Ntr}ES
far
EnglnEEFl ilg
sctEilcE -l
Fslilctatir prunicr Sultqn $hlruddin Adul
,rir
$tdt
lJ-Ir-,
T
UIJIL
FglifrEtmit(
-
I
Frcnier sulfan $alehuddin ^{bdul .*ri: shsh
/1"." ,'
t'/'
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Engineering Science -TL
Topic 1: Physical Quantities and Measurement t4"4{
^
eu+{|M
y-
o't t'Yre *uau'#'i' f )e#nr,srhau *{ , deriv'ed ?*
Dut"oh"n
-sl u*'|{
€ pru*tP*'s
Physical Quantity
f,gr6r'2
r
Physicat quantity is the numerical value of a measurable property that describes a physical system's state at a moment in time. It is a physical property that can be quantified, can be measured with a measuring instrument. Examplei of physical quantities are rnass, volume, length, time, temperature,electric current-
\.il
Quantities can be divided into two types 1. Base quantities 2. Derived quantities
:
Basic quantities
Basic qLJantittes are the fundamental quantities that are not related to each other and that are use to derive all other quantities.
There are seven basic quantities. They are
1. length 2. time 3. mass 4. Thermodynamic temperature - 5. electric current of substance Q "rnornt intensity (7)luminous
'\b
.
Derived quantitigs " Derived quantities are just quantities that are derived from one or more basic quantities. For exarnple: Area is a derived quantity because it is derived from the basic quantity length.
Area=length*length Volume is a derived quantities because it is derived from the basic quantity length. volume = length " length * lengtlt Density is a derived quantity because it is derived from length and mass, two basic quantities. * * density = mass/(length length lengttt)
JMSK/sllee
1
4
88101 fngineering Science -T1
lnternational System of Units All systems of weights and measures, metric and non-metric, are linked through a network of international agreements suppofting the lnternational System of Units. The lnternational System is called the Sl, using the first two initials of its French name Sysfeme lnternational d'lJnitds. It is a scientific method of expressing the magnitudes or quantities of important natural phenomena. There are seven base units in the system, from which other units are derived, This system was formerly called the meter-kilogram-second (MKS) system.
Sl Base Units The Sl is founded on seven S/ base unifs for seven base quanfifr'es assumed to be mutually independent, as table 1.
Table Ba.se
{.
Sl base units
Name
quantity
$5'mbol
Sf base unit
lfftgth
nra.ss
kilo.greu
elech'ic crnreflt
t#fiuddj .'G"g41
',,,
arupere A ,,kei , , K '
amourtofsrlrstatrce roole luminousiat€asitrr .
JMSK/sllee
..,
t,
:
..''r
i
kg
mol
..r.a :,. ,:3.andgla cd ,, i-ra:,,:r.:,' ., :.,.::',
:..:.''
'
88101 Engineering Science
Definitions of the $l base units
Unlt of
length
meter
The meler is the tength of ttre path travelled iry light in vaurum dunng a time interval ot
seconc
lllgg i92
458 cf
a
Unitofmass kilogram lheklogramrstheunloimass:ltiseqra!totrem*ssoftlprr.tematinnalprotolypeofthekrfogram
.
.qe
tltr
*ffi (adar
Unitof
time
second fte selond
is the duration of
g 192 031 770 peiiods
of the raiilation conesponding to $ie
beltireenthelt;o hvperfine levels oftne ground state ofifte cesiilrTi 133
arnpere of electric cunent Unit
atom.
transiiian
,.9rg
Tflffi
The ampere is ihai constanf cunent s'hich, ii mair$ained in lwa straighi parallei concuctors of irifrrute of negltgrbie circular cioss-sec&on. and phced 1 nreter apan in vacium, vrould produce bet*een
[en$h, ..ete
lhese
' H
conductors a foree equal to 2 x 10-i neulion per rneter of lerEti.
Unftof kelvin Tlrcltelvin,unilofiherrnoivnamictenperature,isthefractronl2i3.lti0fthethennoCynamictemperafurecf ..r"p H* thermodynamic tlu triple grornt of nater. temperature
Unlt
of mole of
amoun{
1 The fiole is fre anrftnt of $$slance of a sysiem which cffiiains as ma*y elemenian entlites as tl€re afoins in 0.012 kiloEam ol carbon 12; its
wr*ol
is
'
mo[."
are ..St'
H
substance 2. F/hen the moi,e is useC. the elemeniarr e$rtres must tre speciiled and may be atoms. mo{ecules. .ofis,
other padcles. cr speu{ied groups of
Unit of oandela luminous frquency
The candela is tlte klmtnous intensitv. in a given
intensity
JMSK/sllee
540 x 1012
elediots.
su$ par{i&s.
d #*t" s{eraiian ffiJ
dredion oi a source thai emits rnonochronutic
hert and that has a radlant intensrty in tha{ direstion of 1/683 waft per
radralron
-fr
I
88101 Engineering Science -T1 I I
Sl Derived Units Other quantities, called derived quantities, are defined in terms of the seven base quantities via a system of quantity equations. The Sl derived unifs for these derived quantities are obtained from these equations and the seven Sl base units.
Table
2. Examples of Slderived units SI derirerl onit
..*---.-.-,*-----
.---...-.-.-.---.:-j:>
gea
:
,,
,., .,,
Sdel!{q
-,i
cubicmeter
Sp€E4rdoq'
. ,oetefpq second ",' ,' .
aceleration
metupa . ..,
sas: dcnsi&' qpeiific
rohne
m-
second
.a
tn5
.
u's
sEared
.n'
recipocdneter .\
tEfi'
hlcgrdr[per cubhneter
rubico4qpqirlag!ry
ctrrcdemir.*
allprr€ Pef
Cagnelhfeld sr€oge
SrygrePer:meter
aoou{-of-slb$ancc coac.er8diou
oclcpcr nrbic rcter
!ffi'r..
t1!f]!pq,lge*;i
nusstaction
ldogaru pcr
JMSK,/sllee
::.'
:..
rrriune
ware$rrober
l--:
s,quafr
:':-:
,
{ml
netef
'
hlogu,
r.
:i::'
',,
aobn' .,,,,,.'; ..,:,,,:;
:' ,
-,,
wbich ar4' be represeced
bl
",-,, the
,,
:, :,-ad#
qnbcr
:
1
1
kgfu= I
88101 Engineering Science -T1
For ease of understanding and convenience, 22 Sl derived units have been given special names and symbols, as shown in Table 3.
Derived quantity
Expression in termr of Symbol oiher Sl units
Name
Table
3. Sl derived units with special names and symbols Sl
plane angle
radian {ai
solid angle
steradian
fteqUegCy:.,', ::'i
,
r: r',:: rr.::',,,
;,: :.: ri ,,',,
'rPascal
:,
:
:,
:
joule
power,radiantflux
.,*att,'," ",,
'
electric charge" quantilv of electricity electricpotential difference,
electromotiveforce
:
.
' .
'
''
,i.',.VOlt;t,
capacitance
',:::
,
,
farad
electricresistano: t
i,
siemens i'
,,.,1,,.,,,.ii,,,.,:,,,
-,.',.
, .,,,.,t.',
-.,t,, ,.
.:aeber
,
, .
magneiicflux densig
tesla
indudance
,1gfrv '-,,,,,:,;
,
,
Celsius temperature
iu*itiOiii"
.j,..',.,.,''
,.=;irt.,.' ,:,,ri';,,:.
,,,;1,' ,; ,
'
.
I
ianuA*j"l',-.
'
JMSK/sllee
;tlr1A,
,;,,.,,nftg'-s{A1 '
',a
cA/ v!A
'
s
Ai/
11
lx
,
,.,:'
,a5jbvert
katal
.r
:1:
,'
'"'
m-2.kg-t.s4.42
::
:-
zXjj:{2
#tgs€.A'1
Wbim2
kg's-2-4-1
v,lb,'A
tnz.kg s-e.4-z
.
cd;5rftl
'
'
K rnr-m-.2-Cri=Cd
lmim? 'n?
.,
*.4 66 = 6-216
,.s-1
rBq Gy
J/kg
€Y,,
'idni
kat
,r,'n
rn-2.kg-1.s3,A2
V9'
,wb T
,,*
:,';fggguere!
mr*g s-r
,
sA
',V i,
t,,
'
gray
catalytic activity
m2.kg-s-2
"c
lul ti11,,1,.i,.,,,..::,=',,.',.',
m-r*g s-r
N.m
degree Celsius
l'.l*nen.i
illuminance
ffiVtpr!..i
N,'nf
J
F
ohm,.,::,t:,:,
,,
electric conductance
rna$iiri"*ut
.,
1 {bi
m'kg 5'2
c
:
=
,Pa
Ji's
coulomb
1{bi
,'s'j.
'r1z '
:
mnrl= rp2-r1-2
tl
energy, work. quantity of frcat
:'
.
51 ic)
ne,uton
. '...
derived snlt
rad {a}
.,riierE :,
,
force
pressure,.stress
Expression in terms of Sl base units
'*'"t
"':','1.1.t''I
.
ml:s-z s-1-mot
.
:,
E
88101 Lngineering Science -T1
Sl prefixes A prefix may be added to aunit to produce a multiple of the original unit. All multiples are integer powers of ten. The 20 Sl prefixes used to form decimal multiples and submultiples of Sl units are as below.
Factor Name Symbol FactorName Symbol
1024 yotta y 1021 zetta Z
10-1 ,deii 'd fi2 centi c
101s exa 1015 peta
10-3 rnilli 'm 10-6 micro l,
E
p
tera' T 10e giga G 1012
106 mega M 103 kilo k 102 ftecto h 101 deka.da
10
9
1A-i2
nano
n
pico
p
10-15 femlo f 10-18
atto
a
!o'21 zepto z 10_24
yocto y
Conversion of Units Conversion of units involves comparison of different standard physical values, either of a single physical quantity or of a physical quantity and a combination of other physical quantities Refer some conversion factors, and examples of unit conversion on page 7-10.
