SlideShare una empresa de Scribd logo
1 de 18
Descargar para leer sin conexión
Capítulo 2
ESTÁTICA DOS FLUIDOS
A ausência de movimento elimina os efeitos tangenciais e conseqüentemente a presença de
tensões de cisalhamento. A presença exclusiva de efeitos normais faz com que o objetivo
deste capítulo seja o estudo da pressão. Nesse caso são vistas suas propriedades num fluido
em repouso, suas unidades, as escalas para a medida, alguns instrumentos básicos e a equação
manométrica, de grande utilidade. Estuda-se o cálculo da resultante das pressões em
superfícies submersas, o cálculo do empuxo, que também terá utilidade nos problemas do
Capítulo 9, a determinação da estabilidade de flutuantes e o equilíbrio relativo.
É importante ressaltar, em todas as aplicações, que o fluido está em repouso, para que o leitor
não tente aplicar, indevidamente, alguns conceitos deste capítulo em fluidos em movimento.
Para que não haja confusão, quando a pressão é indicada na escala efetiva ou relativa, não se
escreve nada após a unidade, quando a escala for a absoluta, escreve-se (abs) após a unidade.
Exercício 2.1
( )
N13510101035,1G
Pa1035,1
20
5
104,5
A
A
pp
Pa104,5
210
5,21072,21010500
AA
ApAp
p
ApG
ApAp
Pa1072,22000.136hp
ApAApAp
45
55
IV
III
34
5
53
HII
II2I1
3
V4
IV4III3
5
Hg2
II2HII3I1
=×××=
×=××==
×=
−
××−××
=
−
−
=
=
=
×=×=γ=
+−=
−
Exercício 2.2
kN10N000.10
5
25
400
D
D
FF
4
D
F
4
D
F
N400
1,0
2,0
200F
1,0F2,0F
2
2
1
2
2
BO2
2
2
1
BO
BO
BOAO
==⎟
⎠
⎞
⎜
⎝
⎛
×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=⇒
π
=
π
=×=
×=×
Exercício 2.3
mm3681000
000.136
5000.10
h
hh
Hg
OHOHHgHg 22
=×
×
=
γ=γ
Exercício 2.4
)abs(mmHg3400)abs(
cm
kgf
62,4)abs(MPa453,0)abs(
m
kgf
200.46)abs(atm47,4p
mca10atm97,0MPa098,0Pa108,9
cm
kgf
1
m
kgf
000.1074,0600.13hp
mca2,36
000.1
200.36p
h
bar55,398,0
cm
kgf
62,310
m
kgf
200.36p
MPa355,0108,9
m
kgf
200.3666,2600.13hp
mmHg2660
1
5,3760
p
patm5,3
mmHg760atm1
22abs
4
22HgHgatm
O2H
O2H
2
4
2
6
2HgHg
=====
===×≅=≅×=γ=
==
γ
=
=×=×=
=××=×=γ=
=
×
=
→
→
−
−
Exercício 2.5
kPa35,13Pa350.13025,0000.101,0000.136p
01,0025,0p
1
HgOH1 2
==×−×=
=×γ−×γ+
Exercício 2.6
kPa1,132Pa100.1321000.13625,0000.108,0000.8pp
p8,0125,0p
BA
BOHgO2HA
−=−=×−×−×=−
=×γ−×γ+×γ+
Exercício 2.7
kPa6,794,20100p
kPa4,20Pa400.2015,0000.13615,0p
p100p
m
HgA
Am
=−=
==×=×γ=
−=
Exercício 2.8
kPa55,36103,0500.834p
p3,0p)b
)abs(kPa13410034ppp
kPa100Pa000.10074,0000.136hp
kPa34Pa000.348,0500.83,0000.136p
07,03,07,08,0p)a
3
M
MOar
atmarabsar
HgHgatm
ar
O2HHgO2HOar
=××+=
=×γ+
=+=+=
≅≅×=γ=
==×−×=
=×γ−×γ−×γ+×γ+
−
)abs(kPa55,13610055,36ppp atmMabsM =+=+=
Exercício 2.9
( )
( )
)abs(mca12,17
000.10
000.171p
h
)abs(Pa200.171200.95000.76ppp
Pa200.95000.1367,0p
Pa000.76p000.57
4
p
p
000.57pp000.30p000.27p
000.27pppap
000.30pp
p4p4
A
A
A
A
A
A
ApApAApApAp
2
A
A
kPa30pp
OH
absB
OH
atmBB
atm
B
B
B
ABAB
BCBC
AC
AB
H
2
H
1
1
2
HB2AH1B1B2A
1
2
AC
2
2
efabs
==
γ
=
=+=+=
=×=
=→=−
=−→=−−
−=→=γ+
=−
=→==×
=→−−=
=
=−
Exercício 2.10
)abs(kPa991001ppp
kPa1Pa000.12,010500ghp
m
kg
500
2,0
1,0
000.1
h
h
hh0ghp
0ghp
atm0abs0
AA0
3
A
B
BABBAABB0
AA0
=+−=+=
−=−=××−=ρ−=
=×=ρ=ρ⇒ρ=ρ⇒=ρ+
=ρ+
Exercício 2.11
( ) ( )
( ) ( )
3324
3
o
OH
OHo
OHo
cm833.47m107833,41043,0
6
45,0
xA
6
D
V)c
m45,03,05,0
