Physics: Post your doubts here!

[usama321]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s13_qp_21.pdf

Q5 (c) how do we know that a maxima is observed at point P?

Calculate the path difference between the two waves, which is .15m. Now use the wavelength from part b and divide the path difference with it.

.15/0.025 = 6 waves. This means that wave S2 has travelled 6 complete oscillations more than S1 before reaching the screen. Furthermore, as the number is a whole number, we know that there will be constructive interference. Thus there is a maxima at P.

You can also calculate this by dividing the paths of both the waves with the wavelength. One would be 30, the other 36

 

[sweetjinnah]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_43.pdf Q8b(i) why did they take the rest mass different for proton and neutron when in the data sheet they have given only one value

they said to use data from Fig.8.1 given in the qts not the values from data sheet

 

[RoyalPurple]

Calculate the path difference between the two waves, which is .15m. Now use the wavelength from part b and divide the path difference with it.

.15/0.025 = 6 waves. This means that wave S2 has travelled 6 complete oscillations more than S1 before reaching the screen. Furthermore, as the number is a whole number, we know that there will be constructive interference. Thus there is a maxima at P.

You can also calculate this by dividing the paths of both the waves with the wavelength. One would be 30, the other 36

thank you stay blessed

 

[Hadi Murtaza]

thank you stay blessed

U2

 

[Atif Aziz]

B Part, in detail please, explain any formulas used :/

 

[areebaization]

Somebody please help me with this op-amp graph?http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_43.pdf
Q9 b ii
Also are there any notes for op-amp graphs? I never get the hang of them

 

[usama321]

B Part, in detail please, explain any formulas used :/

The wave from S1 has to travel 100cm. Using Pythagoras theorem, S2 will have to travel 128.1 cm.There is a path difference of 28.1 cm

Using v =f lambda, we know that
lambda= 330/1000 to 330/4000 Thus
8.25cm <= lambda <= 33 cm



Now, we know that a minima is formed when S2 is at its trough, while S1 at its crest. As both waves have the same frequencies, they will have travelled the same amount of distance in their respective first 100cm. However, in the next 28.1 cm, S2 must have travelled at least half, 1.5, 2.5 and so on cycles to cause destructive interference and thus a minima. We will check this with different possible wavelengths.

28.1 = 0.5 lamda
lamda = 56 cm. This is outside our range

28.1 = 1.5 lamda
lamda = 18.73. This is in our range

28.1 = 2.5 lamda
lamda = 11.24 cm This is in our range
28.1/3.5 = 8.02 Outside our range
answer = 2

 

[Ysph06]

Can someone help me out with this ?

http://i.imgur.com/9NZygt0.png

It's a multiple choice question, the answer is D, but I don't know why. I need a clear explanation, anyone?
From Edexcel A-Level Physics May 2009 Unit 3B, Q4.

 

[faizay warsi]

can someone provide the short notes on Xray n MRI .....?

 

[Mohammed salik]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

In Both variants ..
Q 7 b (ii) and (iii) Plz !!

 

[mehria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

In Both variants ..
Q 7 b (ii) and (iii) Plz !!

Variant 1
we know that Kirchoff's law states 2 information
first=> the sum of current entering a junction = the sum of current leaving the junction
second=> the sum of e.m.f= the sum of potential difference

Using the Second Law
the sum of e.m.f=the sum of potential difference
for b(ii)
-E2=-I2R + (-I3R)
-E2=-R(I2 + I3)
the negative signs will cancel out eachother so our ans wll be:- E2= R(I2+I3)

(iii)
E1-E2=I1R+ I1R - I2R
E1-E2 = 2I1R - I2R

Focus on the direction of current n the junctions of cell for the given the loop...

Variant 2
this is solved the same way as the one in Variant 1... all u have to focus is on the directions... n for a given loop only focus on that loop n ignore the rest of the loops...

b (ii) in this part we can see that the direction of the current is the same as the path of the loop so across R, I1 will flow n as the E2's n I3's direction is opposite then they will be negative...
E1-E2= I1R - I3R

(iii) -E2=-I3R - I4R - I4R
E2=I3R + 2I4R

 

[Mohammed salik]

Variant 1
we know that Kirchoff's law states 2 information
first=> the sum of current entering a junction = the sum of current leaving the junction
second=> the sum of e.m.f= the sum of potential difference

Using the Second Law
the sum of e.m.f=the sum of potential difference
for b(ii)
-E2=-I2R + (-I3R)
-E2=-R(I2 + I3)
the negative signs will cancel out eachother so our ans wll be:- E2= R(I2+I3)

(iii)
E1-E2=I1R+ I1R - I2R
E1-E2 = 2I1R - I2R

Focus on the direction of current n the junctions of cell for the given the loop...

Variant 2
this is solved the same way as the one in Variant 1... all u have to focus is on the directions... n for a given loop only focus on that loop n ignore the rest of the loops...

b (ii) in this part we can see that the direction of the current is the same as the path of the loop so across R, I1 will flow n as the E2's n I3's direction is opposite then they will be negative...
E1-E2= I1R - I3R

(iii) -E2=-I3R - I4R - I4R
E2=I3R + 2I4R

Thanx A Million
This really Helped!

 

[mehria]

Thanx A Million
This really Helped!

u're welcum

 

[Mohammed salik]

u're welcum

Just One Thing.
Could YOU Plz Elaborate V1 Part (iii). (What about I3 and How do we know E1-E2 or E2-E1!?)

 

[mehria]

Just One Thing.
Could YOU Plz Elaborate V1 Part (iii). (What about I3 and How do we know E1-E2 or E2-E1!?)

check the junctions of the cell... we know that current moves from positive terminal to negative terminal....

 

[mehria]

Just One Thing.
Could YOU Plz Elaborate V1 Part (iii). (What about I3 and How do we know E1-E2 or E2-E1!?)

n as i told u..for a given junction u only have to focus on that junction n ignore the rest of the junctions...
take that junction as a separate circuit n solve ur question
n ths part we dnt have to make I3 a part of our ans as it is a part of the junction BY ... n we r not asked abt anythng related to ths junction...

 

[Mohammed salik]

n as i told u..for a given junction u only have to focus on that junction n ignore the rest of the junctions...
take that junction as a separate circuit n solve ur question
n ths part we dnt have to make I3 a part of our ans as it is a part of the junction BY ... n we r not asked abt anythng related to ths junction...

Ohhh .. I missed That Part.. FeEL Realy Stupid ..!
Anyways.. Thanks For ur Time!!

 

[mehria]

Alright ..What About I3R?

i already told u that we r nt asked about that part of the loop...only focus on the given loop...

 

[Mohammed salik]

i already told u that we r nt asked about that part of the loop...only focus on the given loop...
View attachment 40565View attachment 40566

!!
Jaka' Allah Sister!!
I read That Part Late!

 

[mehria]

Ohhh .. I missed That Part.. FeEL Realy Stupid ..!
Anyways.. Thanks For ur Time!!

anytime