JMSK/sllee
l
Bro
o+rncrnrtr cIjw
1.2
SGTENCE
colwERSlON OF Ul'llrs
Table 3: Convers[qq
Gonversion factqls-
L'IIdI ILILY
1m= 1606:= 1-o'cm { lzm , tn3 tn - 0-621 mi
Length
1 kg =.1O00 g ! 1 metric ton = 1qq9-!g
Mass
@s r yeir
Time
= 8J6 x 103 h i trort = 6o minulqg-= J.999 sesgnq f raaian (rad) = 57.30o ro - n A1-74\ rad
Angle
Power
t froriepower
\
(hP) = 746 w
1kw=1000 w
Unifs conversion Basic.
tlnit conversion Weight
Time
Lenqth
) 1m + 1cm)
1 km
1000m 100cm
t hour )
60 minutes
10mm
1 minute
)
6O seconds
1 kq
)
1000q
Example I
Thetalleststudentinclassiswithaheightof2.02m.
How tall is he in feet?
Solution: Provide the conversion fa'dor 1m = 3'281 ft 1 = 3.281 ft/m
2'o2'"
Example2
= [3:3i)&1!?' ={r36'Tt -= 6"#++
o,*,
Give this area in square A bacterium has an area of 2.35 square inches. millimeters'
SOIutlon: Zamszuln'
1
FI
I
OO4-TDCHnrrCI]TN SCIENCE
provide the conversion factor 1 in.? = (O.OZ54)2,rnm2
Etample 3 Convert 3.5 kilometre to meter. SoLution:-
1km :101m =1000m l9:0h 3.S km = 3.5 Lr * = 3.5,,x 1000 m =
tkm
\
3500 m
$:-
Convert 3.27 mtligmrn to gram.
*"\ 3
D'
Example 4
"rr'-" /"
,.
.'
I
,,(7
_S_qAtjon;_
"-1fng -1O-39 =0.001 g 3.27 ms 3.27y-
:
;g.z7x 0.001 s
fu
= o,00327 e
Examole 5 Convert 0.087 kW
to
mW.
5"s"fuUon.'
kW*W* 0.087 kw
W* mW
I :
I
,i:
r 'f tr::
mW
:
o.o87
n* *
:
B7w
x
1000w
=
lkw
looomw
lw
=
0.087 x 1000
w:
87 Watt
87 000 mW
Example 6 Convert 670 0OO me to ke.
Solution:
mf)* fl* ka. 67O O00
E, ;r.
mfl :67Oo00 mfl
*
lQ l000mC)
=
670sl,. lkQ = o.67ke 1000c)
i&: d:,.
?amaaui.s
5' &l Dill: &,' €:.
F''L1
f'
I
60
kmlh:
.--
3600s
)F.f
76.67 mls
Example 8
)9
Conveft 9.8 m/s to krn/h. s.B m/s
{ Example
=
n,tl. " l9?0t ls ,1*. 1000m th
= 35.28 --- km/h.
S
9
>{
A speed-trap mission in Utara-Selatan highway has determined the speed * '-* :O lirnit as 110 kilometer/hour (km/h). Express the speed in meter/secoitAirf6f*
(&* // -/.,Y .,''
Solution:
g"i
factor 1 km = 1000 m and t hour = 3800 s
provide the conversion
rewrite the factor of
s
*o
/;,
1
1 = 1000 m/km and
t
hour
11o krn/h = (110 km) (1000 m ) (
m
km
i
U36oA hr/s
t 3G00
hr) = 30.5 m/s s
Exarnrfie 10 'A capacitor is with capacitance 1.2 x 10-6 F. what is its value in picofarad(pF)? Solution: provide the convers.ion factor 1 pF = 10-12 F
rewrite the factor of 'f a= -
r
Tdtr2
1.2 x 10-6 p
pF
F
1 -^r? =1012oF
= (1.2 x 10-6FX
F 1O12
pF) = 7.2x 106 pF F
z..t:r^-=rir.
E 4 tl n:
4 IH rE
BIOO4-TECI{ISI CIA]\I SCIDNCE
ACTIVITY .
1,1r. State: a) base, quantity b) .derive quantity L.2. Convert a) 56km/htom/sM b) 40 N/m to kN/m 1.3. Convert a) 0.37 meter'to.decimete, dt b) 5oo miligram to gram t l*-^ Q, L.4. Convert t' '
1,5.
a)
700 mg/cm3 to g/cm3
b)
900 9/cm3 to kglm3
,l-o\i
Convert 90 km/h to m/s.
t
tn:
l0C6w4: \j
FEEDBACK
1.1. a) b)
Base quantities its length, mass, Derived quantities its area, volume, velocity and pressure.
L.2. a)
56km/h
iooo = 56x-60x60
b) 40 N/m
4o k*/.n = 1000
m/S
= =
15.56 m/s 0.04 kN/m
1.3- ^i-3 0.37m = 9Orn : 0'37x10dm = 3.7dm t0-, \ 1\!y,{st 5oo mg = 5oo x o.ool s = 0.5 dN)) 700 ms/cm3 = J9L sTcm' = a.7 slcm3
.\r'9 ^N \00u
'
b)
1.s
:
,
n,
e00 000 ks/m3 [#U)*, [#) -' = eokm/h = [H#) '" = 25 mls e00 s/cm3
la
'ITEKOU}(IW.
:
lrit
'Pr. Measuring instruments are v-eryimportant in order to measure physicall'guantiiies-: .''- -- -'-''-:--' rEe'rsrrrLJ' Three equipm'ent for measuring length is:
' i; . iii.
migrometer screw
gauge
rt!'
a:
t! f,i -t' 'I: ;1. ' f::
f.
'{t
Micrometer Screw Gauge Vernier Caliper Meter ruler
*
€';
d:
Micrometer screw gauge/ Tolok skru miirometer
ij: ll
Used to measure very small length such as the diameter of a wire Has two scales: (a) Main scale on the sleeve(skala utama mengufuk) (b) Circular scale on the thimble(skata bidal membuiat) . Smallest divisiorr is 0.50 mm Can measure lengUi acolrately up to 0.01 mm
'.
.
RaJche(race0
ry of lass,
f,
Q,
scale
: 35 x O.Oi mm = 0.35 so, actual reading is: (6.5 + 0.35)mm Vernier
Ir..
= 6.85 mm
:isr
aM
b)
Vernier Calipers
es or
a
units
a
a a
Areen
actor
rtity)
a
*; ti: -g:
Use correctly the following measuring equipment
a)
:"{;,
tEi
'
ruler
-
E"
Used in rneasuring length Smallest division is 0.01 cm (0.1 mm) Can measure length accurateiy up to b.ot cm Useful in measuring the inner diameter of an object Divided into main scale and vernier scale jaws (rahang
ctors
et -f l*"'
o stJi).- 3e7"_
II
e' +..
q: 2'i
9' 'F:,:
tabung
uJl
Meter ruter
c)
Used in measuring length Smallest division is 0.1 cm (1 mm) Can measure length accurately up to 0.1 cm Measurement haveto be recorded accurately to 0.1 cm
r--
!.-in'
,*
.1
;ir' .i
.i
'
Length of object = 4.2 cm
ACTIVITY
1.
Discuss
a. Scalar quantities and give 3 examples b. Vector quantities and give 4 examples c. Mornent
d.. Weight
2.
Change the following units: a. 1O0 kg - 50 g (in gram)
b. 1O tan (in gram) . c. 15 micron (in meter) -' d. looTtTj (in m/s) e. 10 9/cm3 (in unit kg/m3)
( +4^
* r.=Oo h5
t2
.
,.1'.r'. E.r'
,,":
.'t
Exercise
t,
-
l.l
t. Convert (a) 3 x 10sm/s
'i
liot x:rol ,
loh^ rA' t> n;-,9
,
l@lh (b) 30tu1h to mls (c) 1.3 kg/m3 to gl*tt
*-t
'ro
{d) 7.g glcnf tn kdr.3 i"i ZSfr2 1,o n"(D 45 d to mm3
to
*
=Li{
:
= |o
'
cm
til;;;""
t ' 3. calculate the volume of a spher-e ofradius l0 cm in V (a) run3 (b) cmt (rl -t ,'
:
'
4.
'
':
.1,
Convert
(a) 60 ms to h (b) 20 pm ro to m \D) zv (c) Is s
(d) 30 pA to mA (e) rwu 1000 km to cm (0 400 JkgrICr to
tops
5''
u I
JgrKr - :
ci,
,ffi
I3>r'ro-{gc*1
Convert the unit of the following quantities and write the final answer in the standard
@.
>o
(a) 13600 kd*t to glctl3 (b) 6400 lsn to m (c) 6x l0-7m to ;rm.
prr *t'r
.s
)o1,m K to- \
'
\
lf".
: )* tO -t'-
l-l^
t tr' '-
,*
\cx
tsoO
nt
i9O c ^^ lO *:t.. .:
1;g
MEASTIREMENT oF. LENGTII
\
Measuring instrrments are very important in order to measurephysical quantities. Three important things should be remembired when using measuring iorn r*rott
l,t-:
(a) All measuring insfiuments have tbe smallest
scale that tliey could measur€. Therefore, the decimal places given by these instuments cannot be s'inaller than their smallest scale.
.t
(b) The zero elTor of the inshuments
must be determined before measurements can be made. Zero elTor is a non-zero reading shown by an instrument while it is not measuring any object
'
(c) To get a true reading, we need to subfact
the zero error from the obsbrved
reading.
Observed reading - Zero
\.o
ewor \=
ir
:
@-'o o)"
2. Cilculatethe *tifi#"^ area of a circle ofradius rvuru l0 in -ii-;;"o':* -'
/....
^q'b'
(1.1)
l
e
;!\
srnn f,
'r:--l^
( qd$**
.i..\
t:..
L
{c:
lg d,0
I
1
It
g;c;]
-f
*
MEASUREMENT
Metre rule
ta
Mefre rule is a measuring insfiument that is commonly av,ailable in the laboratory and its smallest scale is 0.1 cm; Thereforen the reading from a nnene rule must not be more than one decimal place in the unit of centimete. The zero efiot for a mefie rule is the end *0" elTor, wlich is the end portion of a met'er rule which does not give an accurate rcading due to wear and tear.