000.8
6,04,0000.10
x5,0
x2y
D
m3,0
2
4,01
2
yy
xyyx2
x2yx5,0D)b
m4,0
000.10
5,0000.8
y
y5,0)a
2
2
2
=×=××+
×π
=+
π
=
=−−
+
=−−
γ
+γ
=
=
−
=
−′
=→′=+
+γ=++γ
=
×
=
×γ=×γ
−−
Exercício 2.12
( )
( ) ( )
m105
5,11sen
5,4
1
000.8
10
sen
D
d
p
L0Lsen
D
d
Lp
D
d
LH
4
D
H
4
d
L
Pa10001,010001,0p
0LsenHp
3
o
22
x
2
x
222
4
O2Hx
x
−
×=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
−
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
α+⎟
⎠
⎞
⎜
⎝
⎛
γ
−
=⇒=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
α+⎟
⎠
⎞
⎜
⎝
⎛
γ+
⎟
⎠
⎞
⎜
⎝
⎛
=⇒
π
=
π
−=−×=−×γ=
=α+γ+
Exercício 2.13
( )
( )
( )
( )
( )
mca7,3
000.10
000.37
p
Pa000.37000.17000.20000.17pp)b
absmmHg831684147p
mmHg147m147,0
000.136
000.20
Pa000.20p
000.17p10331p104:)1(nadoSubstituin
p000.17p
p4,0000.104,0000.5005,0000.102p
m05,0
4,71
7,35
2
4,0
D
d
2
h
h
4
d
2
h
4
D
h
phhh2p
1p10331p104
0357,00714,0
4
p31
4
0714,0
p
dD
4
pF
4
D
p)a
2
12
abs1
1
1
21
21
2221
21
21
21
ar
arar
ar
ar
ar
3
ar
3
arar
arar
2222
arOHmOHar
ar
3
ar
3
22
ar
2
ar
22
ar
2
ar
==
=+=+=
=+=
====
+×=+×
=+
=×−×+×××
=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=Δ→
π
=
π
Δ
=γ−γ+Δγ+
×=+×
−
π
=+
×π
−
π
=+
π
−−
−−
Exercício 2.14
( )
1
2
11
22
222
111
arar
21
ar
HgO2Har
T
T
Vp
Vp
mRTVp
mRTVp)c
Pa050.12p0000.1361,0000.10155,0p
cm5
1
10
5,0hA.hA.y)b
Pa200.25000.10000.1362,0p
02,02,0p)a
=⇒=
=
=′⇒=×−×+′
=×=Δ⇒Δ=Δ
=−=
=×γ−×γ+
C44K317
100
95
200.125
050.112
373T
cm95105,01010V
050.112000.100050.12p
)abs(Pa200.125000.100200.25p
o
2
3
2
abs2
abs1
==××=
=×−×=
=+=
=+=
Exercício 2.15
3
A
A
A
atmAAabs
atm
OH
A
OH
A
2222
A
212A
m
kg
12,1
293287
576.94
RT
p
)abs(Pa576.94200.95624ppp
Pa200.95000.1367,0p)b
mca0624,0
000.10
624p
h
Pa6240015,02000.8600p
m0015,0
40
4
2
3,0
D
d
2
h
h
4
d
2
h
4
D
h
h2000.83,0000.103,0000.8p
0hhh2p)a
2
2
=
×
==ρ
=+−=+=
=×=
−=−=
γ
=
−=××−−=
=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=Δ→
π
=
π
Δ
Δ×−×−×=
=γ−γ+Δγ+
Exercício 2.16
3
1
2
2
1
12
1
2
11
22
absgásO2Hgás
O2Hgás
absgás
atm
gásO2HHggás
m16,2
293
333
100
95
2
T
T
p
p
VV
T
T
Vp
Vp
)abs(kPa1001090pkPa10Pa000.101000.10z.p)c
m5,0
000.10
000.5
zz.p)b
)abs(kPa95590p
kPa90Pa032.90662,0000.136p
Pa500016,0000.10025,0000.136p16,0025,0p)a
=××==⇒=
=+=′⇒==×=′γ=′
==⇒γ=
=+=
==×=
=×+×=⇒×γ+×γ=
Exercício 2.17
( ) ( ) 2
3
22
2
2
1
2
3
3
2
2
2
12
2
1
1
32
21
3,0p1,05,0p5,0p
4
D
pDD
4
p
4
D
p
000.22,0000.10pp
000.10pp
×+−×=×→
π
+−
π
=
π
=×=−
=−
( )
( )
kPa5,43Pa500.43p3480p08,0
180000.10p33,0p25,0
180p33,0p25,0
000.2p09,0p24,0p25,0
11
11
21
221
==→=
−−=
−=
−+=
Exercício 2.18
3222
2
ct
c
t
t
pGt
o
G
p
22
c
22p
22
c
11p
m
kg
993.10
183,05,010
950.34
LDg
G4
L
4
D
g
G
gV
G
)c
m183,0
5,0210
5,110005,0
L
m0005,0
2
5,0501,0
2
DD
Dv
F
LDL
v
F)b
N5,11FFF
desce196319755,0395030GsenF
cimaparaN196378549817F
N7854
4
5,0
000.40
4
D
pF
N9817
4
5,0
000.50
4
D
pF)a
=
××π×
×
=
π
=
π
==ρ
=
×π××
×
=
=
−
=
−
=ε
πμ
ε
=⇒π
ε
μ=
=−=
>=×==
=−=
=
×π
×=
π
=
=
×π
×=
π
=
−
Exercício 2.19
( ) ( )
( ) ( )
cm8,127m278,1278,01L
m278,0ym0278,0x0600.36x10098,1x000.908000800
2
600.552
0200.735,0x15000.10x98,0800
A
F2
x10yy2,0x2
0200.7330ysen30sen1y000.10y25,0x55,0000.81,0
A
F2
m
N
200.73
30sen1
8,0000.101,0000.8
2
600.55
30Lsen
8,01,0
A
F
030Lsen8,01,0
A
F
6
oo
3oo
21
3
o
321
==+=′
=⇒=⇒=−×−+++
×
