Physics: Post your doubts here!

[JD REBORN]

Q18
Work lost is the unshaded area. you dont put work to unstretch a rubberband
Q23
a is max at the peaks, speed is max at the middle
Q27
wrong direction of motion, its a stationary wave. Motion of particles is reflected.
Q37
If it is drawn into 6 wires for example.
The area of one wire in Q is A/6 so Resistance of one wire is 6R since R is inv prop. to A.
So total R is 6R/6 = R i.e same Resistance as P
Q40
the nucleus with the smallest Z/M Ratio has the lowest a i.e Li ratio is the smallest ( 3/7 = 0.428 )

For Q37 if u say the resistance of one wire is 6R then wont total resistance be 6R *6 cuz there are 6 wires in Q?

 

[JD REBORN]

In Q 18 why do we subtract the pressures shouldnt we add them??

 

[anonymous123]

AoA wr wb

if r=3x+y
the uncertainty in r will be = 3Δx + Δy or Δx + Δy

and if r = ((a^3)(b^2)) / (2a+b)
then ΔR = ??
help plz

 

[SkyPilotage]

For Q37 if u say the resistance of one wire is 6R then wont total resistance be 6R *6 cuz there are 6 wires in Q?

They are in parallel not series! careful!

In Q 18 why do we subtract the pressures shouldnt we add them??

No because the reading of the metre under the surface measures the pressure that is exerted on it, which is due to both the water and the atmospheric pressure.
We just need to find the pressure due to the water. This is why we subtract not add them.

 

[SkyPilotage]

AoA wr wb

if r=3x+y
the uncertainty in r will be = 3Δx + Δy or Δx + Δy

and if r = ((a^3)(b^2)) / (2a+b)
then ΔR = ??
help plz

Im not quite good with uncertainities but here goes:-
**First one I think its only x + y not 3x + y
**Second one, you need to convert to percentages.
When its exponent, you multiply the percentage by the power.
Divided or multiplied you add the power.

P.S:- Would be nice if someone confirms my talk

 

[anonymous123]

ΔR = 3Δa/a + 2Δb/b for the numerator part and for the denominator Δa + Δb?? now add them....ryt?
help me

 

[A.ELWY 7]

Would some 1 pls explain 2 me the relation between intensity and amplitude...i no amplitude square is proportional to intensity, but intensity times 2, is proportional to what amplitude ?????

 

[SkyPilotage]

ΔR = 3Δa/a + 2Δb/b for the numerator part and for the denominator Δa + Δb?? now add them....ryt?
help me

Now now you dont need to cry, we will help! thats what XPC is for
r = ((a^3)(b^2)) / (2a+b)
it will be 3Δa + Δb + Δa + Δb... I think...

 

[SkyPilotage]

Would some 1 pls explain 2 me the relation between intensity and amplitude...i no amplitude square is proportional to intensity, but intensity times 2, is proportional to what amplitude ?????

Intensity prop to amplitude^2
so 2I ----> (Square root of 2 A)^2 ----> Sqr rt 2 times amplitude!
If amplitude is doubled then Intensity will be equal (2A)^2 ---> 4 times Intensity!

 

[elbeyon]

ohhh i forgot about the mass that is why
thanks ^^

Happy to help. WC.

 

[A.ELWY 7]

Intensity prop to amplitude^2
so 2I ----> (Square root of 2 A)^2 ----> Sqr rt 2 times amplitude!
If amplitude is doubled then Intensity will be equal (2A)^2 ---> 4 times Intensity!

but how did u get that 2I is square root of 2A??

 

[A.ELWY 7]

and i may need help in Q3, 13, 33
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf

 

[SkyPilotage]

but how did u get that 2I is square root of 2A??

like I said I ---> A^2
2I ----> 2 A^2
So A = sqauare root of (2A^2) = root 2 A

and i may need help in Q3, 13, 33
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Q3
Use SI units
Q13
torque = force x PERPENDICULAR distance
There are 2 ways:-
1- find the vertical component of the force
2- find the perpendicular distance by trigonometry
Q33
V1 = 5/10 x 2 = 1V
V2 = 3/5 x 2 = 1.2V
V1-V2 = -0.2V!

 

[anonymous123]

Now now you dont need to cry, we will help! thats what XPC is for
r = ((a^3)(b^2)) / (2a+b)
it will be 3Δa + Δb + Δa + Δb... I think...

wouldnt it be 2Δb?
so it will become 4Δa + 3Δb is that correct?

 

[SkyPilotage]

wouldnt it be 2Δb?
so it will become 4Δa + 3Δb is that correct?

I think so! Please check it with the marking scheme and tell me

 

[anonymous123]

there is no ms for it...i made that question myself

 

[SkyPilotage]

there is no ms for it...i made that question myself

.... alrighty then! We can see if someone confirms it. Cz there is a chance of error.

 

[Hussnain]

PHYSICS AS
M/J/2010 (22)
Q2(b)(ii), Q3 (b)(i)(2)
Q6(a),(d)

 

[elbeyon]

PHYSICS AS
M/J/2010 (22)
Q2(b)(ii), Q3 (b)(i)(2)
Q6(a),(d)

For 2(b)ii

Distance traveled at final position= 90cm= 0.9m
Now,
s=(1/2)*a*t^2
or, 0.9=0.5*9.8*t^2
So, t=0.43s [ t= Total time ]
Again Form (ii)
t1=0.40s [ t1= The time the ball fall before the photograph was taken]
Now,
Time interval(t2)= t - t1
= 0.43 - 0.40
= 0.03 s

I hope you are clear now.

 

[JD REBORN]

PHYSICS AS
M/J/2010 (22)
Q2(b)(ii), Q3 (b)(i)(2)
Q6(a),(d)

3b(i)(ii)
Now since the question introduces u with newtons third law of motion,u should realise in (i) it is asking u the same thing.So FA=-FB cuz these forces act in opposite directions.This is simple really.Now for second part it just requires u to understand that these spheres collide with each other for the SAME time.Means this is the time when they are in contact with each other.It has to be equal for both.Now for (ii) the formula for change in momentum is equal to:
change in momentum=resultant force *time.Use this formula for sphere A which will give u FAtA.And for B it will give u -FBtB.Now u equate these momenta and since there is a minus sign with B so it will be opposite in direction to A.
6
6a is tricky and since no numerical value of R or I is given u have to decide which equation of power to use.Now in the beginning of question it is given that R is constant so u have to use P=V2/R.Now in maths when u are asked to find percentage reduction u use the formula:
Reduction/original value multiply by 100.Use the same formula so:
let V1=230 V and V2=220 V
P reduction=V1^2-V2^2 /V1 multiplied by 100.--------->P is proportional to V^2