Vernier Calliper used to measure a length of between 0.10 cm and 12.00 cm. Figure 1.1 shows a venrier calliper which consists of the
Vernier calliper is a measuring insfiilment that is usually following
'
components:
'
?.j
6y Outstde jn+,s to measure the outside diameter or the length of an object, (b) inside jcws to measure the the inside diameter of an object such as apipe, (c) tatl to measure the depth of a hole or a test tube, (d) main scale and (e) verniey scale.
?.
insidejaws
main scale
vcmicrscalc
outsidejaws
0
N
V**
Main scalercading = 1.70 em Vernicr scale rcading = 0.07 cro
Rea
= 1.70+0.07 =l.77cm
Fig. 1.1 Vernier calliper and how to read it the main scale, the length I crn is divided into 10 divisisns. Therefore, a length of I division at the main scale is 0.10 crn. At the vernier scale, a length of 0.9 cm is divided into 1O divisions. Hence, a length of 1 division at the vernier scale is 0.09 cm. Therefore, the length of I division at the main scale is longer than the length of I division at the verniei scale by 0.10 0.09 0,01 cm. Figure l-2 shows the 'length of the main scale compared to that of the vernier scale-
At
-
:
l4-
.'
,10.tcm
0.1-0.09;0.01rn 0.09ca
Fig. t:.2 The Tain and vemier scale on a Vernier Calliper ',
ZeroError for a Vernier Calliper
E
v o
vtzaicrsrb
Zcro qror
-
0
ok
t
(\+q\u-rs's'. rtrnltr Fc>s
i+.\ra ac6 C.?t o ?
;
r-c.96..\rr-.1-< ?
T*roantr = -0,02crn
Fig.
1.3
7*ro etror for a vernier calliper
(a) no zero error, (b) positive zero error, and (c)
negative zero error. ,!],
r5
MEASUREWNT
When the jaws of a vemier calliper are closed without any object in between them, the reading shown by this vernier calliper is its zero error. There are only tbree possibilities of this znta efior, which are: (a) no zero eror (zero e:rpr = 0), when 0 o.n the vernier scale is in line with the 0 on thi main scalg (b) plsitive d*o *ir,when 0 on the vernier scale is on the right of the 0 on the main scale, and (c) negative zaro enor,when 0 on the vemier scale is on the left of 0 on the main scale. Figl.3 shows howthe tbree zero errors ateread.
True Reading for a Vernier Calliper Table 5 gives tlree examples of tue readings from a vender calliper. Table 5 Three examples of tue read'iggs from
W ffi
4cm
'
.X
5 crn
\l(r
Tnrereading
L. 7*roerror :,*0.04 cm
=4.46-0 = 446 om
Obtained reading = 4.46 cm a\
4
0..b
\l'
cm
.
.o5 W
Tme reading
; tl;
o5to
0cm
I
*.'
lcm
Zero enor =O
vernier calliper
Obtaineiiieading
Z.erc enot
Ocm
a
Truereading =4.46-0.04
=4.42cm
Obtained reading= 4.46 cm
; i.
',0"-
\\
lc:n
4crn
5cm True reading = 4.56
- (-0.03)
=4.59 cm
s5 Obtained reading
t0
:
4.56 cm
#re
"
I'EplE 2
FEtrllrnlk Prurdlr $iltm, sdCnffin
.frd ,zir shS
88101 Engineering Science - T2
TOPIC2:LINEARMOT|ON Most of the physical quantities encountered in physics are elther scalar or vector quantities.
Scalar Quantities lsytyquantity is -defined as a quantity
that has magnitude only. Typical examples of scalar quantities are time,6"-Tji"rperature, and volume. For example, the units for time (minutes, days, nouEEtdJ repiesent an amount of time only and tell nothing of direction. Additional examples of scalar quantities are density, mass, and energy.
Vector Quantities
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A vector quantity is defined as a quantity that has both magnitude and direction. --"::' Velocitv. '"'3'v' acceleration, and weight are ve3tor
quantities.
Linear motion Linear motion is motion along a straight line, and can therefore be described mathematicallli using only one spatial dimension. It can be
:
uniform, that is, with constant verocity (zero accereration), or Non-uniform,that is, with a variable veiocity lnon-rero acceleration). The motion of a particle along the line can be described by its position x, which varies with f (ti*:| An example of linear motion is that of a bal tnrown'stlfi;; u[aio talling nact straight down.
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Distance and displacement are two quantities that are used to describe motion.
1'
DISTANCE is the total path lenqth trave quantity and the Sl unit is the Metre (m)
Example 1 : Distance travelled is the length of the path taken by the person from the initial position to the final position.
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It is a scalar
.\ .'ia+.i.,:{r,:;3
I l\
l-
88101 Engineering Science - T2
Example 2
^\ Distance travelled is the length of the path from A to B
\
"-\.\J I
I
I
l" l./ lil,/
Exannple 3 Distance travelled is the length of the path AB+ the length of
the path BC + the length of the path CD
*.
l.'--,-/--
\
thrrrugh all the turns
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DISPLACEMENT is a vector quantity and the Sl unit is the metre (m) The displacement is a vector representing the shortest path from the initial position to the final position. lt is thus a vector quantity. Example 4 Displacement = 10 m to the east
or Displacement = 10 m to the right
Example 5
eF<\
Displacement = 10 m 15o clockwise from the horizontal,
\
\,r"-
)[--^1
t-'\----/r'/' -H--}"
i \/\-/ I
Example 6 Displacement = 15 m 150 clockwise from the horizontal
________l'':---
-1" c
f
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fr
88101 Engineering Science - T2
.g SPEED is the distance per unit time, or, of quantity and the unit is the metre per second (m s-1)
*.lrr**-cc. "G;€It is a scalar
A boy inoves from point A and walks a distance of 50 m to the point B in a time of 20 s. Dista*lg-trwelled
>n??r! = '
Speed
&.
TimetakeFit
of
tlze bay -
ffii
:
Z.C ns
s-l
VELoclTVi*ltisavectorquantityandtheunitisalso the metre per second(m
s-1)
A boy moves from point A to point B performing a displacement of 50 m towards the right as shown in the diagram below. Irelocifr.-
'
V
@
:i =lctiiy
-
D
ispl ace?tlent
Thnetakew 50
f
ne &ay
-=
*it} = 2.5 n, :
-1
io*'cr
55 f ft
: ng ftI
Acceleration It is defined as the rate of chanqe of verocity with time. So, it means that in order for an object to accelerate you need to have a change of velocity over a period of time.
There are 2 ways in which an object can accelerate that is its velocity can change over time, i.e- the velocity can increase or decrease with time.
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88101 Engineering.science - T2 c&eagefn $elorify
arcefej"cfisf*
= *rue fckenfor c*c$geE fnlreptare
arr:eJsraflsl.
_ {flftei
r:ei.acil.y
-
entfmj L'e}sciry,} n--
fimetaire$
?f-f; E
[The acceleration is a vector quantity since it is a vector quantity divide by a scalar quantity.] Deceleration happens when the velocity of an object decreases. ln calculations, a will be negative.
5 EQUATIO": oF
'iulial^(U) .f,rnq, (v) " v(\or*,' , -J 'r Jq
MOTTON "$ A.",J7\a6a,r^4.6
acceleration = (change in velocity)/time taken (finat velocity =
a
^c(t
-
{C-q+tv
^,
,,
*t f('{'l'dr-*'$-rer^
*""'
initiarlrJr"i,vl /time r"nJ;
=(v-u)/t
v=u+at v is the final verocity ; u is the initiar verocity of the object. The area under a velocity-time graph is the distance travelled by the object. Area under graph = 1lz(v + Hence
trtii
u)*t
vdultrtrytl vdqftflrytl
s=t/r(u+ult
v=u+at s = t/r( [u +atJ + u)*t
s=t/r(u+at+u)"t s=tlr(2u+at)*t
Titnefs
\__--)a__J
s =1/z(2ut +at2) G--tr=t
S=ut+1lzat2 fMSKlsllee
BBl0l
S = ut
EngineeringScience
- T2
+tlzafi
v=u+at ,thus
t-'
fr-u)t
s=t/"(v+u)*t,thus
&
,= |t*,
*"1*
rt
Multiplying on both side by za lzcl
*s:
fanl
-]e*
+'l&;s]
Zc;c: {u +:a"}{v-u) 3as
=
Ttrs
-tf *vz - wt
Zgs:tsB-st3
f=u2+2as
'@
,F.r
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u
,':
ii) Total displacement
.,, ,. - ,l
The area under the graph = displacement V (ms r) a c((\(:-c-tlie
o
>
V
A
-1,, : A vetoilS H L = 4aie-rn A f,\^t c+
A.nth.
.r.Ir+t., 4a Onnj
,&
L,--
r(s)
? Soltttion:
=10-0 ^rcaOPl-=zKy
\""o'ck-
aPQ=gms-2(vr,r16"-; = 0-10 oQR--l-=-)ns -2-
son--];g+to)0b1=7so,
s
I
,@
ExamPle 2 the vehicle moving with constant acceleration? which sedion of the graph shows
V(ms-r)
roat,-a-t..rt^.iJ.. -, q= ,,-,^.-.t G i'^s B (P^E+A+ "at"d! +
I
l:
c6.'sfu1* dCcctarrztr9
7s asjt.z"g qcc
n
Solution:
v (ms-l'l
L
G
E.Xampls r
A bus initially at rest rolls down a hill with long will it take the car to travel 150 m, the
:'
I':
u=0mls (from
rest)
Splgtjw. Given:
'
Find: a)v (velocitY)
b)l
of the hill?
:i'
'
(time elaPsed)
a=5-0mls2 s = 1502r
The distance to the bottom of the hill may be found
s=ttt+latz
tr'\'
t5om= o+*(5.0m
,':!@Examoie
t'=
4
J6 =7 '75s
A driver is driving his vehicle at a velocity of 30m/s when he suddenly accelerate his vehicle to 50-sals-Itlakes 3 secontC6r tjr-re vdhiEte to reaclr 60m/s. Determine the amount
&
vehicle in condition mentioned
".rypne
Solutian:
q V g
Initial
velocity -=
velocity = Time taken Final
Q=-
60
30 m/s 50 m/s 3 seconds
-30
ry
a = l0 ns-2'
Example 5 An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff
Solution: Given: Find:
;
a=3.2mls2-
t=32.8s u=0m/s
s=??
s=ut+0-5at2
5 = (0 m/s) (32.8 s)+ 0.5(3.20 m/d)(32.8 s)2
Answer:s=1720m Example6
A car starts from rest and accelerates uniformly over a time of 5,21 seconds for
a
distance of 110 m. Determine the acceleration of the car: ii,
I
' ,.