=×+−×+++
=⇒=
=×+×+−×++−+×+
=
×
×+×+
=
×γ+×γ+
=γ
=γ−×γ+×γ+
Exercício 2.20
( ) ( )
( ) ( )
( ) ( )
( )
kPa50109,39ppp)c
)abs(kPa1,60)abs(Pa100.6039908000.100p
Pa908.39
103,50
150102013,50000.10100
A
FAApG
p
FApAApAApG
cm3,50
4
8
4
D
A;cm201
4
16
4
D
A)b
N15005,008,016,0
001,0
5
8,0DD
v
F
s
m.N
8,0
10
000.810
g
)a
abm
absb
4
4
2
t12a
b
t2bH1aH2a
2
22
2
2
2
22
1
1
21t
2
3
−=−−=−=
==−+=
−=
×
−×−×+
=
−−+
=
++−=−+
=
×π
=
π
==
×π
=
π
=
=×+×π××=+π
ε
μ=
=
×
=
μγ
=ν
−
−
−
l
Exercício 2.21
2
3
p
p
p
p
p
p
2
p
p
pp2
12
m
s.N
8,0
10
000.810
g
m001,0
2
998,01
2
DD
D
vL4
pL
v
4
D
p
LD
4
D
p
pistãonomédiapressãopondephp
000.10pp
=
×
=
νγ
=μ
=
−
=
−
=ε
ε
μ
=→
ε
μ=
τπ=
π
==γ+
=−
−
Exercício 2.22
N33933,0
4
2,1
000.10b
4
R
F
N160.23,02,16,0000.10AhF
22
y
x
=×
×π
×=
π
γ=
=×××=γ=
kPa23,25Pa230.25000.10230.15000.10pp
m
N
230.152000.85,769hpp
Pa5,769
998,0001,0
2,02,18,04
p
21
2p2
p
−=−=−−=−=
=×−=γ−=
=
×
×××
=
Exercício 2.23
m4,02,06,0b
m2,0
6
h
h
2
hAh
I
hh
N920.252,1
2
2,1
000.30hhApF
m2,14,06,0
000.30
000.80
4,06,0h
6,0.4,0.h
2
12
4h
CG
cp
22
p
m
m
=−=
==
×
==−
=××=γ==
=−×=−×
γ
γ
=
γ=γ+γ
N640.8
2,1
4,0
25920
h
b
FFbFhF pp =×==→×=×
Exercício 2.24
N948.59100.115,42,1F
N668.7
2
100.11100.5100.5
2,16,0F
N755.285,46,0
2
100.11100.5
5,46,0
2
100.5
FFF
Pa100.116,0000.10100.56,0pp
Pa100.56,0500.86,0p
f
B
21A
212
11
=××=
=⎟
⎠
⎞
⎜
⎝
⎛ ++
××=
=××
+
+××=+=
=×+=×γ+=
=×=×γ=
Exercício2.25
N500.225,121500.7AhF
m0833,10833,01
m0833,0
5,124AhAh
I
hh
N102,15,124000.10AhApF
F2FF
2o2
1
12
325,1
1
12
3bh
1
CG
11CP
5
1O2H11
22B11
=×××=γ=
=+=
=
××
===−
×=×××=γ==
+×=
×
l
ll
m333,1333,01
m333,0
5,121Ah
hh
2
12
325,1
2
12
3bh
22CP
=+=
=
××
==−
×
l
N105F
333,1500.222F0833,1102,1
4
B
B
5
×=
×+×=××
F
Fp
h hcp
b
h
5m
2 m
A
B
1l 2l
3 m
F1 F2
FB
Exercício 2.26
m736,0
634.7
680.4
2,1
F
F
yxxFyF
N634.73,0
4
8,1
000.10b
4
R
F
m2,18,1
3
2
R
3
2
y
N860.43,0
2
8,1
000.10bR
2
R
F
y
x
CPCPCPyCPx
22
y
c
2
x
=×==⇒=
=×
×π
×=
π
γ=
=×==
=×=••γ=
Exercício 2.27
m65,230cos75,02h
AhApF
o
=×+=
γ==
kN4,991075,365,2000.10F
m75,35,25,1A
3
2
=×××=
=×=
−
Exercício 2.28
( )
( )
( ) ( )
3
oO2H
2
O2H
22
oinfsup
2
2
O2Hinf
2
osup
m
N
000.35
6,0
5,2000.86,05,3000.10
6,0
h6,0h
4
D
6,0h6,0
4
D
4
D
hFGF
6,0
4
D
G
4
D
6,0hF
4
D
hF
=
×−+×
=
γ−+′γ
=γ
π
+′γ=×
π
γ+
π
γ⇒=+
×
π
γ=
π
+′γ=
π
γ=
Exercício 2.29
xCGCG
γ1
γ2
R
R
O
Fx1 F2
Fy1
21 ll =
2
bR
Rb
2
R
F
AhF
FxFF
2
1
11x
1111x
22CG1y11x
γ
=γ=
γ=
=+ ll
6
R
Rb
2
RAh
I
hh 12
3bR
CG
11CP ===−
3
1
22
3
R
2
bR
3
R4
4
bR
3
R
2
bR
b
4
R
VF
2
bR
Rb
2
R
AhF
3
R
6
R
2
R
2
12
1
1
2
2
2
1
2
1
2
11y
2
2
22222
21
1
=
γ
γ
→
γ
=γ+
γ
×
γ
=
π
×
πγ
+×
γ
π
γ=γ=
γ
=γ=γ=
==−= ll
Exercício 2.30
( )
( )
N3,465
1
579,0300.14583,0000.15
BA
brFbrF
FBAFMM
m579,0079,05,0br
m079,0
5,106,1
125,0
Ay
I
yy
m06,156,05,0y
m56,0
000.9
032.5p
h
N300.145,11532.9ApFPa532.9
2
032.14032.5
2
pp
p
Pa032.141000.950321pp
Pa032.5037,0000.136037,0pp
m583,0083,05,0br
m083,0
5,11
125,0
yy
m125,0
12
15,1
12
b
I
Ay
I
yy)b
N000.155,11000.10ApFPa000.10
2
000.15000.5
2
pp
p
Pa000.155,1000.105,1p
Pa000.55,0000.105,0p)a
esqesqdirdir
BBesqdir
esq
esq
CG
esqCP
esq
o
ar
areq
esqesq
esqBesqA
esq
oesqAesqB
HgaresqA
dir
dirCP
4
33
CG
CG
CP
dirdir
dirBdirA
dir
O2HdirB
O2HdirA
=
×−×
=
−
=⇒×+=
=+=
=
×
==−
=+=
==
γ
=
≅××==⇒=
+
=
+
=
