TUGAS SOAL FISIKA Nama: Dhamar Arie
Diket : v = 20 m(3),p = 400 kg/m(3) ta!a : m"""## $a%a& : p = m/v m=p"v" m=400"20 = '000 kg (FLUIa/eit!)
iket = m = 23 kg,F = *00 +" Ta!a Ta!a = a""## $a%a& = F = m"a, a = F/m, a= *00/23, a = 4,3' m/(2) (-k-m e%t. 2)" Nama : Ahmad Anas Subkhan 1.
Biomekanika
Dik : = 2 Kg F = *00 1 Dit : a # $a%a& : F = " a a=F/ a = *00 / 2 a = 4 m/ 2 2.
Fluida
Dik : = 2 m 3 S = *00 Kg / m 3 Dit : m # $a%a& : S = m / v =S" = *00 " 2 = 200 Kg"
Nama: Adi Nugroho
*" .a .a tet teta ag g &i.m &i.meka ekai ika ka Se.ag atit ai 5aak 5a- &eai ega ke6epata 40 m/t etea 4 etik" 7eapaka 5aak !ag itemp- .e atet tee&-t 5ika ipe6epat 3 m/t a ipeam&at 2 m/t#
Jawaban dari Soal :
Diketa-i : t = 40 m/t t
= 4 t
8e6epata a = 3 m/t 8eam&ata a = 2 m/t Dita!a
:
a" St 5ika ipe6epat" &" St 5ika ipeam&at $a%a&
:
*" Geak L-- 7e-&a Teat- ega 8e6epata 0 = t 9 at = 40 9 (3 4) = 40 9 *2 = 2' m/t St = 0"t ; at = 2' . 4 ; . 3 (4) = **2 ; 24 = *3< m *" Geak L-- 7e-&a Teat- ega 8eam&ata 0 = t ; at = 40 ; (2 4) = 40 ; ' = 4' m/t St = 0"t 9 at = 4' . 4 9 . 2 (4) = *2 9 *< = *>< m
.a tetag ?-i e&-a &e5aa !ag tiggi!a 06m &eii pe- ega @at 6ai !ag maa 5ei!a 0,g/6m3" 5ika tekaa -aa ia&aika a g = *0 m3 tekaa i.tati paa aa &e5aa aaa iket : p : 0" g/6m3 = 00 kg/m3 : 0 6m = 0" m g : *0 m/2 ita!a : 8 5a%a& : 8 = p"g" = 00"*0"0, = 2"00 +/m2 Nama: Agus Prihandoko 1.
Biomekanika
Dik : = 30 Kg F = 01 Dit : a # $a%a& : F = " a a=F/ a = 0/ 30 a = *00m/ 2 2.
Fluida
Dik : = 40 m 3 S = 200 Kg / m 3 Dit : m # $a%a& : S = m / v =S" = 200 " 40 = '000 Kg"
Nama : Ahmad Arif Himawan 1.
Biomekanika
Dik : = 2
Fluida
Dik : = *0 m 3 S = 300Kg / m 3 Dit : m # $a%a& : S = m / v =S" = 300 " *2 = 3000Kg" +ama SAIFUL BOCA+ O+TOC SOAL 7IOEKA+IKA .&i A &emaa *000 kg &egeak ega ke6epata 2m* ke kaa, .&i 7 maa!a *200 kg &egeak ega ke6epata 3m* ke kii, &eapa m.met-m!a# a".&i A# &".&i 7# 6"5-ma m.met-m m.&i A a 7#
5a%a&: iket:
maa m.&i A:*000kg ke6epata!a: 2m* maa m.&i 7:*200kg ke6epata!a:3m*(aa &ea%aa)
A" pA = mA"vA
=*000"2
=2000kgm*
7" p7 = m7"v7 = *200"(3) = 3<00kgm* " p = pA ; p7 = 2000 ; (3<00) = *<00kgm* (ke aa kii) O+TOC SOAL FLUIDA Ai megai paa pipa" Apa &ia iketa-i iamete peampag A*= '6m ega ke6epata ai!a 4m*, eagaka iamete peampag iamete A2 = 46m,it-ga: A" ke6epata ai !ag e%at peampag A2# 7" De&it ai tiap 5am # $a%a&: Diamete A* ='6m
B * = 46m =0,04m
Diamete A2 =46m
B 2 = 46m =0,02m
A" * = 4m* A* " * = A2 " 2 (0,04)2" 4 =
(0,02)2 " v2
*
" =A " v = A* " v*
=
(0,04)2 " 4
= 0,020 m3* = 0,020 m3* H 3<00 5am * = >2m3 5am* +ama: B.ie Bi6a B!"#$AN$A
GEBAK LUBUS 7EBU7AC TEBATUB
S.a :
Se.ag atit ai 5aak 5a- &eai ega ke6epata 40 m/t etea 4 etik" 7eapaka 5aak !ag itemp- .e atet tee&-t 5ika ipe6epat 3 m/t a ipeam&at 2 m/t#
$a%a&a ai S.a :
Diketa-i : t = 40 m/t
t
= 4 t
8e6epata a = 3 m/t
8eam&ata a = 2 m/t
Dita!a
: a" St 5ika ipe6epat"
St 5ika ipeam&at
$a%a&
:
Geak L-- 7e-&a Teat- ega 8e6epata