Sotution:
s =ut+0.5atz = (o m/s)(5.21 s)+ 0.5(a)(s.2r 11Qm=(13.57s2)a
110 m
s)2
_
a = (110 m)/(13.57 sr) a = 8.lO m/ s2
/
pcliteltnil( Frenier sulf€il sslahuddiil .4bdul .*eia $rsh
88101 Engineering Science - T3
TOPIC 3 : FORGE
':i
The Goncept of Force Definition of Force It is a quantitative description of the interaction between two physicat bodies, such as an object and its environment. lf is a push or pull that one object exerts on another which produce or tends to produce motionq sfops, or tends fo stop the motion
* Force is a vector. The Sl unit for force is the newton (N). One neMon is equal to 1 kg m/s2. Force is proportional to acceleration. ln calculus terms, force is the derivative of momentum with respect to time.
2 broad catego;'ies of force
Contactforce is defined as the force exerted when two physicalobjects come in direct contact with each other, eg. frictionalforces, tension forces, normal forces, applied force..
Action-at-a-distance force,such as gravitation and electromagne$c forces, cqn exert t*tun theQrolntbracting objects themselves even across the empty vaiuum of space, result "u"-n physical contact with each other. are not in
Weight and Mass Mass is a fundamental property of an object, is a measure of the amount of material in an object, being directly related to the number and typeof atoms present in the object. uJ e ^/\g weight is the force created when a mass is acted upon by a gravitational
field'
Weioht deoend on the acceleration due to oravitv a vector quantitv measured in Newton w=mq
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,1
:tJ
Mass is a constant qlanlitu a scalar ouantitv measured in ko m
-
\o
eu'at,4t
Fp
e
r*^b.
BB1 01 Engineering
Scienc/fS
Newton's Second Law -Newton's second law of motion states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
F= ma
F m a
is the net force in Newton (N) is mass of the body in kitogram ([g] is acceleration of the body in m/sz
The force applied is directly proportional to the rate of change of linear momentum of the object. An object will accelerate in proportional to the net force acting upon it and in inverse the object's mass.
pr"fii;;'i"
i.
*Ir;.
ti
#
E
rn{
Forces in Equilibrium According to Nevvton's First Law of Motion, an object remaining at rest even when !r9es are acting upon it does so, because there is no net forie acting on tne oolect. This means that the resultant or vector sum of all the forces acting o-n tn" object is zero. An object in this state or condition is said to be',in equilibrium.',
Resultant force A force which is the sum of the net force adgd o!_lhe objec! It is a single two or more forces acting on an object. It is the result of two or more forces actiig conjoinily.
rorc
..l#."
Effftf---:r
t:i.
-iieuE t: '2'eclers i: :is
,,
sr*
C**iasl
F;gu€
2.
\,*ic6
thecpNsh€di€ition
a"
fM=Fi+F?
irgure3jPsallelogramconstrlclio[ ;;2 ior 4lding lectors
in
Filue j:\jectors
rn dfierem
L'**
=:il
Ji€alions
fMSKy'sllee
t
k
&
f:
,,..
.--:
.
88101 Engineering Science - T3
Resultant forces can be determined applying graphical method (the sum of vectors), parallelogram method (tor 2 forces only), and component method (for 2 and a ove forces).
Resolution Method (Gomponent Method) Force can be resolved into 2 component vectors, which have magnitudes,
0 FF Fsin g F,= Fcos
il
horizontalcomponent
verticalconiponent
Each force is resoved into its x and y componentsrwith negatively directed components taken as negative
The scalar x-component F, of the resultant force is the algebraic sum of all the scalar xcomponent. The scalar y-component, F, is found in similar way! With these csmponents known, the magnitude of the resultant is given by
ft=
r,Y
*(Ir,I , h
lleafun
ln two dimensions, the angle of the resultant force with the x-axis can be found from the relation
tan9:
B Y
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Zr,
E4
r S 'h deTrcz C')
.
fut
rn
4n
c4
clLie
I
l..l fjf,
&y"
,'r grnol
ttt
+ "il( 4l* f*uo *kn is no np*ftve.e.
b h
"rAj
exuir;rr;s,v1
if
resur/a46y v.edtf Euet * +i 6lird is 2tr
(ratuplzs:
/. bolctt*tt- 7 egt;/rLt;u4,t
g6
tlv^* *,ycl'g ri ll.! $sJuvt cqn 6e gq}J 'a'
-
F^r=fN
L)
o) F>=taN
16o' Fr= 4N
Fs =-2 0t{
@
F=" I
F:F,tFz
t,+F = F.+Fs
= €+to=16N
Frrfs -Fr =(2cor60')tl0-4 = | +1o -+ =f'TN
F=
6) Fr=TN
F,= $N
d)
F>= BN
?s = l6N
Fl{-F++F
F'+Fs
Fs
F=?
F
7+16-8-5 /oN <)
h=toN
(+0,'
=?N
F+F'= Fr+Fs t-- Fr+ft -F t ': 5+t- 3 =?N F-
8N F= 3
Yz=3ON t-
r.a F_
t--
Fg*F, + F
F.-Fs -Tr = 30 -(g
cos
+o')
-
l0
eO- 6.t7 -(O =
l3.sT N
i
I
t'
.: nrf in eTTqlibsu- !
tl,r. rtnl fi,r'cL tw each uanglr-; (
DztrQ.rtwr'r*
:r
a)
-ffiF,=3N l-r=fN rueffo.ct = F, -fz = 3-J= -qN ar 4 N {o tt,e" k++ L)
Fs= 6 cos 80o=S.tq€N
N{t ft}rc.o = F, -Fz +Fg = tt-5+ E.ne = (/../?6 N C +vc
c) \!'7
*o +t,.r
ub[*
!
t'=t'* W I F''+t P'=:N
Notftre: d)
un-enlt9
? :Pt'=
-,
c'vzh^eAr^e
F, =tOS
J*''-ovJs | )
I I
Fr=!0N 200
v
Fs=l?fr :-oN Fr= 3N
Nclf,vcc---Fs-Fr-F=
- 2okq.+-3 =)
=
7- GN C n"a,ouns ,.l,pwo.nd." ! )
.Fr
tO c6s rto
q.+
.
Resol,^hon n^Lfhsd / Csvrpnar,* ,t^"aJks4 Tv det+t wae rest^tk
*io e;rel Q.xqru"Fk
t*
f,rrczs
o{ f
o,,td ehoW N
arl ad;1g
an
.
I:
Cr.J
cltf4f€ {Az resu!fiqn'* *vcz rnSmt24o A6r^r'a I :
I
-. 3N
I
t I t I
60'
t1
2N
hgwe F(N)
I
F" = Fcos $
Fv= Fsvro
3
3cos 6O0 = l-S
5
F
2
3 $rt 60' = )-s?8 _s }rb t&O' = S(o) = 6
cos t8Q' = _S C-D = -g
cos )}Ou
= r(6) s- I
.).
5F^ = -s.sN
Rpsr.llar?*
$''n )|0' --'2Gt)
E' F7 = o.sgg
-l
frvcr, Q = J€F,l'+6Qry =
/crffis?sr 1r.15;;g!l
rane=
=
= 3.ssN
,ffi
€E = *-T = -o-tro?
qd::.
F = lo,rt-{-o.t7o7)
=-9.7. = ffi'
lg0'_ 1-1. -_ tVO.g-
6
BB1O1
Additional.....
Movement with balanced and unbalanced forces A car or bicycle has a driving force pushing it forwards. There are always counter forces of air resistance and friction pushing backwards. You need to know how these forces compare if you are to predict what will happen to the speed of a moving object.
.
lf the driving force is greater than the counter forces, there is a resultant force forurards. This will make the car speed up. lf the driving forde isless than the counter forces, there is a resultant force backwards. This will make the car slow down. lf the driving force is the same as the counter forces, there is no resultant force, and so no change in velocity.
.
.
ilf'f I
s'e#$
F,i::$ffiJ driuirqlorae
redionbrce
I tr
lf the car is already moving, it will carry on at a steady speed in a straight line. lf the car is not moving, it will stay still.
o
fidi,Bl airres{stere
Newton's First Law Of Motion There are four changes that can take place if a force acts on an objects: 1
-
23. 4.
The speed of the object will increase from zero if it was at rest and the object will move in the direction in which the force is acting. The speed of the object will increase if the force is acting in the same direction as the the direction of motion of the object until the force no longer acts on tie object. The speed of the object will decrease if the force is acting in i direction opposite to the direction of motion of the object- The speed of the object might decrease to zero if the force acts for long enough. lf the object is moving and the force is acting in a direction perpendicular to the direction of motion of the object then the object will move in a circular path. When'the application of the for"e stop" ine object will leave the circular path and would move in a straight line in a direction tangential to ihe circular path.
These four changes can be combined together to produce the Newton's first law of motion as shown below:
Unless a force is applied, an object will continue in its state of rest or uniform motion in a straight line. Another way to write this law: lf a force is applied (i) an objects that was at rest will no longer be at rest or (ii)if moving with a uniform will no longer have a uniform or constant speed or lastly (iii) if moving in a straight line will no longer move
f
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in the same direction T
88101 Engineering Science - T3
This law is thus explaining that if a force is acting
ol
?3.
object then the object will experience an
acceleration. However we are officially learn this only when we see the Newton's second law of motion. Newton's Second Law Of Motion lf a force acts on an object, then the velocity of the object changes. The more force is applied, the greater the change in velocity. This must mean that if the mass remain constant then the momentum of the object changes. And hence the rnore force is applied the more the linear momentum will change. A force F acts on an object of mass m for a time t such that before the application of the forcethe object has a velocity v1 hence a.-linear momentum of p1 and after the force is applied the velocity of the object is v2 hence the linear momentum is p2. Before the force is applied velocity = Vr ond linear momentum =
p1
After the force-is applied velocity =V?2 ?rid linear momentum is
p2.
d*b.
