Chapter 9: Fluid Dynamics
Fluid Flow
Density - How much mass is packed
into a given volume of space
Steady Flow - Velocity at any point
is constant as time changes
*
·
=
=
1
Specific Gravity
a
-
9. A, V
,
PrArV
=
Incompressible Fluid Continuity
↳
Pressure - Force applied perpendicular
to the surface of an object per unit area
Unsteady Flow - Velocity at any
point changes as time changes
Compressible Fluid Continuity
Li
=
material
*
A, V
Ar Ve
=
.
Volume Flow Rate - Volume of fluid that is passing
through a given cross sectional area per unit time
&
·
===
p
·
latm
·
=
# At
1
.
P
>
013x105 Pa
sea
>
>
=
Ay
Mass Flow Rate: m
eQ
=
Bernoulli’s Equation
level
Derivation :
·
Q
·
P
-
.
P
===
·
↳ m= f Y
Pr = (v?
P + & f vi + egh
L.
,
,
v
-
) fg(yz y
+
-
,
,
)
Az
P + Efv + fgh
=
>
-
Velocity Heda
A
-
=
Gauge Pressure
P-Patm
=
&
Absolute Pressure
egh p
=
Patm + egh
Case 1: Horizontal
If :
I
y1 Y2
=
Archimedes' Principle - A body immersed
in a fluid exerts an upward force equal to the
weight of the fluid displaced by the body
·
then : P. + zfV +
+gy P +Vi +
=
=
P + V?
,
=
YP A
Vi
W
>
-
V2
PrAr
,
Pr =fV?
+
L
Y 0
=
=
Wobject air Wobject flvid" Wdisplaced
,
,
L Buoyant Force
·
>
-
For a completely immersed fluid,
Fir
·
Y2
Vi
Y1
Pl-Pr egh
↳
V2
↓ Elevation Head
sure Head
Pressure in a Fluid
*
Hewitt - VT Ph
&
Fis
FB
=
=
fluid
Case 2: Speed of Efflux
If
:
=
=
(PBot-Pror)A
y,
+V+
Vi = 0
g
↑
P , Pe Patm,
=
h ; ye
=
0
,
Patm
III
O
then :
&
V =?
O
M++ egy= M Er +
ye
+
&gy = Av
,
=
Ve
=
Ugh
Case 3: Hydrostatic Pressure
If
:
Vi
then :
=
Patm
Ve 8
=
O
P+ egy = P + g
+
=
Pr P
-
,
=
Ogh
↳
PWATER P3
Chapter 10: Oscillations
A
it
Oscillation - Back and forth motion between two positions
*
&
Periodic Motion - Repeats in a regular cycle
O
time
Simple Harmonic Motion - The force acting on an object is:
Let
Proportional to the displacement from an equilibrium position
LE
Always directed towards that equilibrium
-TY
*
A
=
Simple Harmonic Motion
*
Hooke’s Law - An ideal spring obeys this
Fx
=
-
kX
E
,
-
o
runn
M
Y2
= (A)
·
=
Es
=
ERA
w =
=
Ez Es :
=
↳
EmV? +x* RA
?
=
VA
Va
=
WFFX
Since w
X=
X
v=
E
A cost
Y
Xa =
=
Acos(wt)
Adw
Awsin (wt)
as
Acost
X=
AsinG
v=
Acost
a=
-Awcos(wt)
a
- w2X
-
-
Frequency - Number of cycles per
unit time
Angular Frequency - 2π times the
frequency
w=
--V Rw
Aw
Period - Time for one cycle
f=
?
w
RA?
Vm
0
PE
=
Vm
=
M
E = Ez :
- MVm
EmVmax
+
v
was
ret
↳
=
=
=
Fr = Mr 1
| I
·
f
KI
in equilibrium
ummm
T
Amplitude - Maximum magnitude
of displacement from equilibrium
-
=
-
>
constant
Atan Rw2
=
Lif
Chapter 11: Mechanical Waves
Wave - A disturbance that transfers energy through matter or space
with little or no associated mass transport
*
Mechanical Waves
↳
Lat
&
Transverse Wave - Particle displacement is perpendicular to
direction of the wave propagation
Longitudinal Wave - Particle displacement is parallel to
direction of wave propagation
Periodic Wave - Each particle of the medium undergoes periodic motion
(Speed of Propagation)
Xf
v =
Li Sinusoidal Wave Function:
y(x t) Acos(kx wt)
Right
w=
y(x t) Acos(kx wt)
Left
k
=
-
,
+
=
,
rk
=
Lif
=
From
y(x +) Acos(0 0)
+
=
,
-
Initial Angle Pos
Phase
↳
Angle
Speed of a Transverse Wave:
v
F (Speed of Propagation)
=
T
-
Tension
M- Linear Mass
M
Density
=
v = Aw sin
For Umax
*
.
(Kx-wt) (Speed of Particle)
=
An
Standing Wave
y, Acos(kx wt)
=
yz
=
-
Acos(kx wt)
+
&
y+=
2 Asinkx sin cut
*
*
*
*
*
I
I
↳
L
2A Asu
=
·
fn =
·
Nodes
Antinodes
=
y(x +) 0
-
+
,
-
y
f , (Fundamental Mode of Vibration)
f = (2nd Harmonic ; Ist Overtone
(x +) = A
,
↳ Rarefaction
Chapter 12: Temperature and Heat
&
Temperature - Measurement of hotness or coldness
Ly
Scales:
Celsius
Fahrenheit
&
&
·
n Thermal
Tc = (TF
Tr
-Tc
:
-
32 %
·
320
+
=
Tn Tc
+
27 3 15
Th TF
+
67
·
=
&
cloAT
=
.
