Mkuu msaidie kama unafahamu kuliko kumkejeli. Unaweza kumpa hata hint kama alivyofanya jamaa hapo juu kilichobaki atamaliza mwenyewe.Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.
Hio formulae ipo wapiHlo la kwnza ni easy, hapo una apply kanuni ya magnification au mirror formula then kwsha kazi
Haya Matakataka Yanasaidia Nini Maishani..Solution for question number 1,
Let:Magnification=M
M= 2.
Object distance=U
U=25cm..
Image distance = V
V=unknown, let it be equal
to V.
From M=V/U,
M x U = (V/U) x U,
V=M x U
V=2 x 25cm,
V=50cm.
If M=4, making U the subject of the formular (since it is the required physical quantity to be found),
then U=50cm/4.
U=12.5cm.
Therefore, to provide a magnification of four times, An object must be placed at 12.5cm from the convex lens.
Solution question number 2,
Using pressure law,
Let P=pressure
Initial pressure P1 = 1.5 x 10^5Pa
Final pressure=unknown, let it be equal to P2,
Initial temperature=T1=-30°c=(-30+273)K,
T1=243K.
Final temperature=T2=25°c=(25+273)K,
T2=278K
since pressure law states that the Pressure of a fixed volume of a gas is directly proportional to it's absolute temperature,
P=KT., P1/T1=P2/T2=P3/K3=K = ... ... ,
P1/T1=P2/T2.
(P1/T1) x T2 = (P2/T2) x T2,
P2 = (P1/T1) x T2
P2=(1.5 x 10^5Pa ÷ 243K) x 298K,
P2 is approximately equally to
1.84 x 10^5Pa.
Therefore, the new pressure of the gas is approximately equal to 1.84 x 10^5Pa.
Siyo takataka Ndg, haya mambo yanatumika huko viwandani katika utengenezaji Wa kemikali Na bidhaa nyin gine huko.Kwa hiyo ni ya muhimu sana ndiyo maana yanafundishwa shuleni.Haya Matakataka Yanasaidia Nini Maishani..
we dfer bro.sijaona swali hapo.. acha ujinga piga shule
hii ni online educational bro.Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.
Hayo matakataka Na mengine mengi ndo yanakufanya ukenue meno kwny smartphone yako Na kuona Maisha matamHaya Matakataka Yanasaidia Nini Maishani..
i apreciate on yo work bro.Solution for question number 1,
Let:Magnification=M
M= 2.
Object distance=U
U=25cm..
Image distance = V
V=unknown, let it be equal
to V.
From M=V/U,
M x U = (V/U) x U,
V=M x U
V=2 x 25cm,
V=50cm.
If M=4, making U the subject of the formular (since it is the required physical quantity to be found),
then U=50cm/4.
U=12.5cm.
Therefore, to provide a magnification of four times, An object must be placed at 12.5cm from the convex lens.
Solution question number 2,
Using pressure law,
Let P=pressure
Initial pressure P1 = 1.5 x 10^5Pa
Final pressure=unknown, let it be equal to P2,
Initial temperature=T1=-30°c=(-30+273)K,
T1=243K.
Final temperature=T2=25°c=(25+273)K,
T2=278K
since pressure law states that the Pressure of a fixed volume of a gas is directly proportional to it's absolute temperature,
P=KT., P1/T1=P2/T2=P3/K3=K = ... ... ,
P1/T1=P2/T2.
(P1/T1) x T2 = (P2/T2) x T2,
P2 = (P1/T1) x T2
P2=(1.5 x 10^5Pa ÷ 243K) x 298K,
P2 is approximately equally to
1.84 x 10^5Pa.
Therefore, the new pressure of the gas is approximately equal to 1.84 x 10^5Pa.
u dserve more troffy aseeeSolution for question number 1,
Let:Magnification=M
M= 2.
Object distance=U
U=25cm..
Image distance = V
V=unknown, let it be equal
to V.
From M=V/U,
M x U = (V/U) x U,
V=M x U
V=2 x 25cm,
V=50cm.
If M=4, making U the subject of the formular (since it is the required physical quantity to be found),
then U=50cm/4.
U=12.5cm.
Therefore, to provide a magnification of four times, An object must be placed at 12.5cm from the convex lens.
Solution question number 2,
Using pressure law,
Let P=pressure
Initial pressure P1 = 1.5 x 10^5Pa
Final pressure=unknown, let it be equal to P2,
Initial temperature=T1=-30°c=(-30+273)K,
T1=243K.
Final temperature=T2=25°c=(25+273)K,
T2=278K
since pressure law states that the Pressure of a fixed volume of a gas is directly proportional to it's absolute temperature,
P=KT., P1/T1=P2/T2=P3/K3=K = ... ... ,
P1/T1=P2/T2.
(P1/T1) x T2 = (P2/T2) x T2,
P2 = (P1/T1) x T2
P2=(1.5 x 10^5Pa ÷ 243K) x 298K,
P2 is approximately equally to
1.84 x 10^5Pa.
Therefore, the new pressure of the gas is approximately equal to 1.84 x 10^5Pa.