=×+=×γ+=
=×=×γ==
=+=
=
×
=−
=
×
==→=−
=××==⇒=
+
=
+
=
=×=×γ=
=×=×γ=
l
Exercício 2.31
( ) ( ) N6363,06,0
4
3,0000.103,0D
4
hApF
N107,1
4
6,0
6,0000.10
4
D
hApF
2222
MMMMM
3
22
F
FFFF
=−
π
××=−
π
γ==
×=
×π
××=
π
γ==
Exercício 2.32
N230.76
2
083,1000.1205,0000.45
F083,1F5,0F2F
m083,0
412
2
y12by
12/b
Ay
I
yy
N000.1205,12000.40ApFPa000.40
2
000.50000.30
p
Pa000.505000.105p
m3
000.10
000.30p
h
N000.455,11000.30ApF
Pa000.304,0000.1025,0000.1364,025,0p
BCAB
223
CG
CP
BCBCBCBC
O2HC
O2H
AB
ABABAB
O2HHgAB
=
×+×
=⇒×+×=×
=
×
====−
=××=×=⇒=
+
=
=×=×γ=
==
γ
=
=××==
=×−×=×γ−×γ=
l
l
l
Exercício 2.33
Exercício 2.34
m1CBMM
2
CB
bCB3M
3
3
b3
2
3
M
BCAB
BCAB
=⇒=
γ=→γ=
F1
F2
1l 2l ( )
( )
( )
( )
( )
m27,6z
5,1108,225,6z5,2
5,11
5,2z
08,2
5,25,2z
5,2106,4
5,2z
08,2
5,25,2z10
m5,2
N106,4251046pAF
5,2z
08,2
5,2
55
2
53
2
1
=
=+−
=⎥
⎦
⎤
⎢
⎣
⎡
−
+−
××=⎥
⎦
⎤
⎢
⎣
⎡
−
+−
=
×=×××==
−
+=
l
l
Exercício 2.35
2
1
h
x
h
3
x6
h
3
x
2
x
hxb
3
x
b
2
x
2
x
hxbF
3
x
xb
2
x
AhF
FF
2
1
2
2
1
2
22
1
1111
2211
=→=→=
γ
γ
×γ=×γ
=
γ=
=
γ=γ=
=
l
l
ll
Exercício 2.36
kN204H880.218015H
m.kN1805,1120MkN120
000.1
134000.10
V
m.kN880.2
000.1
41126000.10
M
V
x
=⇒+=×
=×=⇒=
×××
=
=
××××
=
Exercício 2.37
O ferro estará totalmente submerso.
N2183,0
4
3,0
300.10h
4
D
VE
22
flfl =×
×π
×=
π
γ=γ=
A madeira ficará imersa na posição em que o peso seja igual ao empuxo.
sub
2
fl
22
mad
h
4
D
E
N1593,0
4
3,0
500.7h
4
D
GE
π
γ=
=×
×π
×=
π
γ==
m218,0
3,0300.10
1594
D
E4
h
22
fl
sub =
×π×
×
=
πγ
=
Exercício 2.38
N625023,0000.25500VGG conconcil =×+=γ+=
F1
F2
1l
2l
( )
m3,02,05,0h
m5,0
1
23,0
000.10
6250
4
D
V/G4
H
H
4
D
VGEG
22
con
2
con
=−=
=
×π
⎟
⎠
⎞
⎜
⎝
⎛
−×
=
π
−γ
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
π
+γ=⇒=
Exercício 2.39
( ) m7,29,08,1BAx:Logo
m9,0
270
6,0080.13,0350.1
F
GE
m6,0
3
8,1
3
BA
m3,0
3
9,0
3
IH
N270080.1350.1GEF:Logo
N080.11
2
6,08,1
000.2b
2
CBBA
VG
N350.11
2
9,03,0
000.10b
2
IHCH
VE
2
BA
IH
FGE
EGF
2F
21
3
2
1
ccc
OHsubOH
321
22
=−−=−=
−=
×−×
=
−
=
===
===
=−=−=
=×
×
×=×
×
γ=γ=
=×
×
×=×
×
γ=γ=
=
+=
=+
l
ll
l
l
l
lll
A força deverá ser aplicada à direita do ponto B, fora da plataforma AB.
Exercício 2.40
( )( )
( )( ) 22
dd
444
3
odo
3
m1036,3A02,0A3,031055103,002,010
12
6,0
AARhGRA
26
D
−
×=⇒−+×+=××−×
×π
−+γ+=γ−γ×
×
π
A B
C
I H
E
G
F
1l
2l
3l
Exercício 2.41
Supondo o empuxo do ar desprezível:
3
c
ccc
3
fl
fl
ap
m
N
670.26
03,0
800
V
G
VG
m03,0
000.10
300E
VVE
N300500800EEGG
===γ→γ=
==
γ
=→γ=
=−=→+=
Exercício 2.42
mm2,7m102,7
005,0
104,14
d
V4
hh
4
d
V
m104,11068,21082,2V
m1068,2
200.8
102,2G
VVEG
m1082,2
800.7
102,2G
VVEG
3
2
7
2
2
3766
36
2
2
2222
36
2
1
1111
=×=
×π
××
=
π
Δ
=Δ⇒Δ×
π
=Δ
×=×−×=Δ
×=
×
=
γ
=⇒γ==
×=
×
=
γ
=⇒γ==
−
−
−−−
−
−
−
−
Exercício 2.43
( )
( )
( )
( )
m8,0hh000.16000.40h000.6000.32
h5,2000.16h000.6000.32
h5,14hp
m
N
000.324000.8p4AApGAp
2Situação
m
N
000.1622A4A
EG1Situação
ooo
oo
ooobase
2basebasecbasebasebasebase
3cbbc
=→−+=
−+=
−−γ+γ=
=×=→×γ=→=
=γ→γ=γ→×γ=×γ
=→
l
lll
Exercício 2.44
m6
000.61009,2
2105,4
x
N1009,2
12
2
10
26
D
E
N105,4135,110AhF
GE
2F
xxE3
3
2
FxG
4
4
4
3
4
3
44
=
−×
××
=
×=
×π
×=
×
π
γ=
×=×××=γ=
−
×
=⇒•=××+• E
G F
Exercício 2.45
( )
( )
( )
3B
B
BAbase
2
b
bc
b
base
bbase
3cAbAbc
m
N
000.25
4,02,0000.15000.13
2,06,02,0p
m
N
000.13