1-
Al
=
AA
=
3 xVoAT
AV
=
AV
BVoLT
B
=
=
=
=
T-Kelvin
35
e-Emissivity (0
Specific Heat Capacity
4190J/kg Co
.
-
AQ-Heat Absorbed
AQ-Heat Released
Ci
=
2000
T/ug co
.
Calorimetry and Phase Change
-
Qost
=
+
Quained
Phase Change
Solid-Liquid
AQ
#Q
=
=
+
mLf
= e=
1)
o Steffan-Boltzmann
mCAT
Cu
=
Vy-A
+
=
= AegTY
A-Surface Area
Quanitfying Heat (Q)
#Q
Radiation - Transfer of heat by EM waves
H
Af A
Volume Expansion
Li
=
LfL
Lu
24AoAT
=
Length
Convection - Transfer of heat by molecular motion
H
Coefficient For Linear Expansion
-
Area Expansion
AA
&
459
Conductivity
A-Cross Section
.
Let
Linear Expansion
AL
↳
Conduction - Bodies in contact
H = = AII
Thermal Expansion
↳
*
⑤
Kelvin
Rankine
Conversion:
·
&
Mechanism of Heat Transfer
La
S
↳
&
Lf
latent Heat of Fusion
=
m Lf( He
,
AQ -mLf
=
334000J
Heat needed to
change
=
the
↳, s
=
phase oflkg He0 at O%
2256000J
-
5 67x10-8m rY
.
.
Constant
Summary
:
P=
9:
II :
⑧
S
y
AP ggh
=
=
A , V ArVe
Q Av
M
=
P + Ev eghe
=
=
-
l
&
kX
f #
=
w
2πf
=
X=
=
Tn
=
Th
=
Emr2 + Exx
=
sinkx sin a
Tc = (TF
O
w= E
E
4
12 :
=
w
=
YA
+
18 :
Fx
VK
=
=
v
=
,
-
=
Fi Of Vf9
P, + &v+ egh
Acos(kX wt)
=
w
⑨
Xf
v=
Al =
Acosut
-
32)
-
@
In Calc
Tc 273 15
+
TF
+
.
459 67
.
aloAT
A= 2 & AoAT
*
AV
AQ
=
:
3 &VoAT
mCAT
AQ mLf
=
Optimized
&
p =
Ap
=
egh
Q Av
= -
W=
&
kX
KAT
H
=
Acol
-
4
"
"
E
=
v = xf
⑨
2f
+
X = Acost
=
.
130
y Acos(kx-wt)
w = VK
* Q =
mCAT
=
=
AQ
mLf
K
Emu ERX2
Tn Tc + 273
# L = lo AT
=
=
w=
g
=
=
F
f=
FB =+ V+
Igh
=
:
f=
P + &Pr +
It
M
=
y 2Asinkx sinut
fn =
t
=
l =
=
2
H
=
=
RA
-
H Acol
=
4
f=
Fx
=
f:
P
Ap
=
egh
-
=
1 if
w =
FB = Vg
w
kX
v=
Xf
y
Acos(kx wt)
=
N =
P + &mu" + Ogh
E
E
=
X=
Emi Ex
+
As Nt
*L
=
a loAT
AQ = mCNT
K
=
AQ = mLf
v
=
H
=
H=
E
:
.
-
w= VK
=
Q Av
Tr = Tc + 273 15
M
y
=
AT
AcOTY
2 sinkx sincut
=
fu =
L
f=
F
Ex
/g4
P
=
Ap=
T
Q
=
=
xX
v
f
=
y
w
=
-
=
=
w =
Av
Pitmu egh
E = EmrtErx
+
y
:
=
Ta Ta + 273 13
Xf
=
=
.
Acos((x wt)
-
A cost
# L = C lo AT
AQ MCAT
W = VK
1f
n
9 Vg
E
=
:
R
=
*Q
v
=
AH :
KAT
=
#H
Aeol
M
y
:
mIf
=
:
-
4
2 Asinkxsinnt
f
2
f=
F - kX
EA
f=
egh
w =
P
:
AP
=
=
1f
Fi = Vg
:
Q
w
P
=
①
Av
, + zmr2
+
egh
v =
Y
:
w
=
K =
v
E Emutex
M=
y Acosut
y
-
:
Tn = Tc + 173 13
Acos(kx - wt)
AL
VK
AQ :
2Tt
T
:
E
=
Xf
M
=
T
2Asinkx sin at
fr
*
2
.
=
AQ
=
@loAT
m
[AT
mLf
AH-RA
AH =
ALoT
"
Ch9
17
:
r= 1
.
n 2
S
P
4m
P
=
P
=
=
Patm
D
.
181 300 Pa
=
719 7
=
.
Pn
RgIm3
+
334(π(1 834)
.
.
# 1203487
783N
=
2m
=
it
.