Now the Newton's second law of motion states that the force applied is directly proportionat to the rate
of change-,of linear momentum of the object. s,ral* du
fsrnse
oy' cfrcnge
.}f :i|n6arflrrilEsr?:iirnr
Frr ' dt. Fr.*,rl s*.crn**:fs'l*_ Fs Pl-Pr f 'l?llfa t?ttrr lu-
easftcl *ns,$*ssfrg.fr
dr
t
m{vr- vr, ^' Fa 'f {}E-$r'
lf r F
=athen
ocn"m
F=
Anna
lfk=1 Then this will reduce the equation to
This where the famous equation F = ma come from. This why it is also considered as the Newton's second law of motion.
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8B101 Engineering Science - T3
Concept of Moment Of Force Just as a force is a push or a pull, a moment can be thought of as atwist. ln physics, measure of the turning effect, or torque, produced by a force acting on a body. lt is equal stance from its line oi action to ihe pc or pivot, about which the will turn. The turning force a
to
The moment of a force can be workedout using the formula
moment = force applied x perpendicular distance from the pivot. lf the magnitude of the force is F newtons and the perpendicular distance is d metres then:
b.
ffi
moment,M=Fxd
FCr
?\J
in
newton metre (Nm)
It is easier t0 und0 a D0[ using a long spenner than a shoft spdnner. This is because more turning force is produced at the bolt (pivot) with less effort. A long spanner is an example of a force multiplier.
ln a simple balanced see-saw, the forces acting on the left- and right-hand sides of the pivot are the same' This is known as balancing moments. Moving the load lt one end will cause the seesaw to become unbalanced. To regain balance, the load on the opposite side must either be increased or its position changed. This is known as the principle of moments.
Definition of modrent of force Moment of force (often just momenf) is a measure of the tendency of a force to twist or rotate an objeg-t about a specific point or axis. lt is the product of the iniensity of the force into the perpendicular distance from the point to the line of direction of the force, moment, M = F x d
b P ' 's
Turrr;r13
{wce
,
ol:noruic
!
{ho,ue^i,
/s.
a tu.e-as.*rQ. o{ hisv,) ynu& o..fuc.e a&,N6 an Ah 6Ljec} cata.ad lArt {o rshre or *anrt . U.n+ in A1rrr fr*d .
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88101 Engineering Science - T3
Principle of moment of force The Principle of Moments, also known as Varignon's Theorem, states that the moment of any force is equal to the algebraic sum of the moments of the components of that force.
Total normalforce (applied force) = Total reaction force
IFil. =
IF{i
lf a body is in equilibrium: the sum of the clockwise moments is equal to the sum of the anticlockwise moments
& lEH
Sum of moment rotating anticlochrvise = Sum of moment rotating clockwise
Irfut
= Ia.r$\
The principles of moment are applied in solving related problems in context of determining the center of gravity for action forces system in equilibrium, the reaction forces , and to determine the distance resulted in specified reaction force in a system.
If o bdV ls ,h <4,FL'bn^wntt, dnder {AtadNn -f o, nun,L*Q fw*t , -14'rz Ngobyoic g,JM af {A,r Motn z'8 *ryd q W {trCls ^Lo*t anf potnt, }s eyAl fo "RO
"' >M(I
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e
= >M)l
to
I .t t
i"',
'
f\4or wvt's Pvtl.lzvt Soln]ict
f,+F,+ = R, t
@ z*J .= ztr? M= Fxd
Apghad '6rce = Rroc*o,. *rrcz
o* 3";A 4t ma;nlain Jlrq rrqclrcn fohq ,"Jho dig*anoe
Dalovnro c-: )hp c+yt/r-r-
tyt*J- ir, sC-aftc egcftb;u*t \ quvd earroL*iaw sfuof
&al^Flc ,: Root
d
pa
suppor.s o{ roec/ion p a,nd gOfV ruspec^r.,c/y. The. ,aao*on&,
l5M
:
Zrut} =
XF{ = zFt 3o KN + SOKN =
G = 6orN
-€xannple
*-;
c)^ ?r,
513KN
Sd'lroa
.
foied v;+h o toqd o{ 30kA/ "i ; is SoKN. F,d ,,a drsla nee x.
SorN
I
toKA/
MOr,r,-crut
5Mt
,
3OKN (ann)
+ SOkN 05-x)r'a = 60 KN Clsrra 9Ofrlnn + l2oo kNru - BO:c.Kt.lrrn gOOrfw, = 8ox t$lr"r = ( go + t)oo - ?oo)fNJv "
+ O, al P
3?o- 9'87sm x^ = - t'd =
a:
F"a
+qq l^
l.Srn S"lu.horr
2.s r!1
3
2rvl
rve
o{ A ond B.
I
:
xFt _ xF+ Ra+Rs= lo+trot40 Rq = 7o -Ru = 7o- 4t.e7 = 29.33 KN
,
Xvr
a
= Zrq9
lo(rs) + 20 C+) + Ae 0) 15t.80+2go=gRu Rr= 37s .il L,1. e7 rN
=g
Ru,
.: 'i
€xan"ple 3
:
'.,-i
bzIero^va
{h,e
o a'\
.
ij\
Ccn*er
*
?Ya.v+-2
FA
+r\.s
Vs{otr x
.=sn eXu;Llo;uaa.
8N B
_x Asswvc€ {A-t cm.ltr op
ga;la
}s 'l+
X
cl-s*a,,rcs
O frro h.rotvro,".* rvra{l^od : 5M(r, = :Mp (ts*-n rs)Cx) = :o c6-x) t 8cg-x) 6.34x = /f,o-)ox +7)-gf 6.3+x + fox- f g r. r?2
{on. pur'^* A
,
'?'m
__
X=. =
9
Rosuilar.t
marvlu.tvL
w
(92
39.3y
s.S?nr
nlqlhad:
(tssnrs')Co) + ro(6)j 8C?) 6.3+ * to +8 +71 = g-39rrt 3+.3+ belerrv,^ne J+< eg^-her- + q,,o!;tu) so J4o,f -tl-e ngstet^ wXI, &. rn
= O+llo
Fg/\^* ,F:
i
e-11Wr(Ila*u*t.
<)-tvl )l 4\,CD
-\
Fer* 'c6vr,rad-ro,..dhod Yir\A) : En4J
.'
(m)cx->) +( l+)(x-3.9 +0s s'rn ns9(x-s-s) =1-sXle-s-x) . S-o)L-ro-o t t+x-49+ 6t.++)c-337 -12-= -17.9t 9x 5(tx+lq-x +6{.?9x -"sx = -_17-s+too+ *9 +3i1.q2
l>0-4*x -- 4o9.42
f,= 3.rf
@ fus.rtlah*
rn,r
srasn^eof
T*JA'J
; )(:#-'
qoq -(|-2 (t0+q4 + 5ir-?z- rr-S -_ lao'9rF l>O'9r} Co)t9 +Q+) (r.9+(:sinu$Gs) - ECre{
5b + l+ +(7s
s1h
Its)-E
._i_.. : _
-c
I'EPIE 4
PElilu{nft
F!d!r $ltm
Sddrd*1,
ftfd, ^Ai'
Sfrh
t
BBl0L EngineeringScience -T5 i I 'l .t*i'3.w
y:'ry*:*
Topic
4:
:r'!':r{'''
WORK, ENERGY & POWER
WORK Definition : The product of an apptied force, and the displacement of an object in the direction o! the applied I
force.
Work = Force x distance in the direction of force W=
F x 6=(m x
;
W=F x
d
a) x d= 1m x!1xd
t
Therefore, the unit
is +
kgx
L s
x
xm
+
kg-' +
Nm orJoule
s
It is a scalar quantity, with unit in Joule. Joule : Work done by
i
force of one Newton acting over a distance of one meter.
1 joule is approximately the energy required to lift a small apple L meter straight up. lf there is no motion (no displacement), then, there is no work is done !
Mechanical work , I W = Fd cosO-l ; more general equation of work when the force the object's displacbnEnr
J
is at an angle 0
to
ENERGY Definition : The ability to do work, or, The capacity ol o system to enable it to do the work, or,The capdcity to do work. tt is a scalar quantity, Sl unit : Joule
The law of 'Conservation of energy' This law states that the total amount of energy remains the same for transformation of energy from one form to another. Kinetic Energy The energy porsesseO by an object because of its motion. Kinetic energy,
rr---T---] |mv, Er.
I t2l =
I
Potential Energy The energy possesse{ Potential energy,
ry.tn !$eracting object waiting to be let loose,
due to its position.
F;*
Gravitational Potential Energy The energy which an object has because of the gravitational interaction.
sllee
1,
Engineering Science -T5
POWER Definition : The rate at which work is done, or the time rote of doing work Sl unit : watt (w) ; joule/second
;
1 watt is, the rate work is done, when an object moves 1 meter per second against a force of 1 Newton. [ln term of power of machine, t horsepower is approximately 746 watts !l
p t
aw -
Fd _; t
-;
-
Fd
c.os
tL_
work -=- time P_
o 1
fy
=|^@ forcl
)(
displaceme nt fime
)
force
velocity
mgh
t
Worker A and B were given a job to lift a parcel from point C to point D. A managed to have this done very quickly compared to B, so, we say A has a greater power than B!
some energy lost because of friction etc, so, not all input energy goes into doing work
Efficiency
= P-, ,lO0 yo P,n
sllee
!
The Lgw of Conselr-ation of f Delg\' Erag.r in a a.saoar.talic mraiu bms (cg. kinaic" pot@ial. cargies in ttu sistm is a co;tat.
fle
hu!
li.$r)- Th. lrtr of
m;m-aiou cf arg'satcs dat
aFrg-t ma:: neirhr bc crcacd nu dcsto-'cd Tbuatac $e
m
sf all $c
mo;t cooo..olv u;ed t:aqpb L Se pcaddru*
a
1\
------+ dIe potanial or31- is:
1!r fomrla to cdorldc
= (to r.6) tc.