1
000.1016,0000.5
A
FA6,0
A
FG
p
FGAp
2Situação
m
N
000.15000.5332,0A6,0AEG
1Situação
=γ
×γ+×=
−×γ+×γ=
=
+××
=
+××γ
=
+
=
+=
=×=γ=γ→×γ=×γ→=
Exercício 2.46
( ) ( ) N171.10
6
12
1085,7132,110
6
D
gG
1085,7
293400.41
200.95
TR
p
m
kg
132,1
293287
200.95
TR
p
Pa200.957,0000.1367,0p
3
3
3
2Har
3
2H
2H
2H
3
ar
ar
ar
Hgatm
=
×π
××−×=
π
ρ−ρ=
×=
×
==ρ
=
×
==ρ
=×=×γ=
−
−
Exercício 2.47
79,0x
21,0x
62
16466
x:Raízes
01x6x6
0
2
x
2
1
x12
1
xFazendo0
22
1
12
0
2
b
2
b
b
2
b
2
b
0
V
I
r
bhbhbEG
2
2
cc
c
c
3
c
12
b
c
c
y
c
sub
2
sub
3
c
4
=′′
=′
→
×
××−±
=
>+−
>+−→=
γ
γ
→>
γ
γ
+−
γ
γ
>⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
γ
γ
−−
γ
γ
γ
γ
−=→>−
γ
γ
=
γ
γ
=→γ=γ→=
ll
l
l
l
l
l
l
l
ll
179,021,00 cc
<
γ
γ
<<
γ
γ
<
ll
Exercício 2.48
estável0m037,00467,0
5,2
103,083.2000.10
r
cm3,083.2
12
1025
12
bL
I0
G
I
r
cm67,433,05cm5yCG
cm33,05,0
3
2
yCC
cm5,0
10
5,2
L
V
h
hL
2
bh
2V
m105,2
000.10
5,2G
V
GVEG
8
4
33
y
yf
im
2
im
im
im
34
f
im
imf
⇒>=−
××
=
=
×
==→>−
γ
=
=−=⇒=→
=×=→
===
==
×==
γ
=
=γ⇒=
−
−
l
l
l
Exercício 2.49
( )
( )
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
γ
γ
−
γ
γ
<→
−
<
<−−→>+−
=
γ
γ
>
γ
γ
+−
γ
γ
→>
γ
γ
+−
γ
γ
→>⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
γ
γ
−−
γπ
πγ
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
γ
γ
−=−=
π
=γπ=
>−
γ
=
γ
γ
=
γπ=πγ
=
ll
l
l
l
l
l
l
l
l
l
l
l
l
l
l
12
1
R
H
x1x2
1
R
H
01x2x2
R
H
0
R
H
2.x
R
H
2
x
1
:RportudodividindoexFazendo
0H2H2R0
2
H
2
H
H4
R
0HH
2
1
HR4
R
HH
2
1
2
h
2
H
4
R
IHRG
0
G
I
r
Hh
HRhR
GE
2
2
2
2
2
2
2
2
222
2
2
4
sub
4
y
2
y
sub
2
sub
2
CG
CC0,5cm
Exercício 2.50
z6
g
g5
1z
g
a
1zp
y
z Δγ=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+Δγ=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
±Δγ=Δ
Exercício 2.51
h
km
2,646,3
s
m
83,17557,3tav)b
s
m
57,320tg8,9a20tgga
g
a
x
z
)a
x
2
o
x
o
x
x
=×=×==
=×=→=→=
Δ
Δ
Exercício 2.52
oo
o
x
4130tg
30cos8,9
45,2
tg
cosg
a
tg =θ⇒+
×
=α+
α
=θ
Exercício 2.53
( )
2x
3
x
3
Hg
s
m
72,1
5,1
257,0
10
x
z
ga
m257,0
000.136
10140175
z
g
a
x
z
)b
m29,1
000.136
10175p
h)a
=×=
Δ
Δ
=
=
×−
=Δ→=
Δ
Δ
=
×
=
γ
=
Exercício 2.54
)abs(kPa106
10
6,010000.1
100ghpp
)abs(kPa7,125
10
6,010000.1
7,119ghpp
)abs(kPa7,119100106,0
2
5,10
000.1p
s
rd
5,10
60
100
2n2pr
2
p
3atmC
3AB
32
2
A
atm
2
2
A
=
××
+=ρ+=
=
××
+=ρ+=
=+×⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
××=
=×π×=π=ω→+Δ
ω
ρ=
−
Exercício 2.55
2x
x
s
m
78,2
10
6,3
100
t
v
a
g
a
tg)a ===→=α
140
175 Pa
zΔ
( ) ( )
( ) ( ) Pa600.314,05,0000.10h5,0p
Pa400.614,05,0000.10h5,0p
m14,0278,05,0h
5,0
h
tg)b
5,15278,0
10
78,2
tg
O2HB
O2HA
o
=−×=Δ−γ=
=+×=Δ+γ=
=×=Δ→
Δ
=α
=α→==α
Exercício 2.56
2
o
x
xo
oo
o
4
3
dir
dir
4
3
esq
esq
s
m
8,530tg10a
g
a
30tg
m73,1
30tg
1
30tg
h
L
L
h
30tg
m11011hm11
10
10110p
h
m10
10
10100p
h
=×=⇒=
==
Δ
=⇒
Δ
=
=−=Δ⇒
×
=
γ
=
=
×
=
γ
=
Exercício 2.57
s5
4
6,3
72
a
v
t
t
v
a
s
m
4
5,0
2,0
10a
g
a
tg
x
x
2x
x
===→=
=×=
=α
Exercício 2.58
( ) kN6,13N600.131010006,31000GmaFmaGF
s
m
6,31
000.10
200.27600.13
g1
z
pp
a
g
a
1zpp
Pa600.131,0000.1361,0p
Pa200.272,0000.1362,0p
2
12
y
y
12
Hg2
Hg1
−=−=×−−×=−=⇒=+
−=⎟
⎠
⎞
⎜
⎝
⎛
+
−
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
Δγ
−
=⇒⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+Δγ=−
=×=×γ=
=×=×γ=