1000
N 0 3 m
=
.
600
no 0 9 m
=
.
. G oil
5
.
a)
0 60
=
.
b) P =
(2 3)111492
=
F+
=
.
KgIm3
Po-Patm
=
kgIm3
Poil
600 (9 8) (0 9)
=
.
.
Po-Parm 5292 Pa
=
FI
Pw -Po
668952N
Pw -(5292
1000 (9
=
+
81(0 3)
.
.
1013007
=
4908
Pw 111492Pa
=
3)
V
2200m3
=
P
m 900N
=
mo
=
mp
=
1 23
.
170ON
FB
3200N
FB
=
=
=
1
.
kg/m3
23(2200)(9 8)
.
26518 8
.
m FB=50N
8 8
.
a
ma
=
2114
·
Kong
Ple
9
4)
=
Win
0
=
963kg/m3
.
63
.
x
10
m
Q
=
Av
Q = (6 5 x 10 3) (1 27
-
Hales 12
v = 1
.
.
=
0 60
3m
:
.
118227 334 Pa
.
=
A
.
.
16927 344
181300pa +
F 118227
2)
719 7 (9 8) (2 4)
P=
?
=
101300pa
=
2 mis
.
.
Q
=
0
AiVi
.
=
V
=
0001593 mys
AzVz (12)
19 97 mis
.
.
an 10
1)
=
Y
0
++
12 m
.
1
L
Om Is
V
, =
at t
0 8
=
=
y
.
,
0
.
a)
v=
f 0 625Hz
v=
.
1
m=
x(t)
a) T
0
=
074as[1
.
- =
2f =
=
w=
b) 4 162 (1 3)
.
.
k
c) V
V
=
V
=
2
.
42)
=
.
528m/s
b)
4 times
+
a) (Aw)" 2(x)
1 51s
.
12
At = 2x?
34
=
X
k
96
dat =
x
EwA2 EMv #XEmwAt
v + (Aw)
(A)2
V = wX
V
=
=
0 308 m/s
.
5)
Fmax RA
(25 96)(0 074)
=
a=
KE
PF
GRE
FPE
.
An sin (wt
-
b)
-
-
-
o)
X= 0
.
V= 0
.
a = (0
0742s(4 16
-
.
2
-
074(4 16) sin (4 16-2
.
.
.
074)(4 16720s (4 16
.
-
.
.
2
42)
.
a=
0
0
383m/s
.
216 m/s2
.
=
F= 1
F
0
=
.
600mm
=
ma
.
(0 216)
3
0
323N
.
V
=
An
w
V
=
0
.
=
2πf
0006 (2 (3921)
V 1 48 m/s
=
.
=
V (m/s
=
=
.
=
E(20) (1)
105
b) at
m
=
=
0
000027ug
.
=
t (0 00027)(1 48)2
.
.
RE
=
2 96
.
5
x
10 J
=
T
=
0 373Hz
.
f
2
b)T
EKA
?
=A
0270g
.
0
f
A = 0 426m
.
.
68S
2
105
=
.
T=
2 35 rads
=
=
RE = E mu + Ex
9) Va ?
3f
↳
RE
.
0006m
.
10(2) 10(0) v(20)
W
f = 392Hz
A = 0
w = 2 if
=
+
w
F
110
v(m +m)
+ mu
42)
.
-
10g
v= 0
2 m/s
a) mu
42)
.
0 0125m
=
v =
m
=
R=
2Awcos(wt
-
long
.
-
-
+
1 92N
Acos(w+ b)
x =
m =
v =
.
=
v=
+
=
.
.
F
3)
.
440(2)(0 6)
[m(Aw)" em(wx) Em(wX)
=
X
L = 0 45m
=
=
=
587
587He
0 074(4 16)
Fmax
e)
.
-
Aw
=
m
d)
23
=
=
+
for
E = Emu2 + ERX
4)
5kg
.
0024g
:
f
6 s
=
2)
?
=
x
.
.
.
=
0 8s
T= 1
0
=
f 448Hz
12 m
.
=
Sm
.
2g
m =
A = 0 12 m
T
0
=
=
1 34s
.
Ch1l
1) T 2 3
=
X
.
=
6m
3)
L = 1
2H = 0 62m
v
a) v
9) X
.
?
=
v
6( )
fin
1 2 m/s
f
=
.
4)
=
.
# 0 31 m
=
c) If
2A = 0
v
=
0 3m
T
.
=
m
0 0125N
w
T =
X
.
=
T
4
=
.
0
=
t=
00
=
2 Au
0
V
1 96 m/s
AN
=
t
=
0
5
.
m
2nd Harmonic
c)
v =
e)v
Xf
v= 0
.
V= 5
.
0534s
n
X 0 03% 3 m
=
.
+(0 5)
.
x = 0 0123m
.
.
.
=
3
3(10 42)
.
21 m/s
.
=
m=
5)
12 57
2Awisksiwt
=
67 N
.
x=
m/s
V=
.
T
V
10 42H
=
b)
(28 08)2 (a)
=
Pos2 :
=
W + 0 0125N
k=
.
V =2Awsinkx sin wt
T = 0 096s
X
.
=
:
v=8
f
8 5 mm
Xy 0 75m
=
.
00
.
1
R=
5 m
k
63 478d
PosI
1-3
=
.