PE = ngh Jlre mass ofdre bal =
Substitute
i
-O,
6r ulues
irno
=!-$ 6 5'11 6e fonula ad yw gcc g
*. Or. bal is +prochins
thc
r*s-) Lo'zd
=lq.L 9fld 6Joule v S2-
i0ke
ThcLeiglt.h=0.3m The acelentioa dne to gnr-itr'.
8'
guplr ball posirim iie P= is dccruiq
liic
$c
X! ! bgca;iag: At
lhlc
the Pot:oial
ciacdi haFsa;'
bttrm
dc bht
ad puplc
ball
positim de ?E
-
fiI
I
i...-..
i:;t i t'
ttr< rositio of
*r pupl:
ball is
drae
the Kirctic E:ilrg;..'is at its raariom
haar
@!) -
|
1
i
-tt *is poic- tbcs:dcaXr.
all
Sc P! ba; randornrn irao
K!> Iluefse mn rh:
K-E
= I9.6J n'hle
*r
Pi
= 0-
I
i iln. |
1\'e
ca
p*irioo Adrcpinli nos'
''aaaod
ball is
rtse&r Porrtial
Eoag;,(PD is osce agah ariis oarimrn
ad&c iibaic !o
und:r:raad tb:a
?I+Kf,=0 PE =
ibc sulr of
Pi ad
KE
i tlc total
-XE
re chaliel eoerg': Total Mcc[adcal Enera' = PE = IE
\OT1: Tlis is uith
fsrrs brro:
of
iLc abscm:r of outside foces ;uc! as
$gigq
dr tn' .ri6.r65g17d6 qf dlirg'r-
CONVERNNG POTENTIAT ENERGY TO KINETIC ENERGY
Roller Coaster Marbles
Slow and clanking, the string of cars is pulled up to the crest of the r:allest point on the roller coaster. One b'' . . -, the cars start downhlll on the oiher siOe, rintit gravity takes over and the full weight of the.train is careening down into cL:'ves, twists. and tums. The roller coaster is a great exa;ple of converstons between potenlial energy (stored energy) and kinetic energy (the energy of motion). As the cars are being puiled up to the top of the first hill, they areicquiring potential energy. The chain that pulls them up the hill works against the forie'of gravi'ty. At the top of the hill, the cars' potential energy is at it's maximum. When the cars start down the otherltde, this potential enerdy ts converted to kinetic energy. The cars pick up speed as they go doerfihill. As the cars go through the n€xt upiill section, they slow down. Some of the l.inetic energy is now being converted to potential energy, which will be be releBsed when the cars go down the other side.
T BBtoI.lLSL
3
PRINCIPLE OF CONSERVATION OF ENERGY :
a
a a
The principle states that energy can neither be created nor destoyed. Energy can be converted from one form to another. Exarrple of energy conversion is electric energy to kinelig energy.
KE
The ram of a
pile-driver possesses mechanical energy - the
ability to do work-
Hhen held at a height,
it
possesses mechanical energy in the form of potential ener.gy. As it falls it possesses mechanical energy in the form of kinetic energyAs it strikes the spike, it applies a force to displace the spike - i.€., it does work on the spike_
'
Energy transfer from potential energy to kinetic energy and the opposite Object of any mass falting verlically downwards from static its loss in potential ,
€nergy
is equal to the amount of kinetic energy gained, assuming that no energy is tost during the
process.
mgh
%mf
o If object is rebound
to any height after hitting the ground and left it, its maximum kinetic energy while movrng upwards will be equal to its potential energy at its maximum height
lrmv2 '
mgh
For an object that slides down a friction free slope, its potential energy is equal to its kinetic energy at the bottorn
mgh =
Yrmv2
,+
Exanplel: height of 8 m from the ground' On hitting the ground" the ball rebounds to a height of 3.2tb. Assuming there is no other energy leavls or enters the system during the process, .
A
steel ball
of
a rnass 2
kg is released from
a
find:
a) bi --ci
the kinetic euergy of the
,
^etl
E
ball a+it+edre+the
ground' J ad^^e-vc
rhekinetic
of the ball onrebound. the velocity"""tfiyoftheballasitl@d.
r^anx hl^'(ht nikn
solation a)
the kinetic energy of the bail
3
rt rebcrui"J
it reaches tbe ground.
as
t:*+zas
I t/,
1s6.96
b)
the kinetic energy ofthe ball
as
f
(2) 156.96
J
= (o)2 + (2) (e.81) (8)
=
l-56.96
it leaves the ground on rebound.
Kinetic energy of the ball as it leaves the ground is equal to its maximum
potential energy on rebound E6
mgh 2 (e.81) (3.2) 62.784 J
c)
the velocity of the ball on rebound
Es 62.7841 62.7841
v
Yrmv2
%Q)i t? 7.9236 mls
Example 2:
An object of 30 kg ru$s was lifted as high as 3 m from
the ground and was let to fall
under the gravitational reaction, Calculate the gravitational potential energy aud the kinetic ettergy possessed by the object under these situations:
a) before it was let to fall b){ meter under free fall c) right after its touched the ground Solution
a)
before it was let to fall Ep
mgh
30x9.81x3 882.9 J Eg
%ml % (30)(o)
5
i*x*;tr:iig+.+lsi:e+i.€1!*i;--::,F
6
b)
I meter under fue fall
Dp F
mgh
=
30x9.81x2 588.6 J -
Er
.
%(30)(v':) ObJain
v ffi-e
the formulae
I
0-+2 (9.81)
: E6
=
c)
= u2 + 2as
(l)
Vrmvt %(30)(v2\
";f.l','nn' \--1
right after its touched the ground
Er
mgh
:0 "'Er
30 (e.81) (0)
Yrmvz (30) (v,)
Obtain v from the formulae
v-t
u2
=
rl
= u2 + 2as
+ 2as
0+2(9.81)(3) 58,86 n/s
E-DK
/2 Vrmvz Uta
%(3qfh :
88219
J
\--1
Example 3: A boy drops a 1.2 kg of vase from height of 6.4 rr. (a) What is the kinetic energy-of &e vase when it is at a height of 4"2 rrfl (b) With what speed will the irase hit ihe groundt(Neglect air resistance) *-SoIution (a)
First" find the total mechanical energy Since this quantity is conperved wliile it is falling. Initially, with vo = 0, the vase's total mechanical energy is all poteutial.
Taking the groqnd as the zero reference poing we bave
c
i
't
E-K*U=0+mgho p = (t.2kgfu st* t t'\u.o^) E:75.34J wehaveK
=E-(J,
K:E-mgh K :75.34J -(t.z*gft.swt
s'?\+.Zm)
K:25.9J tb)
Just before the vase strikes energy is all kinetic.
fte grouid,@ = O t
:
O), lhc total mchanical
E = K =lmvz Thus,
."\-_, (O
. ':.:
':,...; '.itj: ,'!.
{ i'
1
Topic 4 : Work, Energy and Pourer
a (l) E
o C)
P
) o-
uo C,
At the end
The Law of Conservation of Energy
of this lesson, you should
states that
able to Energy cannot be created or destroyed; therefore, the sum of all the energies in the system is a c6nstant-
r'statelhe ltw of Cooservalion ol Energy
/calculate kinelic energy appling fornulas
The totalimount of energy remains the for transformation of energy from one form
and polentbl enetgy
l-
(o
(l)
J
Changing Forms of Energy o Energy is most noticeable as it transfbrms from one form to another.
o What are some appliances where electrical energy .undergoes transformation?
Kinetic and Potential Video Points
of lnterest
E
O ln many sitrlations., there is a conversion between potential and kinetic energy.
....Efficiency, ... .Energy Conversion.
t"
3
8
Pendulum System
Mechanical Energy
Ep- Polentbl Ek- Kinetic
o The mechanicalenergy DOES NOT CHANGE because the loss in potential energy is simply transfened into kinetic energy.
Ep
Ep max Ek min
o The energy in the system remains
EK
Ep min
7
Erample 1: Calculation of Energr
Trythis out... A coconut falls frorn a tee
Epmax=?
-\
fom a height of 20 m.
'Whatis lhe'Velocity of cooonut just before hitting
.Ek min = ?
q
=.E*
mgh = 14 mv2
tn(9.81x20) =l!mv2 vo
= 4O0
v = 20
Calculation ofEnergy and Velocity ED- Polential
ms'l
Action question..... Amarkerpen of mass 100 gnamis from a height of 2 meter. Calculate the energy possessed by the maiker right at the moment it is being released! [Assume that the acceleration dueto gravity is 9.81msl
Ep=E16=2 Whatis thevelocitv ol 12
I
i.
..
:.
t
t
Summary 1. What
isihe Law of Conservation of
?
Enersy
23hatis
,
tfl
tne foimuta ior
.:::.1:. . r::
,:,.,..kin!tic energy ?
. potential energy ? 2
b
,g
' ' Thank you
'
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to
llJ-II-'.
L
t ulJlL 3
pfililcknit( Frcni$en $ultan Sglshuddin
Adul
eir
shsfi
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BB101 Engineering Science-T5
TOPIC 5: SOLD & FLIIID SOLID Solid is a sample of matter (substance) that retains its shape and density when not confined, it is firm or compact of definite shape and volume, tends to keep its form rather than to flow or spread out likea liquid or gas. Example : Rock, glass, wood , ice, most metals, etc.
LrQrrrD A state of matter, has
a definite volume but not a definite shape; able to move and change shape without separating when under pressure; particles easily move and change their relativJpositions without separation of mass. Example : Water, mercury, oil and etc.
GAS
A
state of matter, consists of particles, without definite shape or volume; conforms to the shape a container in which it is held.
Example : Oxygen, helium, carbon dioxide and etc.
Properties of solid, fluid & gas in term of characteristics of particle (Particles arangement; force between'particles; shape & volume) Solid : Particles close together, arrange properly; bond with strong force, vibrate at a point; have certain shape & volume Fluid: Particles loose and not arranged; bond with weak forces, glide freely; shape change & follow container but maintain its volume Gas : Particles separate far from each other, distribute randomly; very weak bond with negligible forces"move freely & randomly; shape and volume changes follow container.
MASS DENSITY The measurement of,moss content in s sabstance, os mass per
wit
volume; unit is in kgm-3.
