Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
$!$RatJumper$!$:
W08 P1 Q 10, 27, 31, 36, 37 please :)
Sorry for all my questions! :(
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 10:30:59 pm ---W08 P1 Q 10, 27, 31, 36, 37 please :)
Sorry for all my questions! :(
--- End quote ---
Q10
Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)
Q27
Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3 (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010
Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1
Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
$!$RatJumper$!$:
--- Quote from: TJ-56 on November 13, 2010, 10:49:27 pm ---Q10
Relative speed of approach = relative speed of separation
u1 + u2 = (-)v1 + v2
so:
u1+ u2 = v2 - v1 rearranged (v1 i given a negative sign due to its direction)
Q27
Between 2 maxima is 15 mm , so wavelength = 2 * 15 * 10-3 (remember its a stationary wave)
so speed c = 3 * 108
c=f * lambda
and f = c/wavelength
we have the 2 values required, so the frequency is calculated as 1 * 1010
Q31
Wire p has twice the diameter, not twice the area, its quadruple the area because the diameter is squared,
current I=V/R V is constant, so its a ratio of resistances, which is 4 to 1
Q36
Looked confusing to me at first, but turned out to be rly easy:
V=IR
V=7.5 (1.5 volts are lost to internal resistance)
R=15
Go for I to give 7.5/15
which is 0.5A
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
--- End quote ---
Thank you that did help a lot :)
I just have a few questions
For Q10, why cant we say u1 + -u2 = v1 + v2
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 13, 2010, 11:00:14 pm ---Thank you that did help a lot :)
I just have a few questions
For Q10, why cant we say u1 + -u2 = v1 + v2
--- End quote ---
Well the original equation is
u1 - u2 = - (v1 - v2) Which is derived by using both the principle of conservation of momentum and energy, where it is assumed all directions are the same (for example both to the left before and both with velocities to the left after)
since u1 and u2 are in different direction , its now u1 - -u2, which is u1 + u2
and on the right hand side, -(v1 -v2) which is v2 - v1
Thats all there is to it i guess, hope that cleared it up.
Hypernova:
--- Quote from: TJ-56 on November 13, 2010, 10:49:27 pm ---
Q37
If we have the emf of cell E2, we could calculate pd per unit length along XY.
Im still not completely sure why i cant be response A, but B made much more sense.
Hope that helped.
--- End quote ---
Q37)
The potentiometer!!
In this circuit,
E2 = E1 x resistance of XT
resistance of XY + R
since the galvanometer is reading a null result, we know that the pd across XT is equal to E2.
The galvanometer is basically an ammeter. 0 amps means 0 p.d which means that the p.d across XT exactly nullifies the emf produced by E2
So we know the length of XT. All we need is E2 and divide it by the length to find its pd per length.
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