.
=
W = Lif
=
.
Xand
:
d)
X5 Im
50
:
a) T ?
A = 0 13
.
Flash
48Hz
=
fu 64Hz
w =
2m/s
=
3m
=
a)
1N
1
2A
1
=
Ymax
.
X
.
xf
v =
=
3m
.
F
16 Hz
=
.
X=
1 = 0 3m
T
.
fy
and fr
,
xf
b) 2A 0 62m
A
f =
48m/s
=
v =
v=
2)
W
b) X , and fe
Sm
.
At
b) zi
fe
=
A
0
c)a
V = Aw
sink* sinut
Vmax
Awsinkx
-
004m
.
= (t))
=
.
240Hz
=
.
T= 0 00208s
v = 192m/s
f
8184kg
0
=
i) O mis
a) Find Antinodes
4180 it rad/s
=
i)
=
y
ii) y
.
Livf
N=
y
0
=
Asinkx
0 004 sin (2 3· 0 4)
.
=
.
.
=
0 004 sin (2 5 +. 0
.
.
iii) y 0 004 sin (2 3r
=
.
.
.
0
.
.
2)
1)
=
=
Om
0
0
-
Amax
Awutsinkx
Ausinkx
=
O m/s
0 Um6.003s
=
=
.
.
004 m
00283m
9095 83 m/s
?
.
6431
.
72m/s
Ch
12
17 s 2
:
.
B: 2
.
in
3
?
Al
=
LdAT
=
2 3
-
.
.
.
T=
=
+
0
Tf
(1 2 x 103) (T+
.
-
20)
=
446kg
.
0
%
=
0 C
095kg
.
035
.
Ti
kg (ms) and latm
?
=
Qu + MLfi + Qu
= -
Q
. 446(390)(Tf) + 0 093(334000 +
0
86 67 %
f
Ls
=
mi
a)
=
2 502 2 3
m 0
:
at 20 %
302 in
a) Ty
4) Cu
20 %
at
.
.
0 033(2256000)
=
Li
-
.
Los + Los & sAT
Lob
=
+
17 3
LoXBAT
↓
STOBIT
T
b)
50
- 5(1 2x153)
2
.
.
#99.8
=
.
398 0357
.
=
.
Ong
=
0 035 + 0
.
ma
ms
31730
.
+
=
=
0
.
.
095
130ng
Oug
80
Tf
2)
M:
V
& T 55 %
=
F: V =
@T
10 m3
=
=>
.
0%
0
=
=
=
10 (18x10 · 3) (33)
0
.
for
=
1
7 x 10 34
:
.
25kg &75%
40 %
(2000) (20) mi(334000)
+
=
900 0
,
mi
+
mi
10(B)(55)
&-20%
mi =?
mi
-
--
B
=
0895m3
AVHg-AVglass
⑳0893-18x10
Mw
.
%
=
. 0895m3
8
3)
-79 8 %
@T 0 C
10m3
Overflow
V
=
=
=
0 073 kg
.
334000+ 16
60
+
mi
(4190)(40)
=
-0
.
.
095(4190)(Ty)
0 035(4190)(T+
86 13%
mi =
Mw
945f
+
0
25(4190) (40-73)
-
100)
778960
-
146 63 14665]
.
Summary
Fluid Mechanics
Oscillations
o
f =
Pr-P
FB
=
=
L
,
En Emu + Ekx2
=
Vf9
,
m
zovi fgh
+
?
AzVe
=
=
E
Pz zv eghe
+
+
w
G
w
PF
w
=
Fi (P2
=
FB
&
=
=
Ogh
Av
=
,
Az
P, & fvi fgh
+
+
3
+
=
=
cos(wt)
=
a=
a=
=
-
w2X
u
=-
1=
E
E = mu +
fVg
ExF
w
=
X=
Acos(wt)
"
KE + DE
④
⑧
P=
P
E
=
P2-p
Fi
FB
=
=
f=
,
w
-
,
,
,
=
=
E
Emv2 + x
w=
k=
OVg
Q A V
+
w = 2πf
(Pe P )A
=
P,
Ogh
=
A2Ve
Ev egh
+
E =
Au
v
Lif
k=
P, )A
-
=
=
w
:
X= A
P=
.
Es : Um
=
f=
-
=
Er Es
7
O
Pr P
In
=
0Q
=
,
Ez EKA
Optimized
1
E = - Mr
.
Q A Vi
+
F
-
If
=
k= #
=
,
2if
=
w
Ogh
(Pz P ) A
=
P
w
+
X=
?
Acos(wt)
=
k
F
=
- KX =
mgsint
X
E
general
=
:
Emv2 Ex
+
?
Bing
a
Given : h
a) Pn-Parm
10
=
.
fr
92km
=
1023
egh
=
P (1023 6) (9 8) (10920) + 1
=
.
.
Po
.
64g/m3
45 8 x 10 "Pat
R=
.
013 x 103Pa
.
19
1 096 x 10 Pa
=
AP
fo
=
Pact
=
* P= 1
.
P
-
16 x 100
= (0 2) (0 2) (0 5)
* P = 6 4 10" Pa
x
.
Ap
(1030)(43 8x10")" (6x10 %
=
V
0 01 m3
=
.
m
=
m
=
OV
1000 10 011
.