Más contenido relacionado

La actualidad más candente

Resolução do capítulo 1 brunetti[1]
Resolução do capítulo 1   brunetti[1]Resolução do capítulo 1   brunetti[1]
Resolução do capítulo 1 brunetti[1]Ídilla Kaenna
 
Exercicios resolvidos de resmat mecsol
Exercicios resolvidos de resmat mecsolExercicios resolvidos de resmat mecsol
Exercicios resolvidos de resmat mecsolDanieli Franco Mota
 
2051 exercicios_mec_fluidos_nova versao - manometria exercicio 25
2051  exercicios_mec_fluidos_nova versao - manometria exercicio 252051  exercicios_mec_fluidos_nova versao - manometria exercicio 25
2051 exercicios_mec_fluidos_nova versao - manometria exercicio 25SHEILA VIVIANE MARIA DOS SANTOS
 
Problemas resolvidos e propostos
Problemas resolvidos e propostosProblemas resolvidos e propostos
Problemas resolvidos e propostosFranck Lima
 
Exercicios hidraulicaprova1
Exercicios hidraulicaprova1Exercicios hidraulicaprova1
Exercicios hidraulicaprova1Stefanny Costa
 
Solucionário Capitulo4 FOX
Solucionário Capitulo4 FOXSolucionário Capitulo4 FOX
Solucionário Capitulo4 FOXMarilza Sousa
 
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016Afonso Celso Siqueira Silva
 
Mecanica exercicios resolvidos
Mecanica exercicios resolvidosMecanica exercicios resolvidos
Mecanica exercicios resolvidoswedson Oliveira
 
Funcoes de varias variaveis calculo 2
Funcoes de varias variaveis  calculo 2Funcoes de varias variaveis  calculo 2
Funcoes de varias variaveis calculo 2Kassiane Campelo
 
Fenômenos de transporte MecFlu.
Fenômenos de transporte MecFlu.Fenômenos de transporte MecFlu.
Fenômenos de transporte MecFlu.Ailton Souza
 
Exercicios resolvidos de_hidraulica
Exercicios resolvidos de_hidraulicaExercicios resolvidos de_hidraulica
Exercicios resolvidos de_hidraulicaSérgio Lessa
 
3 lista-exercicio manometria
3 lista-exercicio manometria3 lista-exercicio manometria
3 lista-exercicio manometriaMarinaldo Junior
 
Resmat ii material de aula com exercicios da av1 até av2
Resmat ii material de aula com exercicios da av1 até av2Resmat ii material de aula com exercicios da av1 até av2
Resmat ii material de aula com exercicios da av1 até av2Douglas Alves
 
Mecânica dos fluídos i capitulo 4
Mecânica dos fluídos i   capitulo 4Mecânica dos fluídos i   capitulo 4
Mecânica dos fluídos i capitulo 4Bruno Gava
 
Anexo c problemas resolvidos e propostos
Anexo c    problemas resolvidos e propostosAnexo c    problemas resolvidos e propostos
Anexo c problemas resolvidos e propostosGiuliano Alves Santana
 

La actualidad más candente (20)

Resolução do capítulo 1 brunetti[1]
Resolução do capítulo 1   brunetti[1]Resolução do capítulo 1   brunetti[1]
Resolução do capítulo 1 brunetti[1]
 
Capitulo 4 livro
Capitulo 4 livroCapitulo 4 livro
Capitulo 4 livro
 
Exercicios resolvidos de resmat mecsol
Exercicios resolvidos de resmat mecsolExercicios resolvidos de resmat mecsol
Exercicios resolvidos de resmat mecsol
 
2051 exercicios_mec_fluidos_nova versao - manometria exercicio 25
2051  exercicios_mec_fluidos_nova versao - manometria exercicio 252051  exercicios_mec_fluidos_nova versao - manometria exercicio 25
2051 exercicios_mec_fluidos_nova versao - manometria exercicio 25
 
Problemas resolvidos e propostos
Problemas resolvidos e propostosProblemas resolvidos e propostos
Problemas resolvidos e propostos
 
Exercicios hidraulicaprova1
Exercicios hidraulicaprova1Exercicios hidraulicaprova1
Exercicios hidraulicaprova1
 
Solucionário Capitulo4 FOX
Solucionário Capitulo4 FOXSolucionário Capitulo4 FOX
Solucionário Capitulo4 FOX
 
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016
Resistência de materiais.pdf exercícios resolvidos em 26 mar 2016
 
Mecanica exercicios resolvidos
Mecanica exercicios resolvidosMecanica exercicios resolvidos
Mecanica exercicios resolvidos
 
Funcoes de varias variaveis calculo 2
Funcoes de varias variaveis  calculo 2Funcoes de varias variaveis  calculo 2
Funcoes de varias variaveis calculo 2
 
Fenômenos de transporte MecFlu.
Fenômenos de transporte MecFlu.Fenômenos de transporte MecFlu.
Fenômenos de transporte MecFlu.
 
Exercicios resolvidos de_hidraulica
Exercicios resolvidos de_hidraulicaExercicios resolvidos de_hidraulica
Exercicios resolvidos de_hidraulica
 
3 lista-exercicio manometria
3 lista-exercicio manometria3 lista-exercicio manometria
3 lista-exercicio manometria
 
Resmat ii material de aula com exercicios da av1 até av2
Resmat ii material de aula com exercicios da av1 até av2Resmat ii material de aula com exercicios da av1 até av2
Resmat ii material de aula com exercicios da av1 até av2
 
Respostas Incropera 6ed
Respostas Incropera 6edRespostas Incropera 6ed
Respostas Incropera 6ed
 
Rm exerc resolvidos
Rm exerc resolvidosRm exerc resolvidos
Rm exerc resolvidos
 
Mecânica dos fluídos i capitulo 4
Mecânica dos fluídos i   capitulo 4Mecânica dos fluídos i   capitulo 4
Mecânica dos fluídos i capitulo 4
 
