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q.21 :
pressure due to oil = 830* 9.81*x
pressure due to water = 1000*9.81*(2000-x)
so now the equation is ,
17.5x10^6 = (830*9.81*x) + (1000*9.81*(2000-x)
solve it and u get the ans 1270 (D)
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AS Physics P1 MCQs Preparation Thread.
- Thread starter mujtabashahnawaz
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pressure due to oil = 830* 9.81*x
pressure due to water = 1000*9.81*(2000-x)
so now the equation is ,
17.5x10^6 = (830*9.81*x) + (1000*9.81*(2000-x)
solve it and u get the ans 1270 (D)
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
In qs 11, whenever an elastic collision occurs the relative speed of approach will be equal to the relative speed of seperation. this only applies to elastic collsions or to inelastic collisions
In qs 11, whenever an elastic collision occurs the relative speed of approach will be equal to the relative speed of seperation. this only applies to elastic collsions or to inelastic collisions
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I think, only elastic.
If I'm wrong..plz let me know..
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ONLY for elastic..
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can anyone plz explain me the question of paper MAY/JUNE 2003 Q.10
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and for qs.25,
distance b/w X and Y = 1.5 lambda
and we know dat a path diffrence of 1 lambda(wavelength) = a phase diff of 2 pi radians.
it means , 1 lamda = 2pi rad
so 1.5 lamda = 2 x 1.5pi => 3pi
so n = 3 (C)
so in inelastic it the speed of seperation will be less then ???
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F= ma
Totl mass = 4m
hnce a = F/4m...., Fx= Fy
Fx = ma
3m * F/4m = 3/4 F---->D
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F=ma
so, F =(m+3m)a
F=4ma
hence a=F/4m
block x in exerting a force on block y , so Fx=3m(F/4m)
so Fx=3/4F!
HOPE IT HELPED
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
how do we do q33. is the circuit parallel?
how do we do q33. is the circuit parallel?
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not xctly less..bt kinetic energy will not be the same..it may be lost in some other form.. for eg: as sound or heat energy...
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10 ms = 10^-3 = 0.001s.
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http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
pls sm1 help me wid q.s. 18 amd 25
pls sm1 help me wid q.s. 18 amd 25
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particly ya..cz KE is less..in sum case da object dosnt move..they stay still...
like in the case wen a lump of clay is droped on da floor..al KE has been transformd into wrk done in flattein da lump n into sum sound energy..
i hope i am claer enough!!
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
how do we do q33. is the circuit parallel?
how do we do q33. is the circuit parallel?
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_1.pdf
how do we do q33. is the circuit parallel?
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minimum work done will be to lift up the barrel up with out doing any work to overcome friction... so it will be mgh.. 9.8*50*1.6...
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wt abt q.25