10
m =
&final
Fi = T +
524kg/m3
.
283 524
+
1830
=
Ifinal
pug
.
.
-83
.524x100
↳
3kg
m =
.
21N
3(9 8)
+
frg
=
.
29 4
=
-
T
SON
Wat
=
30 N
=
?
Fi
=
35N
=
+
T
20 =
20
V
.
=
b) Fi
fgV
1000(9 87(V)
=
2
.
04x18-3m3
15 N
=
=
13N
p (9 8)(2 04 x 18 3)
-
.
m3
#
3 (9
T
=
c) mg
=
Dug
+ T
-
28
.
-
1000(0 00086)(9 8) T
=
.
.
ma
fug
3 (9 8)
3(2 5) 1000 10 00083)(9 87 t
+
-
=
-
.
.
.
.
T
=
13 S
.
9 758
=
9
-
1
15
. 0199
8
5
+
=
2
ma
8) + 3(2 5)
.
.
.
00086
.
=
.
58 - 30
egV
9808
b) mg
10 (9 8)
FB = 28 N
v
=
=
-
.
-4
Wair
Wonk
↳
1000 v(9 8) + 21
=
.
:
a) V
fug
2)
+
-
.
.
.
.
d) mma = Vg
↓ mg
T
.
180 (0 01) (9 87
=
80 36
L
%o
27 3
1) Given
*
mg = vg
10(9 8)
.
T
t
a) V = ?
T
=
mg
mq
180 (0 0179 8 = T +
1313 524 kg/m3
=
T+
=
C%
=
kg
.
Al = 283
.
.
.
al
.
180 R9m 3
=
V = E Bhd
VT + mg
1 096x 108
-
:
Fo
AI
+
97) Given to
↑
.
b) Pfinal
.
.
3
kg/m3
=
&=
4x10-3
=
2500kg/m3
2) Given
4)
:
I
A, V
n = 1
Pw-Po
=
+
+
+
+t
+
eg(z)
2
=
.
.
=
Sigh
SG = 1
2
Pe
=
+
Vil
P, =
P2
+
+ (1000) (9 92
P De
+
-
.
-
2 475 %
917kg/m3
t
=
0 24s
.
.
=
.
6) Given
:
1 = 0 2m
V=0
.
.
.
m
008 m3
GKg
a) E +gv
=
0
+B
.
78 4 N
=
.
b) 78 4 Fjet W
=
+
.
Fet
Fret
f
# set
0
=
.
=
.
1079 33
.
=
=
Kgm
x
91 79E
=
100-91 7
.
3%
#
.
P
n
=
+
6(9 8)
.
-5 8 8 N + 78 4
.
.
=
19 GN
.
==
c)fgh
.
.
.
.
FB
00218 m3
1000(9 8)(0 008)
=
f = 85)
+ 0080 85)
5
.G
1 5
y = (9 8)(0 24)2
m =
917
.
26 m/s
.
0
V=2
S G
=
.
28
↳
24g
a) f 1
=
(9
8)(2)
.
.
=
V
Vi ?
;
t=
45637 1875 Pa
%0 = 85%
=
4
=
b) X
EgV
= (vi
P=
.
.
+
m =
-
n = 2m
V= 6
Pr
:
800 (9 8) (1 3) + 1
1
v
P1 =
=
+
=
g
9 9m/s
t fV
+
+
a)
475m/s
.
2m
&wghw + Pogho Parm
1000 (9 8) (2)
.
P,
b)
=
9 9
=
V
,
3) Given
Pw
5) Given :
Ve =
g
=
=
Igh
↳
+
=
b)4V
=
=
=
+
(5) Ve
Vz
Pw
b)v
=V
=
n
3m
132660Pa
#
+
=
=
.
1000 ugm3
=
+Xi egn Pe V ent
# 0 f9(5) 1 + Vi 0
( +g(z) ( tfV
P,
=
a) Po-Patm Ogh
Veo
=
.
Water
=
V
4V
Oil
8 800kg/n3 P
Ar Va
=
,
Given :
=
Pret
=
09 80
.
.
05m
.
.
.
013x103Pa
ving 10
&
:
Wair
14 18) Given
.
=
a
0 5
m=
:
.
Wyl
Reg
(3 6ams] sin [(4 71s"It
-
-
-
.
E]
FB
T=
f
.
-
-
=
(4 71)2(0 0076m)
a=
.
.
169m/s
-0
P
.
=
P
4x10"
,
05(4 7172
.
#
.
09
13) f
, + PV
2
=
=
X
,
a
=
w
=
w
? at t
=
=
=
1
.
I
013x105
+
egh
=
Pe
1 1 c
P
13CM/s
Pe
.
-
=
09h
+
.
107800
=
(11)
+
135300
+
155300 + 1 013x10"
.
P-1)5 4x10"Pa) Patm]
2f
P
2π(2 3)
P2-Pain
St
a=
171 4142
117
.
.
m =
k
at
t
a)A ?
=muc
+
.
(5]2 (1 1 cm)
2g
=
=
300m
0
v=
,
-12 OmIs
.
b) + Mu2 ERX" * A
+
=
=
EKA
=
?
96m
#
0
.
h
155300
1000 (9 8)
=
P2
0
a=
>3
zev"tegh
+
w"X
=
-
-
.
