PHYSICS QNS. lets get back in class kdgo

[Mathew kileo]

ni topic ya form three na hyo maswal ymenishnda na naamin kuwa kuna mmbers genious humu who can help me
unaweza ukalifanyia pmben then ukapiga pcha ukatuma
thank u kwa kutumia mda wko kusoma na shukran sana kwa wale wataojitolea kunisaidia asanten

 

[GANG MO]

Ungepeleka jukwaa la elimu ingekua pouwa sana. Just ushauri

 

[Itzmusacmb]

Hlo la kwnza ni easy, hapo una apply kanuni ya magnification au mirror formula then kwsha kazi

 

[Sang'udi]

Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.

 

[mwaminifuhalisi]

Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.

Mkuu msaidie kama unafahamu kuliko kumkejeli. Unaweza kumpa hata hint kama alivyofanya jamaa hapo juu kilichobaki atamaliza mwenyewe.

 

[the locksman]

Hlo la kwnza ni easy, hapo una apply kanuni ya magnification au mirror formula then kwsha kazi

Hio formulae ipo wapi

 

[The End..]

nimeyacheki haraka haraka ila la pili linaenda kirahisi mno... ila lazima ubadilishe centigrade kwenda kwenye kelvin...!!!!!

 

[trigeminal]

sijaona swali hapo.. acha ujinga piga shule

 

[The End..]

tumia P1 T1 = P2 T2

 

[Ntahandinkimuhila]

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.

 

[Smokey D]

La pili ukipata Pressure law ni easy sikumbuki frsh ila ni P1T1=P2T2 at constant Volume

 

[msafwa93]

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.

Haya Matakataka Yanasaidia Nini Maishani..

 

[Ntahandinkimuhila]

Haya Matakataka Yanasaidia Nini Maishani..

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.

 

[mansuly]

1)hi/ho=v/u
2)apply pressure law
P=kT, p1/T1=p2/T2 then proceed as required.

 

[Mathew kileo]

sijaona swali hapo.. acha ujinga piga shule

we dfer bro.
dats y kuna jamii forums na hii ni moja ya part ya kaz yke

 

[Mathew kileo]

Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.

hii ni online educational bro.
am using my time the bext i can bro

 

[kabon14]

Haya Matakataka Yanasaidia Nini Maishani..

Hayo matakataka Na mengine mengi ndo yanakufanya ukenue meno kwny smartphone yako Na kuona Maisha matam

 

[simplemind]

Rigid container is an abstraction? Is it?

 

[Mathew kileo]

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.

i apreciate on yo work bro.
thnk u kwa sna

 

[Mathew kileo]

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.

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