V1 ¼ Wρ 1
2

W2 ¼

W1ðρ2 — ρ1Þ
ρ
2

p1260

Solve for ρ2
ρ2 ¼ ρ 1

o0330
p1270
p1275
p1280

W 1— W 2
7-8
...
7-20 to substitute for h
we obtain
P ¼

p1285

W1

2T cos θ
R

With θ ¼ 0 R ¼ 10–3 cm
T ¼ 72:8 dyn=cm
2 × 72:8
P¼
¼ 1:46 × 10 5 dyn=cm 2
10—3

Page 24 of 43

o0335

p1295

7-11
...
Using the analogy of the rubber band, the maximum force generated by
the reduced surface tension is
F ¼ ðT1 — T2ÞL
¼ ð73 — 50Þ dyn=cm × 3 × 10—1
¼ 6:9 dyn
F
6:9
2
2
¼ ¼
¼
2:3
×
10
cm=s
m
3 × 10—2
Speed at ¼ 2:3 × 102 × 0:5 ¼ 115cm=sec
a

¼ 1:15 m=sec

Page 25 of 43

s0045 CHAPTER 8
o0345
p1315

8-1
...
51 × 10–4 torr
From Eq
...
Q ¼ 8 × 10 cc=min ¼ 133 cc=sec

η ¼ 0:04 R ¼ 0:5 cm
L ¼ 30 cm
Using Eq
...
1 torr ¼ 1
...
33 × 105 dyn/cm2 ¼ ρgh
1:33 × 105
h ¼
¼ 129 cm
8-4
...
(a) Q ¼ πR ð8ηL
Flow is proportional to R4
4

o0370
p1355

Q1

¼

Q2
(b)

o0375

R1
R2

4

R1

4

R2

¼ 0:1

R1
R2

o0385
p1375
p1380

¼ 0:41

Therefore

p1365

o0380

4

0:08
0:1

¼

8-6
...
The blood flow is constant throughout the body
...
3cm3/sec A × v × no
...
From Eq
...
As shown in the text the K
...
of the blood at a flow rate of 5‘/min is
3
...
E + P)/cm3 ¼ 83 + 160 ¼ 243 x 103 erg/cm3
Q ¼25‘/min ¼ 417 cm3/sec
Power ¼ Q × Energy/cm3 ¼ 417 × 243 × 103 ¼ 10
...
1 W
8-10
...
33 × 103 erg/cm3
Pressure at the right ventricle is 1206 torr ¼ 20 torr ·
Note 1 torr ¼ 1
...
33 × 103dyn/(torr cm2) ¼ 26
...
E
...
33 + 26
...
3 cm3/sec
Power ¼ E × Q ¼ 30 × 103 × 83
...
25 × 10 erg/sec
¼ 0
...
E
...
33 × 103 × (5)2 ¼ 83
...
7 × 103 erg/cm3
Energy ¼ K
...
+ P ¼ (83
...
7) × 103erg/cm3 ¼ 110 × 103 erg/cm3
Q ¼ 25‘/min ¼ 417 cm3/sec
Power ¼ Q × E ¼ 417 × 110 × 103 erg/sec ¼ 4
...
5 W
8-12
...
That is,
L ¼ (Volume of the liquid) / (surface area of volume)
...

Therefore, L ¼ (l x w x d) / 2 (l x w) ¼ d/2
(b) The Reynold’s number R ¼ vρL/ η
The values provided in the Exercise are: v the average flow speed¼
1 cm/s, ρ ¼ g/cm3, η fluid viscosity ¼ 0
...

With these values, we obtain, R ¼vρL/ η ¼ (1 x 1 x 5x10—4)/10—2 ¼
5x10—2
This is well within laminar flow
...
Boyle’s law (Eq
...
That is as pressure changes so does volume so that
P1V1 ¼ P2V2

p1560

At 40 m depth the additional pressure above the surface pressure P2 is
ΔP ¼ h × ρ × g h ¼ 40 × 102 cm
ρ ¼ 1 g=cm3

p1565

Therefore
ΔP ¼ 40 × 102 × 1 × 980
¼ 3:92 × 10 dyn=cm
6

2

1 atm ¼ 1:01 × 106dyn=cm2
ΔP ¼ 3:88 atm
V1 ¼ 6‘ P1 ¼ ΔP + 1 atm: ¼ 4:88 atm
V2 ¼ ? P2 ¼ 1
P1V1
4:88 × 6
V2 ¼
¼ 29:3 ‘
P2 ¼
1
o0430

