Physics: Post your doubts here!

[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

 

[A.ELWY 7]

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

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!

well thanx...but for quest 3 i no that we must use THE SI units...but i also couldn'y figure it out, can u explain it pls???

 

[elbeyon]

P reduction=V1^2-V2^2 /V1 multiplied by 100.--------->P is proportional to V^2

I guess it would be:
P reduction=V1^2-V2^2 /(V1)^2 multiplied by 100. [As P ∝ V^2].

 

[SkyPilotage]

well thanx...but for quest 3 i no that we must use THE SI units...but i also couldn'y figure it out, can u explain it pls???

g x h = ms^-2 x m = m^2 x s^-2
Square root = ms^-1 = velocity SI units --> A

 

[hassam]

pls tell mo how u solve it

 

[A.ELWY 7]

g x h = ms^-2 x m = m^2 x s^-2
Square root = ms^-1 = velocity SI units --> A

thanx very much

 

[SkyPilotage]

pls tell mo how u solve it

An ultrasound is emitted to reach the fetus front edge of the head and is reflected back to the source and shown as the first pulse on the graph. The distance travelled is from the source to the front edge back to the source.
Another ultrasound is emitted to reach the fetus rear edge of the head and is reflected back to the source and shown as the second pulse on the graph. The distance travelled is from the source to the rear edge fo the face and back to the source. It covers the distance travelled by the first ultrasound and travels twice over the fetus's head.
There is a 5.5 x 20 microseconds difference between the reflection of the ultrasounds . i.e size of the fetus head x 2
The distance is = speed x time = 1.5 x10^3 x 5.5 x 20 micro seconds
Therefore the fetus head size is the distance diveded by 2!
This is my interpretation of the question! Hope that helped