It is a characteristic of a body, usually used to determine whether an object or material floats or sinks if placed in a fluid.
'ilp-l+iil
]MSK/Sllee
.
Unit in
kgm-3
m-masinkg p
-
v
-
the Greek letter 'rho' volume in m3
of
l\
BB1O1 Engineering Science-T5
Relative Density Sometimes known as specific gravity - older literature ! Usually means RD with respect to water. It is the ratio of lhe density of a subtance to the density of a given reference material.
For example, suppose an object has a density of 4000 kg/m3' To calculate the Relative Density with respect to water which has a density of 1000 k1/m', the RD is 4000 kg/m3
:4
1000 kglm3
PRESSURE Define as the perpendicularforce per unit area acting on a surface. SI unit is Pa @ Nm-2
Unit in Pascal F
A
-
-
Force in Newton Area in m2
It describes the influences upon fluid behavior.....
PASCAL'S IJAW transmitted
Pascal's Law states thatpressure applied to an enclosedfluid causesforce to be equally in all directions;@orce acts at right angles to any surface in contuet with (the walls of the container! )
thefluid
Application The Pascal's Law states that the liquid is spread to all directions and equal transmission in confined. The weight is changed in between the elevations. The hydraulic press principle, water towers, dams.and scuba divers are application of Pascal's Law. We can use the vessels for
verification of liquid. It uses the pressure as main aspect in application.
IMSK/Sllee
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BB 1O 1
Engineering Science-T5
Explanation for Pascal's Principle Application The hydraulic machines eg: hydraulic press hydraulic, hydraulic brakes and hydraulic based in Pascal's law principle.
lift are
Suppose there are two cylindrical vessels A and B fitted with pistons. The vessels are connected by a horizontal tube C. Liquid is fitted in vessels. The cross-sectional areas of vessels are Ar and A2 respectively. Such that Ar < 42. Suppose a force F1 is applied on piston A by putting a weight on it. Then pressure applied on piston
Accordin! to-pascal's law principle the pressure exerted on piston A is transmitted to liquid in B. Thus pressure exerted on piston B upwards: If Fz is the upward force exerted on piston B, then pressure on piston B
Therefore,
-.
:
-
-
As Az> Ar ,Fz> F1. This indicates that a smsll force Ftapplied on smaller piston A can be used to exert a large force Ft on bigger piston B.
Hydraulic Press Application Hydraulic press is based on Pascal's law and was inverted by an engineer, Bramah. So it is also called Biamah press- It is used.for compressing bales of cotton, paper, cloth etc.
A, Weight of load, F1
t:v
F2,
P=L
D-t2
Al
tr A2
Pump plunger
p :L-
Press plunger
F'
4,, Az'.
n, therefore "
A,=',Fl
Design with Az>Ar will cause smaller force F applied can be used to exert fMSK/Sllee
Upward force
a large force F
2
88101 Engineering Science-T5
It consists of two cylindrical vessels A and B, fitted with water tight smoothly moving pistons and connected together by means of a pipe C. The water is filled in vessels. The cross-sectional areas of pistons fitted in vessels A and B, are Al and A2 respectively such that Al
hydraulicjachs.
Other examples of applications
. . . .
The underlying principle ofthe hydraulic jack and hydraulic press Force amplification in the braking system of most motor vehicles. Used in artesian wells, water towers, and dams. Scuba divers must understand this principle. At a depth of 10 meters under water, pressure is twice the atmospheric pressure at sea level, and increases by about 100 kPa for each increase of l0 m depth.r'r
Additional.... ln the physical sciences, Pascal's law or the Principle of iransmission of fluid-pressure states that "pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure ratio (initialdifference) rernains the same."[l] The law was established by French mathematician Blaise Pascal.t'r
AP
-
Bstxrh)
where
AP is the hydrostaticlessure (given in pascals in the SI system), or the difference in pressure at two points within a fluid column, due to the weight of the fluid; p is the fluid density (in kilograms per cubic meter in the SI system); g is acceleration due to gravity (normally using the sea level acceleration due to Earth's gravity in metres per second squared); Aft is the height of nuiC uUou, the point of measurement, or the difference in elevation between the two points within the fluid column (in metres in SI). The intuitive explanation of this formula is that the change in pressure between two elevations is due to the weight of the fluid between the elevations. Note that the variation with height does not depend on any additional pressures. Therefore Pascal's law can be interpreted as saying that any change in pressure applied at any given point the fluid is transmittedundiminished throughozr the fluid.
Equation: (Pr)ff r) jMSK/Sllee
:
(Pz)fVz)
of
t:1,:::.-1
BB
f0t
Engineering Science-T5
ARCHIMEDE'S PRINCIPLE A body immersed in aJluid is buoyed up ful aforce equal to the weight of the disptaced jluid. p
-
fluid density in kglm3 ; g - 9.81 m/s2, acceleration due to gravity h - height in meter, depth of fluid.
The Principle An iron needle sinks in water but
a
hug; ship floats on the surface of water.
All
objects whether inside the liquid (or fluid) or partially inside, appear to be lighter than in air. This is because liquids (or fluids) exert upward force on the immersed object. This is called the upthrust, the Buoyant force or buoyancy. The upthrust due to liquid pressure, pushes the object from all sides, but is greatest on the bottorn where the liquid is deepest.
Llp.thrrut
B$d!.
Statement of Archimedes' Principle When a body is immersed wholly or partially in a liquid (or fluid) at rest, it appears to lose some of its weight. The apparent loss in weight of the body is equal to the weight of liquid displaced.
fMSK/SIIee
BB
101 Engineering Science-T5
Proof Consider a solid cylinder of height 'h' and area of cross-section A, to be completely immersed in fluid of constant density r. The horizontal thrusts on the cylinder balance each other.
P
a
& p=f
= !9|l"
F=aP =
A1lg[v)
..=::a-:
l:.:-c:::
Consider the vertical thrust on the cylinder.
Total downward thrust on the top face of the cylinder = (P + hrpg)A Total upward thrust on the bottom face of the cylinder : (P + h2pg)A
<- AP+ Qgl",) I + AP +qeU) t
Resu'[an'i'|hrus'[.
HI;!il;;:r* =
mg
:
= M3
Weight of the displaced ffuid
[t#r]
Mg:Mass of liquid x 5 Volume of liquid displaced x density x g) upthrust: weight oftilE displaced fluid. {.1-.
:>?
Applications of Archimedes's Princinple
r o
Floating ship displaces water equal to its own weight, icncluding that of cargo. A submarine sinks by taking water into its buoyancy tanks. Once submerged, the upthurst is unchanged, but the weight of the submarine increases with the inflow of water and it sinks faster. Cornpressed air is used to blow water out of the tanks when it has to resurface.
Balloons filled with hot air or hydrogen, weight of cold air that it displaces. Thereforece, the upthrust is greater than its weight and the resultant upward force on the balloon, courses it to rise.
fMSK/Sllee
BB1O
TUTORIAL l.
1 Engineering Science-T5
5
State 3 differences between solid, liquid and gas.
2.
Define (include with formula and unit) a) density b) pressure
3.
Calculate the density of a rectangular block using dimensions as below: Length :27cm, width : 10cm, height : 30 cm, mass : 9509.
(t1. >t 4.
b5/u 7
Thedensity of airl.29kg/^t.Calculatethemassof airinaroomwithdimensionof 15m 20m x4m.
x
'-,J3,,:-,r, L'
5. Alan uses a knife
with a sharp edge and a cross-sectional area of 0.025cm2to cut open
a watermelon.
a)
Ifthe force applied on the knife is l5N, what is the pressure exerted by the knife on the watermelon.
(' fiapa* b) After that,tr?cuts open a pineapple using the same knife by exerted a pressure 3X105Pa. Calculate the migintude of foice applied to cut tire pineappll.
@
of
0-K/v 6.
Calculate the pressure and force on an inspection hatch 0.75m diameter located in the bottom of a tank when it filled with oil of density 875 klm3 to a depth of 7m.
{=
xts
lr-- 4 rw
'A P= tr01^ = ,,/ ,/ -. 600Ed.
:g-fi=g=75-y^
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: ;4s:23_*u'
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BB1O1 Engineering Science-T5
7.
The figure I shows hydraulic press. The cross-seetional area of piston B iS 20 cm2. The cross-sectional area of piston A is 100 times the cross-sectional of piston B.
350N l00X20cm2
B
Figure
I
a) What is the pressure on piston A that is required to raise a load of 350N on piston B ?
b) If position A is pushed down a distance of 3 cm, how high would piston B can be raised ? c) Calculate the cross-sectional area of piston A.
3nr
0 8.
ffsuoo
2w\
The cynlindrical piston of a hydraulic jack has a cross-sectional area of 0.5m2 and the plunger has a cross-sectional area of 0.003m2. .
i
*, A=
I.l'*
+
-:
f*
E{ls*N
I
0.003n0t I < 1..-
I
I
A**'5nr1
Fignlc! a) The upward force for lifting a load placed in top of the larger piston is 2000N. Calculate the downward force on the plunger required.
ilN b) If the distance moved by the plunger is 50cm, what is the distance moved by the larger piston ?
0 bolwt - 0 -3 cfMSK/Sllee
ft
I'EF]C E
ruilrlclt( Pfinlq: $liil Sdlhdffr frrhrl, fdr,$rfl
tt-
'|'
a.l
+,
BB 1O1 Engineering Science-T6
TOPIC 6 : TEMPERATURE & HEAT WIIAT IS HEAT? Heat is energy that can be defined as the sum total of the kinetic energies of all the constituent molecules of abody, or as the intemal energy of a body. Heat, like other forms of energy, can transform the matter it toucheq heat it and cause molecules to move, or cause a reaction like burning. The Universe is made of matter and energy. Matter is made up of atoms and molecules. These particles and molecules me continuously moving around or vibrating back and forth. The movement of the atoms and molecules create energy; this energy is also known as heat or thermal energy. If more energy is added to an object its molecules and atoms move faster inueasing its energy of heat or motion. It does not matter if an object is cold, it still has some heat energy because its atoms are still in movement.