9-3
...
As an analogy consider the particles to be on a conveyer belt rather than in a
beam
...
As the conveyer starts moving the number of
meters of conveyer belt that pass by per second is VD
...

o0445
p1590
o0450

—5

4

9-5
...
5 ‘/hr ¼ 4
...
02 cm3/sec × 2
...
08 × 1020 molec/sec
9-6
...
5 ‘ of O2 per hour
...
7)
Each breath under resting conditions consists of 1/2 ‘ of air (see text)
The fraction of O2 exchanged is 0
...
163 ¼ 0
...
3) No
...
5 × 0
...
0232 ‘ of O 2
...
of breath per min ¼ 14:5
¼ 10:4
60×0:0232

Page 30 of 43

o0455

o0460

9-7
...
Therefore oxygen needed is
14:5‘
4
3
¼ 2:07 × 10— ‘=cm hr
70 × 103

p1610
p1615

Maximum radius of animal is obtained from
2
...
71 × 10–5 × area of animal
4
2:07 × 10—4 × πr3 ¼ 1:71 × 10—5 × 4πr2
3

p1620
o0465

r ¼ 0
...
5 cm
9-8
...
87 × 10–5 atm
ΔP ¼ 29
...
12 × 103 × 9
...
87 atm
9-9
...
Therefore if lung area
can decrease by 10, r can increase by 10
...
Kleiber’s law stating that the basal metabolic rate (BMR) is proportional to M3/4 can be expressed as an equality using a pre-multiplicative
constant C
...
e
...
75) with an intercept of log C
...
As stated in the text, 4
...
The volume of the room is 27 m3 ¼ 27 × 106 cm3 ¼ 27 × 103
liter of this volume, 27 × 103 × 0
...
4 × 103 liter is oxygen
...
83 cal/liter ×
5
...
1 × 103 Cal
From the text the basal metabolism of the person is 70 Cal/hour
3
×10 ¼ 373 hours
Therefore the person can survive at most 26:170
11-5
...
70 ¼ 1
...
83 Cal
(See Section 11
...
08
¼
× 10 liter
Container volume required
V¼

4:08 × 105

100
¼ 4:05 m 3

o0490
p1715
p1720
p1725
p1730
p1735
p1740

3

¼ 4:08 × 10 ‘

11-6
...
70 ¼ 476 Cal
16 hour rest ¼ (16 × 40) × 1
...
Surface area of person from Eq
...
202 × 600
...
40
...
47 m2
Difference in caloric consumption between sleeping and sitting upright
(50 – 35) × 1
...
Heat flow from Eq
...
5 cm ¼ 4 × 10 m
ΔT ¼ 38 – 25 ¼ 13°C
18 × 4 × 10—2 × 13
H¼
¼ 18:7 Cal=h
0:5
Typically a person with a surface area of about 1
...
(a) Multiplying out the right hand side of the equation we obtain
T4 + T3Tr + T2T2 + T3Ts — T3Tr — T2T2 — TsT3 — T4
s

4 s

4

¼ Ts — Tr
o0510
p1825
p1830
p1835
p1840
p1845
p1850

p1855
o0515
p1865
p1870
p1875

2

s

r

r

s

s

r

r

r

(b) At Tr ¼ 40°C ¼ 313 K, Ts ¼ 35°C ¼ 308 K
Ts3 + Ts2Tr + Ts Tr 2 + Tr3
¼ 3083 + 3082 × 313 + 308 × 3132 + 3133
¼ 2
...
97 × 107 + 3
...
07 × 107
¼ 11
...
92 × 107 + 2
...
26 × 106
+ 2
...
8 × 107
11:98—9:8
Percent change ¼ 9:8 ¼ 0
...
11
...
5 m2, e ¼ 1
TS – Tr ¼ 32 – 25 ¼ 7°C
63
2
K¼
r
1:5×1×7 ¼ 6
...
Volume of breath/h ¼ 20 × 0
...
24 × 1016/cm3
...
24 × 1016 ×
103 × 24 ¼ 7
...
32 × 10–2 × 600 ‘/h × 0
...
07 Cal/h
11-12
...
11-4
Hc0 ¼ Kc0 Ac ðTS — Ta Þ
¼ 10 × 1 × ð299 K — 233 KÞ
¼ —660 Cal=m 2h

o0530
p1915

p1920
o0535

p1935

p1940

o0545

p1950
p1955
p1960
p1965

11-13
...
8 mole
The amount of heat required to heat the air from –40°C to + 37°C
(ΔT ¼ 77C°) is
H ¼ 6
...
8 mole/h) × 77°C ¼ 14
...
(a) As stated in the formulation of the exercise, the basal metabolism of
a 70 kg person is about 70 Cal/hour¼ 1,680 Cal/day
...
75 ¼ 1,680 x 2
...