Anexo c problemas resolvidos e propostos
Anexo c    problemas resolvidos e propostosAnexo c    problemas resolvidos e propostos
Anexo c problemas resolvidos e propostos
 
Tabela de integrais
Tabela de integraisTabela de integrais
Tabela de integrais
 
Aula18(3)
Aula18(3)Aula18(3)
Aula18(3)
 

Similar a Resolucao de-exercicios-cap 2 - franco-brunetti

Exercicios Resolvidos Série MIPS Embarcados
Exercicios Resolvidos Série MIPS EmbarcadosExercicios Resolvidos Série MIPS Embarcados
Exercicios Resolvidos Série MIPS EmbarcadosElaine Cecília Gatto
 
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片Chyi-Tsong Chen
 
Variable charts
Variable chartsVariable charts
Variable chartsHunhNgcThm
 
Granular Material 5. Table 2 shows the results from a resilient modul.pdf
Granular Material 5. Table 2 shows the results from a resilient modul.pdfGranular Material 5. Table 2 shows the results from a resilient modul.pdf
Granular Material 5. Table 2 shows the results from a resilient modul.pdfgopalk44
 
Mechanics of fluids 5th edition potter solutions manual
Mechanics of fluids 5th edition potter solutions manualMechanics of fluids 5th edition potter solutions manual
Mechanics of fluids 5th edition potter solutions manualGloverTBL
 
CASIO 991 ES Calculator Technique
CASIO 991 ES Calculator TechniqueCASIO 991 ES Calculator Technique
CASIO 991 ES Calculator TechniqueMark Lester Manapol
 
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...Julio Banks
 
Pushover analysis of simply support concrete section beam subjected to increm...
Pushover analysis of simply support concrete section beam subjected to increm...Pushover analysis of simply support concrete section beam subjected to increm...
Pushover analysis of simply support concrete section beam subjected to increm...Salar Delavar Qashqai
 
Solution Manual for Mechanics of Flight – Warren Phillips
Solution Manual for Mechanics of Flight – Warren PhillipsSolution Manual for Mechanics of Flight – Warren Phillips
Solution Manual for Mechanics of Flight – Warren PhillipsHenningEnoksen
 
Design of machine elements - Helical gears
Design of machine elements - Helical gearsDesign of machine elements - Helical gears
Design of machine elements - Helical gearsAkram Hossain
 
coordenadas normales y tangenciales
coordenadas normales y tangencialescoordenadas normales y tangenciales
coordenadas normales y tangencialesWASHINGTONSTALINPILA
 
Solutions_Manual_to_accompany_Applied_Nu.pdf
Solutions_Manual_to_accompany_Applied_Nu.pdfSolutions_Manual_to_accompany_Applied_Nu.pdf
Solutions_Manual_to_accompany_Applied_Nu.pdfWaleedHussain30
 

Similar a Resolucao de-exercicios-cap 2 - franco-brunetti (12)

Exercicios Resolvidos Série MIPS Embarcados
Exercicios Resolvidos Série MIPS EmbarcadosExercicios Resolvidos Série MIPS Embarcados
Exercicios Resolvidos Série MIPS Embarcados
 
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片
Ch 02 MATLAB Applications in Chemical Engineering_陳奇中教授教學投影片
 
Variable charts
Variable chartsVariable charts
Variable charts
 
Granular Material 5. Table 2 shows the results from a resilient modul.pdf
Granular Material 5. Table 2 shows the results from a resilient modul.pdfGranular Material 5. Table 2 shows the results from a resilient modul.pdf
Granular Material 5. Table 2 shows the results from a resilient modul.pdf
 
Mechanics of fluids 5th edition potter solutions manual
Mechanics of fluids 5th edition potter solutions manualMechanics of fluids 5th edition potter solutions manual
Mechanics of fluids 5th edition potter solutions manual
 
CASIO 991 ES Calculator Technique
CASIO 991 ES Calculator TechniqueCASIO 991 ES Calculator Technique
CASIO 991 ES Calculator Technique
 
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...
T liner simulation parametric study of a thermal-liner by Julio c. banks, MSM...
 
Pushover analysis of simply support concrete section beam subjected to increm...
Pushover analysis of simply support concrete section beam subjected to increm...Pushover analysis of simply support concrete section beam subjected to increm...
Pushover analysis of simply support concrete section beam subjected to increm...
 
Solution Manual for Mechanics of Flight – Warren Phillips
Solution Manual for Mechanics of Flight – Warren PhillipsSolution Manual for Mechanics of Flight – Warren Phillips
Solution Manual for Mechanics of Flight – Warren Phillips
 
Design of machine elements - Helical gears
Design of machine elements - Helical gearsDesign of machine elements - Helical gears
Design of machine elements - Helical gears
 
coordenadas normales y tangenciales
coordenadas normales y tangencialescoordenadas normales y tangenciales
coordenadas normales y tangenciales
 
Solutions_Manual_to_accompany_Applied_Nu.pdf
Solutions_Manual_to_accompany_Applied_Nu.pdfSolutions_Manual_to_accompany_Applied_Nu.pdf
Solutions_Manual_to_accompany_Applied_Nu.pdf
 

Último

me3493 manufacturing technology unit 1 Part A
me3493 manufacturing technology unit 1 Part Ame3493 manufacturing technology unit 1 Part A
me3493 manufacturing technology unit 1 Part Akarthi keyan
 
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...Amil baba
 
Test of Significance of Large Samples for Mean = µ.pptx
Test of Significance of Large Samples for Mean = µ.pptxTest of Significance of Large Samples for Mean = µ.pptx
Test of Significance of Large Samples for Mean = µ.pptxHome
 
Power System electrical and electronics .pptx
Power System electrical and electronics .pptxPower System electrical and electronics .pptx
Power System electrical and electronics .pptxMUKULKUMAR210
 
Basic Principle of Electrochemical Sensor
Basic Principle of  Electrochemical SensorBasic Principle of  Electrochemical Sensor
Basic Principle of Electrochemical SensorTanvir Moin
 
