PHYSICS QNS. lets get back in class kdgo

Mathew kileo

Mathew kileo

Member
Jan 22, 2017
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1
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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
 
Hlo la kwnza ni easy, hapo una apply kanuni ya magnification au mirror formula then kwsha kazi
 
Hivi watoto mnafungua lini shule. Home work zenu badala mzifanye nyumbani huko mpo busy na JF.
 
nimeyacheki haraka haraka ila la pili linaenda kirahisi mno... ila lazima ubadilishe centigrade kwenda kwenye kelvin...!!!!!
 
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.
 
La pili ukipata Pressure law ni easy sikumbuki frsh ila ni P1T1=P2T2 at constant Volume
 
Giriad.A.Birusha said:
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.
Click to expand...
Haya Matakataka Yanasaidia Nini Maishani..
 
1)hi/ho=v/u
2)apply pressure law
P=kT, p1/T1=p2/T2 then proceed as required.
 
Giriad.A.Birusha said:
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.
Click to expand...
i apreciate on yo work bro.
thnk u kwa sna
 
Giriad.A.Birusha said:
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.
Click to expand...
u dserve more troffy aseee
 
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