Considering typical lion activities, the actual energy consumption
is likely to be twice that of basal metabolism, that is about 49,640
Cal/week
...
Such an
antelope provides 20 kg of food at 3 Cal/gm ¼ 3,000 Cal/kg
...

Thus, the number of antelopes required by one lion per week is
(49,640 Cal/week)/60,000 Cal/antelope ¼ 0
...

This calculation of course entailed several gross approximations
...
At 1 m the intensity of riveting is 10–7 W/cm2
Threshold of hearing is 10–16 W/cm2
Let R be the distance at which the sound intensity of the riveter is
10–16 W/cm2
R2

p1990
p1995
o0555
p2005
p2010
o0560

10—7

¼ —16
I
10
2
∴ R ¼ 109 m2
R ¼ 3
...
6 km
12-2
...
From Eq
...


p2030

Therefore P0 ¼ (I2ρv)1/2 ¼ (10–9 ×2 2 × 1
...
3 × 104)1/2
–4
¼ 2
...
In 70 msec the
sound pulse must travel 2D (There and back)
...
5 m
2 ¼
2

12-8
...
1, threshold of hearing is 10 —16 W/cm2
...
70 m x 3dB/m¼ 2
...
21
p2055
(Sound level at the source)/10—16 W/cm2 ¼ 100
...
62
p2060
(Sound level at the source) ¼ 1
...
62x10—16
W/cm2 ¼ π r2 x 1
...
14x0
...
62x10—16 ¼ 2
...

o0570

Page 35 of 43

o0575
p2085
p2090
p2095
p2100

o0580
p2110
p2115
p2120
p2125
p2130
p2135
p2140
p2145

12-9
...
That is,
f x λ ¼ v with f ¼ λ ¼ v/f
with v ¼ 1,540 m/s, and f ¼ 5x106 Hz
λ ¼ v/f ¼ 1,540/5x106 ¼ 3
...
308 mm
This is approximately the smallest the size of objects in soft tissue that
can be detected
12-11
...
15 cm3 tissue is:
12 W x 1 s ¼ 12 J ¼ 12 J x 0
...
87 cal
...

m ¼ 0
...
15 g
...

ΔT ¼ (Heat input)/ (heat capacity x mass) ¼ (2
...
15 g) ¼ 19
...
1) oC
o
¼ 56
...


Page 36 of 43

s0065 CHAPTER 13
o0585
p2160

13-1
...
02 × 1020 ¼ 9
...
5 × 10–12/meters3 ¼ 78
...
of Na+/meter of axon ¼ 78
...
03 × 1018 ¼ 7
...

13-2
...
13-4 by RT + Rm
R2T + RTRm ¼ 2R RT + 2RRm + RTRm
R2 + 2RRT — 2R Rm ¼ 0

p2190
o0600
p2200

p2205
o0605

o0610

Solving for RT yields Eq
...
Using the voltage divider relationship with
R0 ¼ R i ¼ R
Rm RT
VA
Rm + RT
VB ¼
RmRT + 2R
R m + RT
Multiply by Rm + RT and divide by Rm RT yields Eq
...
Substituting in Eq
...
V2 ¼ V1 1 — Δxλ
V3 ¼ V2 1 —

p2220

RT ¼ rΔx + r2Δx2 +

Δx
λ

¼ V1 1 —

Δx

2

λ

Proceeding section by section we obtain Eq
...
The expansion shown follows from the binomial theorem
ð1 — yÞn ¼ 1 — ny +
—

p2230
p2235
p2240

o0620

o0625

p2260

nðn — 1Þðn — 2Þy3
+⋯
3!

with y ¼ Δx
,
λ
with nΔx ! x
and n – 1 ≈ n – 2 ≈ n etc
1—

p2245

Δx
λ

n

¼1—

nx
x2
x3
+
—
λ
2!λ2
3!λ3

This is the series for e–x/‘
13-8
...

Therefore, to generate 500 volts the number of cells connected in
500
series is —1
10 ¼ 5,000 cells
(b) From Problem 13–1 number of ions entering per meter of cell per
pulse is 3 ×
10–8 coulomb
Therefore current flowing per pulse into one cell is
i¼

p2265
p2270

p2275
p2280

nðn — 1Þy2
2!