W
+
5 H2
.
v =
a)a
.
-
.
133300 Pa
=
P
=
at t 0
5
=
P
k=
5 4 x 10"Pa
, -Parm
muz
.
.
↳
76m
#
wiX
14
.
4 2
.
b)
.
(1000) (v) (9 8)
0 000423m3
#
.
14
=
800
334s
[1
=
.
V=
T
a=
8
.
4 2
. 2
t
d)k
10
=
=
4
a) T = ?
9)
13N
=
(12)
2
+
X =
=
150x2 = 130 (0 96)
.
0 196m
.
. 196
0
0
=
=
Acos(*)
1 365
.
2
-
Parm
=
107800
=
+
135300
263100 Pa
L
=
0
.
A
5m
=
0 0015
.
m
F= 1N
Flash
=500
=
alt 5 x2 ?
-T
,
=
,
b) 2nd Harmonic
4x mindvV
Xf
=
0
=
.
3 (10 42)
.
#min
21 m/s
109bs
faus
42 He
#
.
L
=
n
↓
) V
=
-2 Awsin Kxcos we
#
=
w
=
R
0 015 m
.
=
X
(10 42)
1π
·
=
4
= 0 125
=
.
m
t= 0 te =
;
=
0 024s
.
#letro
statiFor
it
EX I
.
.
·
-
-Ion
a Gmmq
S
=
F2
=
Fi
X=
20mm
*
b
=
&
4 mm
=
44[
[Fx Fix
+
=
Ge = + In C
·
=
=
-
F2x
F Cost , + Fr cost2
,
-8
25x10")
.
+
x
4
.
33
#
x
-
d
2
#
Y &
L
X
dF
-
↑LinearChangeMeni,a dy
=
=
/dEx JdFcos
=
↑
JdFx)
=
g
F
dr=
[F Ex
Ex
=
Garss' Law
Field
-lectric
m
M
M
Fgrav =
earth
g
M
=
=
my
=
g
=
due
g acceleration
=
to
gravity
gravitational field
9 Olson
.
E
-
>
Q
--
Fe
?
M
>
-
-
Electric Potential
V
Ex What
electrical field at point P
is the
figure
Let
qu
E
F
=
M
Fo
**
qE 2C(9x1097
=
-
x10"N
=
1 4
.
V
4
=
.
-
*
FQIE
Ed
d
↑
gd
E,
=
[Ex
=
E
E
=
,
+
0 7
+
.
,
Es
Unet
-0.
=
5NC
-
=
V
,
at P!
V2
+
-
Electric Potential
Energy
E , Fe
t
=
=
·
7
W
E, -Essin 430
qz
3x10
=
[Ey
=
.
(10)
T
#
0
7
2NC
=
9) Total Electric Potential
&
+
com
*
El
cos0 Es cos 450
P
T
*
WEz
E
(202)
4x18 3 m
sum
Y
E,
&
+
2 iC
=
5x10t
,
g
d
V
#50
Ex 1.
o
V
9x100
v:
=
=
=
u
Q 20 N C
b)
the
5mmE
t
9)
in
=
=
M
Mz
=
=
-
=
kQg[-]
S#
Let
.
Fd
SFedr
KQq
M
Ve =
q
Wel
Ex
23 19
.
B
.
0
08r
. 06m
0
Find
work done in
.
A
q
*
<
· Osm
q
.
q
=
=
2
-
.
4nC
6
.
5n[
G2
0 03
.
mi
W
=
-
moving
[Un-Ui]
Q
=
2 5
.
C
from B to A
Unq
)(24))
#845x1
ge
M
Mr
,
(2)
-
76x105/
#-1
.
WBA
=
-
T
-
1
845x10 1-1 76x10-9]
-
.
t
8 3 x 10-85
.
.
Ara
Chapter 14: Electricity
+
e-resistivity
-
I
R- Resistance
V Potential
-
i
+
-
L
R
Difference
=
-
I
p
Voltage
-
Current
p-power
&
crits t
R
·
·
-
Current
I
=
Flow of
I
i = = - A-Amplitude
t
V
#
iR
=
ilI
S V
:
R
Series
Mi
Re
R3
I
=
in
iz
i3
V
=
ill
=
=
V , + Ve + Vz
[Req i B
Reg
=
+
icRz + 13 Bs
IR ,
+
[Rz + I Be
R
R2 + R3 +...
.
,
,
+
Ex
R , 100 -
↓
I
=
V
=
v , + 12 + 13
Vi Ve V
=
=
↑=
I
=
300-
Re
=
2001
a) Reg =?
I
R2
R3
=
=
1
E
100 &
V 100 v
I,
111
10V
=
-
Resistance in Parallel
R3 is
temp
.
I
*
constant
↓
=
in
=
=
ii = in 23
=
Ri
Ohm's law
↳ Constant with
-
Resistance in
Charge
time
1
!
=
i , + iz + 13
=
= N+
Reg
=
=
b) I
=
100 1 + 2001 + 3001
600s
?
I
= Fe =
I
=
0 17 A
.
c) Vz =?
Vi IM
=
V3 izEg (ii)
=
V3
V3
=
=
0
.
17 A(3002)
50v -
,
V , = 17 A
Ve
=
IR2
V=
=
34 A
↓
V
=
101V
Ex 2
.
