Physics: Post your doubts here!

[Silent Hunter]

any method for the estimations type questions?

 

[Tasnim Akhter]

hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii

 

[hussamh10]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q2 c part how to determine what is the pivot to be taken

 

[hussamh10]

hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii

K.e=1/2 mv^2
=692550

power=workdone/time
=692500/8.1
85500 W
there is a resistive force acting which causes the acceleration to reduce or p=fv=mav
P and m are constant se when v increase a decrease

 

[hussamh10]

hey can anyone help me with ON'06/02/1.c.ii and 2.c.iii

K.e=1/2 mv^2
=692550

power=workdone/time
=692500/8.1
85500 W
there is a resistive force acting which causes the acceleration to reduce or p=fv=mav
P and m are constant se when v increase a decrease

 

[hussamh10]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_23.pdf
Q1 d part can any one show it how to do it in vector triangl way not mathematically....thanx plzzz

 

[hussamh10]

Hey plzz help me in how much sf must be the answer for eghttp://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20%289702%29/9702_w11_qp_23.pdfin q3 ii)2 part
mgh
1800 x 9.81 x 15=264870 how much s.f must be it taken and how to determine it???

 

[bamteck]

www.xtremepapers.com/papers/CIE/.../9702_w07_qp_2.pdf

HELP NEEDED for no. 5 (a)

 

[geek101]

www.xtremepapers.com/papers/CIE/.../9702_w07_qp_2.pdf

HELP NEEDED for no. 5 (a)

okay remember that the 2 waves have the same frequency...so the wavelength remains the same, which is 30 boxes. you can use proportionality to do this.
If 15 boxes >> 180 degrees
x boxes >> 60 degrees
x = 5 boxes so that means, wave X is 5 boxes ahead of wave W

now it says wave X has 1/2 the Intensity >> 1/4 the amplitude of W.
the amplitude of W >> 10
10 x 1/4 = 2.5
amplitude of X >> 10 - 2.5 = 7.5 boxes in the vertical direction

 

[bamteck]

okay remember that the 2 waves have the same frequency...so the wavelength remains the same, which is 30 boxes. you can use proportionality to do this.
If 15 boxes >> 180 degrees
x boxes >> 60 degrees
x = 5 boxes so that means, wave X is 5 boxes ahead of wave W

now it says wave X has 1/2 the Intensity >> 1/4 the amplitude of W.
the amplitude of W >> 10
10 x 1/4 = 2.5
amplitude of X >> 10 - 2.5 = 7.5 boxes in the vertical direction

how the graph looks like ? oO'

Please help me for no. 6(b)(i) too

 

[unknown22]

anyone has specimen paper for AS Physics(9702) for oct/nov 2012.it's urgent

 

[geek101]

how the graph looks like ? oO'

Please help me for no. 6(b)(i) too

this is how mine turned out...

 

[bamteck]

this is how mine turned out...

why does it leads the wave W ?

 

[geek101]

why does it leads the wave W ?

this is what the mark scheme says:
correct phase (ignore lead/lag, look at x-axis only and allow ±½ square)
see so you can either show it ahead by 60 degrees or behind. = )

 

[TeKnOzOr]

why does it leads the wave W ?

Hmmm...its because they have diff phase difference...thats why it leads wave W!

 

[TeKnOzOr]

how the graph looks like ? oO'

Please help me for no. 6(b)(i) too

6(b)(i)
It is given that the voltage of the main supply is 230V.And it is said in the ques that "The potential difference across the shower unit must not be
less than 225 V"....that is,the minimum voltage of the main supply is 225V
The minimum voltage of the cable=220V-225V=5V
Therefore....the P.d in the cable=5V
The current in the wire= 46A
For Max. Resistance,Voltage must be Min.
Therefore, v=5V,I=46A
Using V=IR
R=V/I
=5/46
=0.10869
=0.11 ohm
This is the max. resistance of the cable!

 

[Silent Hunter]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf

question 1 (b)(ii)

 

[bamteck]

6(b)(i)
It is given that the voltage of the main supply is 230V.And it is said in the ques that "The potential difference across the shower unit must not be
less than 225 V"....
Therefore....the P.d in the cable must be= 230V-225V
= 5V
The current in the wire= 46A
Therefore,
Using V=IR
R=V/I
=5/46
=0.10869
=0.11 ohm
This is the max. resistance of the cable!

Thanks mate

Would you mind helping me with this one : http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s08_qp_2.pdf

no. 4 (b) (c)

 

[hussamh10]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_23.pdf
Q1 d part can any one show it how to do it in vector triangl way not mathematically....thanx plzzz
bamteck TeKnOzOr
geek101

 

[TeKnOzOr]

Thanks mate

Would you mind helping me with this one : http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf

no. 4 (b) (c)

(b) In this question, you must look for the sizes of the expression
for e.g, it is said that ∆p is measured in Pa,
So ∆p must be large...which is shown on the right hand-side where there is the value 2.2 × 10^9 which is Very Large
It is also said that "water is assumed to be incompressible" ..which means that the change in volume is Very Small..
which is shown by ∆V/V which is very small.
Hope u've understand dr!!
(c) In this ques,we must use ∆p = hρg
It is given that the "approximation of h=10"
Using the eqn to find the actual h.
∆p = hρg
1.01 × 10^5= h x 1.08 × 10^3 x 9.81
h=9.53m
Therefore to determine the %error,
we take, ∆h / h x 100
∆h=10-9.53
=0.47m
Therefore %error= 0.47/9.53 x 100
= 4.9%
(You can also take the "h" as 10)
Hope it helps!!