ΔQ × ‘

Δt
‘ ¼ length of cell ¼ 10–5 m
Δt ¼ duration of pulse
3 × 10—8 × 10—5
—11
amp
i¼
¼
3
×
10
10—2
To produce 80—2ma ¼ 8
...
7 × 109 cells connected in parallel
3×10—11

Page 38 of 43

s0070 CHAPTER 14
o0630
p2295
p2300

14-1
...


¼ 11:1 × 10—6F:

1
2E
400 J
E ¼ CV 2 ;C ¼ 2 ¼
2
V
36 × 106
6
Q ¼ CV ¼ 11:1 × 10— × 6 × 103 ¼ 66:6 × 10—3C
Q 66:6 × 10—3
i¼ ¼
¼ 13:3 amp:
Δt
5 × 10—3

Page 39 of 43

s0075 CHAPTER 15
o0635

p2315

15-1
...
A 3-6
1 1 1
+ ¼ f ¼ 1:5 cm
p q f
q ¼ 6 m ¼ 600 cm
1 1
1
¼
—
p 1:5 600
1
p

p2320
p2325

o0640

¼

600 — 1:5
1:5 × 600

¼

598:5
¼ 0:665
900

p ¼ 1
...
5 cm
...
004 cm
15-3
...
A 3-10
1
nL — n1
nL — n2
¼
—
f
R1
R2

p2335
p2340
p2345

For the cornea: From Table 15-1
nL ¼ 1
...
33
R1 ¼ +0
...
0073
1
1:38 — 1
1:38 — 1:33
¼
—
f
0:0078
0:0073

p2350
p2355
p2360
p2365

¼ 48
...
8 ¼ 41
...

nL ¼ 1
...
33
Minimum power R1 ¼ 0
...
006 m
1
1:40 — 1:33
1:40 — 1:33
¼
+
f
0:01
0:006

p2370
p2375

¼ 7 + 11
...
7 diopters
Maximum power R1 ¼ 0
...
0055
1
1:40 — 1:33
1:40 — 1:33
¼
+
f
0:006
0:0055
¼ 11:7+ 12:7 ¼ 24:4 diopter

Page 40 of 43

o0645
p2385
p2390

15-4
...
A 3-10
n1 ¼ n2 ¼ 1
...
0078 R2 ¼ 0
...
41 – 6
...
39 diopter
15-5
...
33, nL ¼ 1
...
002 m R2 ¼ –0
...
Using Eq
...
5 cm ¼ 15 mm

15-7
...
15-6, the separation between point a and b on the
retina is 4μm ¼ 4 × 10–6 m (one cone that is unexcited plus 2 ×
radius of the excited cones)
—6
From Figs
...
tan—1 2θ ¼ D x ¼ the size of the whites of the eyes D is the distance from
–4
which they can be seen θ ¼ 5 × 10 rad for small θ
x
x
1 cm
θ¼
D
¼ ¼
D
θ 5 × 10—4
+4
¼ 0:2 × 10 cm
¼ 20 m
15-9
...
15-6, we obtain Eq
...
(a) Magnitude of force F ¼ K l ¼ 1 N/m × 0
...
5 10–9 N
...
R is the
distance between charges measured in meters
...
5 10–9 N
Q2 ¼ (0
...
6 × 10–38 C2 and
Q ¼ 2
...
4 10–19/1
...

This is effectively 1
...
5 protons facing each other,
distributed over and area defined by the sampling tip-surface overlap
...
Radius of particle ¼ R Radius of atom ¼ r
number of silver atoms on particle surface ¼
2
(areaofparticlesurface)/(crosssectionalareaofatom) ¼ 4πR2/πr¼
4R2/r2
number of silver atoms in particle volume ¼
(volume of particle)/(volume of atom) ¼ (4/3πR3)/(4/3πr3)¼ R3/r3
number ratio of surface to volume atoms ¼ (4R2/r2)/(R3/r3) ¼ 4r/R
(a) R ¼ 10 nm; r ¼ 0
...
Number ratio of surface to volume
atoms ¼ 4r/R ¼ (4x0
...
12
(b) R ¼ 1 mm ¼ 106 nm; r ¼ 0
...

Number ratio of surface to volume atoms ¼ 4r/R ¼ 4x0
...
2x10—6
To be inserted in the in the “Answers to numerical Exercises"
...
(a) Number ratio of surface to volume atoms ¼ 0
...
2x10-6
18-7
...
23 x 10—16 cm3
With density of particle ¼ 1g/cm3, mass per particle ¼ 5
...
23 x 10—16 g/particle) ¼ 2
...
of particles entering the lung one breath ¼ (2
...
15x107

Page 43 of 43



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