Combination of Resistors
R
=
1001
B3
R2
=
111I
=
2002
300m
By 300
=
E 100V
=
Reg =?
Mic
=
R , + Re
=
300 -
Riz
↳ 1+ 1
Vy = ?
E3-
Ris Bie R3
V
Ris
DesTo
=
-
-
Bu
=
150 -
Vi
Reg Rizz 14
Reg 15 01 3001
Vz
Reg
iz
Rizz
=
+
=
=
I
I
=
+
4501
=
=
0 22A
.
[Ry
=
=
0
.
22A(3002)
66V
=
100 v
=
34V
-
66V
3
↳
=
mz =
in I-iz =
=
i
iz
=
=
-
Vi i R ,
=
,
Ve in Re
=
Kirchoffs Current Law
KcL> Junction Rule
->
[1
:
0
Sign Conventionin
m
V2
E 30V
=
is
2
=
+
0 1A
,
LoopI
.
[V
KVL :
To
>
=
+
E
loop
=+
t
" loop E
=
-
-
·
V
=
RCL
M
RVL
V
111l
-
.
-
loop
t
-
,
0
V=
=
.
3
1
0
loop
-
4
=
i
2
.
zor
=
Kirchoff’s Voltage Law
-
!
:
.
Loop Vivo
>
1
0 033A + 0 066A
30V-10V - 201
+
.
-
Ex
ic
i
-
.
[ =0
,
I
.
=
=
+
1A 2A
-
-
I1
-1A-Towards the junction
I = 1A
IR
-
Loop1
[V
E
Loop I
[v
=
,
=
Er
(1) (32)
0= +
=
0
=
-
El -1A( > a) + 201
18V
=
-
7V
2(3)
-
Ez
+
E
-
,
1(5)
Chapter 13: Electrostatics
Electric Charge - Attribute, positive and negative
Ag
Ly
3
Negatively Charged Electron - 9 10 x 10 - kg
.
Lt
Positively Charged Proton -
1 67 , 10. Rg
L
Uncharged Neutron -
1 67 x10
.
.
Rg
Ionization
*
&
L
Positive Ion - Electron removed
↳
Negative Ion - Electron gained
Principles
LX
Conservation of Charge - Sum of all electric charges in any closed system is constant
↳
Quantization of Charge - Magnitude of charge of electron or proton is a natural unit of charge
9
Conductor - Material that permit electric charges to move easily
*
Insulator - Material that does not permit electric charge to move easily
↳
&
Semi-Conductor - Intermediate in both the properties
Induction - Charged object is brought near a neutral conducting object, its presence will force electrons within the
conductor to move
&
Coloumb’s Law - Point charges are charged bodies that are very small in comparison with the distance between
them
FEk1q)
1
F
or
=
qoE (N)
M2
t
*
7
+
-
n=
9x109 Name
>
-
k=
(Electric Force Constant)
I
Note :
450
Eo
=
8
.
854x10"
vectors X and y
+
Electric Field - Force per unit charge exerted on a test charge at any point (Test charge is small enough that it
doesn’t disturb the charges)
E
&
==
Electric Potential - Electric potential energy per unit charge
V=
&
Force needs
(v)
Electric Potential Energy - For work
'Wie-(U2- Ul
U
:
K
T
W
=
1
.
W K
N
5
f=
(Uz - U )
=
=
R
=
V
=
1R
Chapter 14: Electricity
Current - Time rate of flow of charge through a cross-section area per unit time
b
1 =
&
5
=
Resistivity - Ratio of the magnitudes of the electric field and currents density
f
=
E
=
rm
Reciprocal
&
,
o=
=
+
f
=
y
=
z
Ohm’s Law
V
E
*
is conductivity
Resistance - Ratio between voltage applied across a piece of material to the current
R
*
(Ampera)
A
=
=
=
IR
(5-Potential Difference)
IM
Kirchoff’s Rules
LT
Junction Rule - Sum of currents in any junction is zero
[1
LE
=
0
Loop Rule - Sum of potential differences in any loop is zero
Iv
=
0
Voltage
A= B + C
C
A
&
v
7
B
Chapter 15: Magnetic Field and Forces
7
.
=
B
-GE
=
1
- Right Hand Rul
>
Lt
Ex
#xB ABsino
B
<A
=
30m
20 NW
%
;
* B (20)(30s in 160° T
=
3
28m : E
=
203
.
Moving charge force
2m2T
magnetic field
=
B
Tesla
=
Motionof Charga
T
-
↑
EM force
a
N
7
Moving
Static
1)
thereisa field
If
charge
If charge is moving
1)
E field will exert
2) E force on
·
Electromagnetics
Electrostation
,
musmagnetic
2)
any charge
G
a
=
Fi
Or
DE
10-V
2) I Fe
①
F2
=
.
B
=
①
yFB
0
.
1T w
FB
,
g
=
v =
2
=
:
3x1065 ; N
=
gr JBs in fr
=
[2x10"] [3 10357 To 1T]
+
0 64
.
-
sin
; UP
The magnetic field is B= 0 ITI
.