In a simpler way, heat can be defined
thermal enerry that transfers from a body to another due to temperature difference between the bodies. as the
The S.L unit of heat isjoule. A calorie ofheat is equal to 4.18 joules.
WHA'T IS TEMPERATI]RE? The temperature of an object is referred to as the degree of internal energy of a body, which determines the level ofwarmth or coldness we feel when contact with it. Thus temperature is a measure of the amount of heat present in a body. Or, in a simpler way, temperature is a
measure of the degree of hotness or coldness, and this physical quantity is measured by using thermometer. Temperature is measured in Celsius (C) or Kelvin (K). The Kelvin scale is equivalent to Celsius except that it begins at absolute zero andthe Celsius scale begins at the freezing point of water. Absolute zero is the point at which an object contains absolutely no heat energy. At 0 K or -273.15"C, the net internal energy of a body is defined as 0. The S.I. unit of temperature is Kelvin (K). One unit in Kelvin scale is equivalent to one unit in Celsius scale, therefore zero Kelvin or absolute zero is equivalent to -273 "C. Another example -lVhat
:27"C:
(27+273)K
:
300K,
or xoC: (x+273)K
is o Thermometer?
A thermometer is an instrument that measures the temperature of a system in a quantitative way. The most simple way of doing it is by finding a substance that has a property that changes in a regular way with its temperature. One of these substances is mercury. Mercury is fMSK/sllee
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', BB 1O1 Engineering Science-T6
liquid in the temperature range of -38.9 C to 356.7 C. As a triquid mercury expands as it gets wanner, this expansion is linear and can be accurately calibrated
HEAT TRANSFER The second law of thermodynamics states that energy or heat transfer can only take place in one direction, from a higher grade to a lower gade state. This means that heat will only flow from hot to cold, never the other way around. This tendency will always achieve an even distribution of heat energy in an environment with no additional heat input. Heat frartsfer between objects occurs in three ways: conduction, convection, and radiation.
Conduction When one end of a rad is heoted. the atoms Iin the rod above the flome] vibrate faster. bump into the neighbouring atoms and startthem vibrating. In this way, the atoms csnduct heat from the hot end to the cosl end. But during the process, the ntoms themselves do not move frsm one end of the rod to the other.
5imilorly. heot can be transfemed befween two bsdies e,g,. from o hot drink into and along a metal spoon.
{onduction is the tronsfer of heat as a result of the direct contsct of rapidS moving atoms through a rnedium nr frsm one medium to onother. without movement of the media. Ittaterials that allow hent to travel through them in ihis way are colled condu*t$ru, lfietals are good conductors of heat. Non-mefals such os plastic. clay, wood and paper ore poor conductsrs of heat they are also called insulntorx,
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Convection
Convection occurs when a liquid 0r gas is in contactwith a solid body ot a differenttemperoture and is always accompanied by the motion of the liquid or gos. In the kettle, hotwater rises and cold water descends until ollthe water is at the snme tempersture. In the atmosphere, convection results in the movement of hot ond cold oir ,,. winds!
f;cntmction is the transfer of heot by the physicol movement of the heoted medium itself. Convection occurs in liquids ond gases but not in solids.
Radiation The sun connot transfer heot to eorth by conduction [becouse there is no physicol contactwith the flflrfh] or convection fbecouse there is no liquid 0r gfls between them], The sun heats sur earth by rndiation which does not require contoct or the presence of any rnotter between them,
Rqdiution is the trsnsfer of hent in the form of waves fhrough spdce fvocuum]. Dull block surfaces are betfer than white shining ones at nbsorbing radiated heat, ,""
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4n electric heater trcnsfers heat by rndiatiorrfdirectly rodiating its heat energyl as well as by convection. fn convection, heoted air near the heater rises and is replaced by cooler air and the cycle repeats thus warming up the entire roam,
TIIER]VIAL
E
QUILIBRIT]M
if
Heat can be transferred from one body to another colder body if they are placed in thermal contact. The process continues for some times until both bodies achieve the same final temperature. At this stage, these two bodies would not exchange energy by heat again, , and there is no net heat transfer of flow! This situation is called thermal equilibrium.
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IIEAT CAPACITY (C) The heat capacity of a substance is the amount of heat needed to raise the temperature of the subtance byone degree Celcius (l"C or lK). The S.I. unit of heat capacrty iJJ'C-t or Jl(r
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"-ao Q:C4O SPECIFIC HEAT CAPACITY (c) The specific heat capacity of a substance is the amount of heat needed to raise the temperature of I kilogram of a subtance by one degree Celcius (1"C or lK). The S.I. specific heat capacity is Jkg-r"C-r or JkgrK-l.
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l. Define 3.
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2.
"x
aI
unit of
a) Temperature
b) Heat
State and explain 3 ways of heat transfer.
Calculate the heat quantity needed to increase 2 kg of copper from 20oC to 85"C. Given: specific heat capacity of copper: 0.35 KJ/kg"C.
4.
How much heat does 30g of aluminium gives offas it cools from 120oC to 60oC ? Given: C aluminium : 880J/kg"C.
5.
A l00g silver
spoon at20.C is used to mix tea at 95oC. After a few moment, the spoon and tea achieved 88"C. If the mass of the tea is 3009,
a) Calculate the thermal absorbed by spoon. b) Determine the specific thermal capacity for tea.
Given : specific heat capacity of silver spoon :0.23 KJlkg'C.
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BB1O1 Engineering Science-T7
TOPIC 7 : BASIC ELECTRICITY Charge(Q) Electric charge is made up of tiny particles calted atoms; unit in coulomb.
Q: It Current (I) The rate of charge/elections flow pass a given point in a circuit. The direction of a currenl is opposite to elections flow direction.
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Voltage (V) An electrical force that causes current to flow in a circuig measured in volt. It is equal to the amount of work per electric charge that an electric source can do.
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Resistance (R) The opposition of a conductor to the flow of electric current. It is the ratio of the voltage to the resulting current flow; SI unit is ohm (A).
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p - resistivity of material in Qm; L - Length;
A
-
Cross-sectional area
Capacitance (C) A measure of the amount of charges stored on each plate for a given potential difference or voltage appears.between the plates; gI unit is Farad (F). t-',
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Q
V
Farad
:
I coulomb of charge
stored due
to lV applied across the plates.
The energy stored in a capacitor is equal to the amount of work need to cause the voltage across the capacitor, given as :
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Ohm's Law The main basic electrical law, defines the relationship between voltage, current and resistance, with one ohm is the resistance value through which I volt maintain I ampere at current.
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Cbnncection of Circuit Elements Connection of circuit can either in series or in parallel.
Series Connection
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R1o61=R1
Rtotal Rr
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Rn
Direct Current (DC) Unidirectional flow of electric charge, no change on the direction and the magnitude of the flow of electric charge.
Alternating Currpnt (AC) Electric current which direction of the flow of elechical charge reverses cyclically (opposed to direct current), and the magnitude changes in cycle.
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Gives TWO (2) types of electric charge
(ii)
(2 marks)
Define the definition ofelectric charge, current, voltage and gives SI units for each (9 marks)
(3 marks)
,
2-
v,.
.
l-:..^ firrn /.\ r--J ^3-r-^r:-:-^--:r- and - J draw Give TWO (2) electic circuits t1@es of treir simple schematic
diagrams.
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If
(6mark)
50 C of an electic charge flows through a wire in l-5 minuteg what is the
current in the wire?
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ts:'
A battery with a potential difference of t.Sv is connected to a bulb with a resistance of 90 O. Calculate the guantity of current flowing through the bulb.
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, .
(4 marla)
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5]
(3 marks)
A 20V potential difference is used to charge a 4 pF capacitor. Calculate the total
qge
charge. ICLO 5]
7:
Give TWO (2) differences between direct current (DC) and altemating current (Ac). lcl,o s] (4 marks)
8-'.
A 25 voltsuppry is used to chargea capacitor of3.5pF. Determinethe
f " ' ,.
.
charge stored in the capacitor
plates.
A
duration of 6 minutes using a dry.cell. Assume
larnp X. is lighted
'p
for
a
(4 marks)
that the dry cell provides a steady clurent during that duration-
(i)
How much charge flows through the lamp reading
(ii) .
of
if the ammeter shows a
0.2 A?
(3 marks)
Determine the number of electons that flows through the lamp the charge of one electon is l.g
x
l0rt?
i:.'_'
.t
lA..
X
(4
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.
X if
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+ /.'8 -:
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A torch bulb has a resistance of I O when cold. It drarvs a current of 0.2 A from a sour.ceof 2 V and glows.{ClO 5l
riiiii.
circuit. Calculate the resistance ofthe bulb when glowing. explain the reason of the difference in resistance. Sketch the flow of curreniintne
(2 marks)
t3 marks)
(2marks)
4 X tO," e
t-
lf.
In Diagram 6(c), three resistors 6f 22
gL, I 5
O and 10 O are connected in
parallel to a 6V battery. Calculatc
6V
I'1,
F
Ir
22n t+t lz
150
13
IOJL
r@
.
t"T"t ."E
a
+.ir
.
The total resistance ofthe circuit
0- >73,ct,t, o .6 Aq;
ii.
t'2+\
.
The curreirt
' 11
, 12
and
(3 marks) (6 marls)
13.
'Itre ioial curent flowing in the circuit
Ih.
(2 marts)
,
12.
A resistorof2 O.F c6'nn{dtao( paf.rcrrr-pornf
f)
are
in parallel
be.tw-een
A and B. If
a Uattery of 4
8* "-Ro,stors
\i is connected
3 Oand 7
betwee,n
A
and
.C as shorvn in Figure 6(e), calorlate [CLO 5] i,tt
4tJ>
the total resistance
1.
ll.
oqs"\ .'..... -
. -lii. I
tt : 13.
(4
tlre total current
a
A potential difference of r[ v is appried to u n"tn*t oiieristors the figure
l.
below.
as st
o*n in
,-,
ata 4c,
o.t(A
(i)
rgV
(iD
(-
o.ls+A
(iii)
R,:fa
Wn-at is the reading on the ammeter
A?
(j
what ii th.e pitential difference a cross the paralrer 'what is rhe current flowing.through go the
marks)
nenuort
resistor?
(z marts)
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