0 = t 9 at
= 40 9 (3 4)
= 40 9 *2
= 2' m/t
St = 0"t ; at
= 2' " 4 ; " 3 (4)
= **2 ; 24
= *3< m Geak L-- 7e-&a Teat- ega 8eam&ata
0 = t ; at
= 40 ; (2 4)
= 40 ; '
= 4' m/t
St = 0"t 9 at
= 4' " 4 9 " 2 (4)
= *2 9 *<
= *>< m F-ia *" $ika kita &eii tea- ekat ega e keeta api, kata!a kita &ia kee.t kaa- aa keeta api !ag eag &egeak 6epat paa e tee&-t it- i6 k.. kata!a" e--tm &agaimaaka # &t%, 5aga pake i&-ktika !a ta &ia ga%at a-at :gee: 2" D-a m.&i !ag &egeak e5a5a paa aa !ag ama aka tetaik at- ama ai" G-aka pegeta-am- megeai piip .m 7e.-i -t-k me5eaka keaea ii (eve *) 3" K.. kata!a a5- agi &etam&a teaap ketiggia i ata pem-kaa &-mi" 7eaka # 5ika &ea, megapa &ia emikia # (eve *) 4" Semaki tiggi 6e.&.g aap, emaki &aik ik-ai aap !ag &e&..g&..g ka&- ai tempat peapia 5eaka pe!ataa ii (eve *) (m.e) +ama: 7a!- 8atama *"&eika 6.t. peeapa ?-iaaam &iag keeata# 5a%a&:p.e i5eki 6aia keaam t-&-
2"(ke6epata eati?) e&-a pea-" aam pe5aaa !a pea- mee%ati e&-a &.t.(titik A ) at- 5am kem-ia pea- &e&aik aa a &epapaa ega &.t. tai paa 5aak H &eat &aa = */> H '0='0/>=**,43kg" 2) e&-a &a. memiki v.-me 00m3 !ag &eii ei-m(maa 5ei 0,*'kg/m3) ega &eat g..a 200kg,&eapa &eat !ag apat i aggk-t &a. tee&-t# 5a%a& : D* : = 00m3 8(.) = 0,*'kg D2 : m"""""""""""""""# D3 : m = . H v = 00m3 0,*'kg = *<2kg Nama: %a&ih "us&ika
.t. .a F-ia : *" (a) Se.ag peaga%ati ega &eat 40 + megg-aka epatak tiggi ega -k-a ak 0, 6m 0, 6m" Tet-ka tekaa !ag i&eika peaga%ati tee&-t paa atai ketika ia meagka a e-- &eat &aa!a it-mp- .e aa atepat-!a (&) Se.ag pia ega &eat &aa '00 + (&eat ai peaga%ati) megg-aka epatega -k-a aa epat- aaa ' 6m 2 6m" Tet-ka tekaa !ag i&eika pia tee&-t paa atai ketika ia meagka a e-- &eat &aa!a it-mp- .e aa at- epat-!a"
8e!eeaia: Diketa-i: •
7eat %aita (peaga%ati) F = 40 +
•
L-a ak epat- = 0, 6m 0, 6m = 0,2 *04 m2
•
7eat pia F = '00 +
•
L-a epat- pia = ' 6m 2 6m = 2 *02 6m2 aka tekaa: (a) 8eaga%ati ega ak epat- tiggi paa atai aaa: 40 N
F
0,2 × *0
A P =
=
−
4
m
2
×
= *'
*0< +/m2
(&) 8ia ega -a epat- 2 *0 6m paa atai aaa:
F
'00 N
A P =
2 × *0
−2
m
2
=
×
= 4
*04 +/m2
K.meta, mekip- &eat peaga%ati e&i iga i&aigka ega &eat pia, tetapi tekaa !ag i&eika peaga%ati teaap atai ekita 40 kai e&i &ea i&aig tekaa !ag i&eika pia teaap atai" Ca ii ie&a&ka -a pem-kaa epat- peaga%ati *2, *04 kai e&i ke6i i&aig -a epat- pia" Dega emikia peaga%ati ega epat- ak tiggi e&i me-ak atai ai paa pia"
.t. .a &i.mekaika : 8eempa 6akam megakii p-taa!a iap -t-k meepaka 6akam ega ke6epata ataata 2,23 aia/etik" 7ea!a 5aak ataa p-at 6akam ampai ega p-at p-taa (6.-ma vete&ai) aaa 0 6etimete (ai-)" 7eapa ke6epata iie 6akam# = 2,23 0 = 22,>* m/t $ika ai- ipepa5ag me5ai *00 6m, maka: = 2,23 *00 = 2,23 m/t •
Dega emikia ke6epata iie &e&aig -- ega pa5ag ai-!a, apa&ia ke6epata p-taa (ke6epata --t!a) tetap
•
Oe kaea it- peempa 6akam a- &e-aa meetagka ega epa5ag m-gki aat 6akam iepaka"