-
A charge of q= 2MC is
-
15
>
moving at 3x180 . Find FB
-
·
FB qV x B
=
Fis
=
Fi
=
=
qVBsi- 90 on
(-2Mc) (3x10") (0 1) sin 90T
·
-0 3 N T
.
↳
Index
=
v
Middle
=
B
thumb = FB
=
guBsin
.
BR
mUR
=
[
90[p)
Thompson Apparats
mr
R
Fi =?
Ex
=
g
FB gixB
- - w
=
megB
Moq
①B
B
=
2)
gE
grBi
F ma
=
=
= 2
[]
=
15 :
Chapter
-
20
Magnetic Field and Magnetic Forces
B
#xB
a
=
ABsinG
? Defined
Ex 1
By Right-Hand Rule
.
#
B
=
:
20m
:
E
30m : 20 NW
JAxB
=
=
20 (30) sin 160
B
=
(Away +)
,
205 2n2 *
N
.
wi
B
X
S
N
"
N
F
E
S
S
Magnetic Field (T)
Electrostatics
1) If there is
1) If the
charge
.
there is electric field
2) Electric field will exert
an
electric force on
in
charge
1)
the vicinity
charge is moving,
magnetic field
there is
2) Magnetic field will exert
magneticTo re
any
on
any
moving charge in the vicinity
,lisual
Right-Hand Rule
1)
E
--
Index
+
Middle
2)
I
2)
=
gE
X
X
X
%
>W
t
-
Fe
1FB
Fe
X
X
FB
=
=
qv x
qB sin Di
=
v
=
B
Thumb = FB
Ex 1
-
B
0 1 T; W
=
FB
.
g
=
v
=
FB
2MC
qvBsin Gr
=
(2 10 "c)(3x10m) (0 /T) sin90 (Up)
°
-
3x10 ; N
=
=
=
x
O
.
.
GN ; up
?
Ex 3
.
~
The
Y
magnetic field is B 0 1TT
:
A
.
charge q -242 is moving at
3 x 10s Find FB
=
>
.
FB qv- Bsin Or
=
( - 2m2)(3x10= )(0 15) singoy
=
Th
.
FB
=
-
0
.
6Nj
Page 201#
q
electron
=
B
=
B i + 3
,
r =
.
=
-1
.
6 x10
0Bx]
(2 0m/s)i (4 0 m/s)i
+
.
.
F (6 4 x10- 19N)k
1
=
.
F
=
Y
i
Ve
Vy
Vz
By
By
By
Y
Magnetic Flux
F
qE gVxB
+
=
Motion of a
0B
=
=
=
Charge in
magnetic field
a
Magnetic Flux
B A BA cost
=
.
SB . dA
B
:
A
=
&
Magnetic Flux
Area
Angle Between B and i
=
Unit Vector
=
Magnetic Force
Current
on a
Fi
1
X
Carrying
to the surface area
Wire
X
O
(
> v
t
I
1
h
X
I
I
X
FB gv Bo
E
=
=
l
=
g EB
=
IlB
=
IlBsint
FB
S
M
↑
-
OF
I
ord
Intothe
;
=
@
=
Angle between I and B
FM
:
[A
=
I ABsinG
B
+
=
T
=
(90)
MB(IA M)
=
I
B
IN
Ex
1) 27 39
.
X
X
X
1
Y
O
F = IlB
X
30 an
I= 4 5 A
.
>
Y
-
60cm
X
C
YX
X
60 cm
X
X
B
X
=
0
.
Y
60cm :
Foo
X
245
=
(4 5 A)(0 6n) (0 245)
=
0
#30
=
F
=
.
.
.
6484(Vp)
.
(4 5 (0 3m) (0 247)
.
.
.
=
0
.
324N
(0 648)
.
=
0 724H
.
(Left)
"
+
10 324)
.
O
:
tan"
=
63
0 648
0 324
.
.
.
438
Workbook
:
Tik
6
.
4
+
x 10
9N
q
:
By
T
0
T(0 - 0) j(0 0) + y(2 3Bx 4Bx)
=
-
3Bx
i xi
=
4x10" (-1 6
(6Bx
.
Bx
-
-
=
-
4Bx(Y
(2Bxm)
=
.
Y + Y
0
=
6
-
x
2T
10 9) (2Bx]
.
-
Chapter 15: Magnetic Field and Magnetic Forces
Magnetic Field
1. A moving charge of a current creates a magnetic field in the surrounding space
2. Magnetic field exerts a force on any other moving charge or current present in the field
Magnetic Field Strength
↳
B = T (iesta
·
-
=
IG (Gauss)
↳
=
·
10 "T
X
-
-
Out
In
Magnetic Force
·
↳
F=
=
gr xB
quB sing (N)
Magnetic Force + Electric Force (Lorentz Force)
·
F
=
q(E vx)
+
Magnetic Flux
?
N
1@
Note :
(WD
BA cost
=
B
S
=
Tmi)
If
<
t
7
↳
Equal to 0 in
B
=
A
=
O
:
a
closedsurface
Magnetic Field
Area
Angle between B and in
Motion of a Charge in a Magnetic Field
qv- Bi
F=
quB
:
Magnetic Force on a Current Conducting Wire
F=
IT B
x
IlBsinG
=
Force and Torque on a Current Loop
E
:
[ * x
But
T
=
M
mx
=
It so,
then
,
F =0
-w
t
AB
①