Landsman converter for power factor improvement
Landsman converter for power factor improvementLandsman converter for power factor improvement
Landsman converter for power factor improvementVijayMuni2
 
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdf
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdfsdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdf
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdfJulia Kaye
 
Multicomponent Spiral Wound Membrane Separation Model.pdf
Multicomponent Spiral Wound Membrane Separation Model.pdfMulticomponent Spiral Wound Membrane Separation Model.pdf
Multicomponent Spiral Wound Membrane Separation Model.pdfGiovanaGhasary1
 
Modelling Guide for Timber Structures - FPInnovations
Modelling Guide for Timber Structures - FPInnovationsModelling Guide for Timber Structures - FPInnovations
Modelling Guide for Timber Structures - FPInnovationsYusuf Yıldız
 
Lecture 1: Basics of trigonometry (surveying)
Lecture 1: Basics of trigonometry (surveying)Lecture 1: Basics of trigonometry (surveying)
Lecture 1: Basics of trigonometry (surveying)Bahzad5
 
The relationship between iot and communication technology
The relationship between iot and communication technologyThe relationship between iot and communication technology
The relationship between iot and communication technologyabdulkadirmukarram03
 
Dev.bg DevOps March 2024 Monitoring & Logging
Dev.bg DevOps March 2024 Monitoring & LoggingDev.bg DevOps March 2024 Monitoring & Logging
Dev.bg DevOps March 2024 Monitoring & LoggingMarian Marinov
 
Mohs Scale of Hardness, Hardness Scale.pptx
Mohs Scale of Hardness, Hardness Scale.pptxMohs Scale of Hardness, Hardness Scale.pptx
Mohs Scale of Hardness, Hardness Scale.pptxKISHAN KUMAR
 
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....santhyamuthu1
 
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
Engineering Mechanics  Chapter 5  Equilibrium of a Rigid BodyEngineering Mechanics  Chapter 5  Equilibrium of a Rigid Body
Engineering Mechanics Chapter 5 Equilibrium of a Rigid BodyAhmadHajasad2
 
Gender Bias in Engineer, Honors 203 Project
Gender Bias in Engineer, Honors 203 ProjectGender Bias in Engineer, Honors 203 Project
Gender Bias in Engineer, Honors 203 Projectreemakb03
 
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docx
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docxSUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docx
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docxNaveenVerma126
 

Último (20)

me3493 manufacturing technology unit 1 Part A
me3493 manufacturing technology unit 1 Part Ame3493 manufacturing technology unit 1 Part A
me3493 manufacturing technology unit 1 Part A
 
Litature Review: Research Paper work for Engineering
Litature Review: Research Paper work for EngineeringLitature Review: Research Paper work for Engineering
Litature Review: Research Paper work for Engineering
 
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...
Best-NO1 Best Rohani Amil In Lahore Kala Ilam In Lahore Kala Jadu Amil In Lah...
 
Test of Significance of Large Samples for Mean = µ.pptx
Test of Significance of Large Samples for Mean = µ.pptxTest of Significance of Large Samples for Mean = µ.pptx
Test of Significance of Large Samples for Mean = µ.pptx
 
Power System electrical and electronics .pptx
Power System electrical and electronics .pptxPower System electrical and electronics .pptx
Power System electrical and electronics .pptx
 
Basic Principle of Electrochemical Sensor
Basic Principle of  Electrochemical SensorBasic Principle of  Electrochemical Sensor
Basic Principle of Electrochemical Sensor
 
Landsman converter for power factor improvement
Landsman converter for power factor improvementLandsman converter for power factor improvement
Landsman converter for power factor improvement
 
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdf
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdfsdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdf
sdfsadopkjpiosufoiasdoifjasldkjfl a asldkjflaskdjflkjsdsdf
 
Lecture 4 .pdf
Lecture 4                              .pdfLecture 4                              .pdf
Lecture 4 .pdf
 
計劃趕得上變化
計劃趕得上變化計劃趕得上變化
計劃趕得上變化
 
Multicomponent Spiral Wound Membrane Separation Model.pdf
Multicomponent Spiral Wound Membrane Separation Model.pdfMulticomponent Spiral Wound Membrane Separation Model.pdf
Multicomponent Spiral Wound Membrane Separation Model.pdf
 
Modelling Guide for Timber Structures - FPInnovations
Modelling Guide for Timber Structures - FPInnovationsModelling Guide for Timber Structures - FPInnovations
Modelling Guide for Timber Structures - FPInnovations
 
Lecture 1: Basics of trigonometry (surveying)
Lecture 1: Basics of trigonometry (surveying)Lecture 1: Basics of trigonometry (surveying)
Lecture 1: Basics of trigonometry (surveying)
 
The relationship between iot and communication technology
The relationship between iot and communication technologyThe relationship between iot and communication technology
The relationship between iot and communication technology
 
Dev.bg DevOps March 2024 Monitoring & Logging
Dev.bg DevOps March 2024 Monitoring & LoggingDev.bg DevOps March 2024 Monitoring & Logging
Dev.bg DevOps March 2024 Monitoring & Logging
 
Mohs Scale of Hardness, Hardness Scale.pptx
Mohs Scale of Hardness, Hardness Scale.pptxMohs Scale of Hardness, Hardness Scale.pptx
Mohs Scale of Hardness, Hardness Scale.pptx
 
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....
SATELITE COMMUNICATION UNIT 1 CEC352 REGULATION 2021 PPT BASICS OF SATELITE ....
 
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
Engineering Mechanics  Chapter 5  Equilibrium of a Rigid BodyEngineering Mechanics  Chapter 5  Equilibrium of a Rigid Body
Engineering Mechanics Chapter 5 Equilibrium of a Rigid Body
 
Gender Bias in Engineer, Honors 203 Project
Gender Bias in Engineer, Honors 203 ProjectGender Bias in Engineer, Honors 203 Project
Gender Bias in Engineer, Honors 203 Project
 
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docx
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docxSUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docx
SUMMER TRAINING REPORT ON BUILDING CONSTRUCTION.docx
 

Resolucao de-exercicios-cap 2 - franco-brunetti