Physics: Post your doubts here!

[Usman Chughtai]

i got it look
the horizontal line given is 7 cm
so scale is 4ms^-1 = 7
so 1 ms^-1 =1.75 cm
vertical velocity is 6.2
draw the resiltant it will be 7.4
do u want me to post the diagram but its not neat ?

Oh thanks a lot
would appreciate if u could post it here or Inbox me

 

[TheJDOG]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q5 b complete

Hello there, for B I)
We know that there is a stationary wave in that tube.
At antinodes which are maximum displacement, the wave pushes the heaps apart.
At nodes which are minimum displacement, the heaps are formed
This forms the dust heaps.

For B Ii)

Look at the figure, you'll see that the distance between 6 heaps is 0.39 meters

We know that this stationary wave's wavelength at these heaps is 2.5 Lambda.
Because it has Compression-Rarefaction-Compression and so on.
One wavelength is from C to C or from R to R.

We know Freuquncy F= 2140 Hz

We know speed of wave equation V= Lambda x F


We need 1 lambda and not 2.5 lambda so we need to convert.

0.39 m-----> 2.5 lambda
? ------------> 1 lambda

It's 0.156 m

So V= 2140 x 0.156=334m/s


Iii)
Stationary waves is formed by interference of incident and reflected waves
the speed is the speed of the incident waves

 

[Thought blocker]

I am on!

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

Question 6 complete PLEASE!!!
Especially the b part!!! :$

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_23.pdf. Q6 ci pleaseeee help

 

[omaaaar]

Hello there, for B I)
We know that there is a stationary wave in that tube.
At antinodes which are maximum displacement, the wave pushes the heaps apart.
At nodes which are minimum displacement, the heaps are formed
This forms the dust heaps.

For B Ii)

Look at the figure, you'll see that the distance between 6 heaps is 0.39 meters

We know that this stationary wave's wavelength at these heaps is 2.5 Lambda.
Because it has Compression-Rarefaction-Compression and so on.
One wavelength is from C to C or from R to R.

We know Freuquncy F= 2140 Hz

We know speed of wave equation V= Lambda x F


We need 1 lambda and not 2.5 lambda so we need to convert.

0.39 m-----> 2.5 lambda
? ------------> 1 lambda

It's 0.156 m

So V= 2140 x 0.156=334m/s


Iii)
Stationary waves is formed by interference of incident and reflected waves
the speed is the speed of the incident waves

Can you explain how did you get the wavelength as 2.5 lambda

 

[_Ahmad]

Thought blocker

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Can you please explain Q5 a ii) and iii)

 

[kitkat <3 :P]

Oh thanks a lot
would appreciate if u could post it here or Inbox me

 

[TheJDOG]

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

In question paper 21 can anyone explain Q)5) especially part b)?

Hey, for part b)

This is the key for this question: For minima to occur: Path difference= 0.5Lambda, 1.5 Lambda. 2.5 Lambda and so on...

If you look at the figure, you'll see that it's a right angled triangle, use paythagors theorem. You'll get S2M= 128 cm

Path diff= 128 - 100= 28 cm


Your range is from 1khz to 4 kHz

Use v= lambda x frequency
Speed = 330 m/s

Range of lambda= 8.25 cm to 33cm

Now, start trying, remember path diff= nlambda

So 28= 0.5 x Lambda, 56 not in allowed range
28= 1.5 x lambda, 18.6 allowed range
28= 2.5 x lambda , 11.2 allowed range

So 2 minima occur

 

[Ahmed Aqdam]

http://papers.xtremepapers.com/CI.E/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_23.pdf. Q6 ci pleaseeee help

For maximum light intensity, the resistance will be lowest i.e. 1.2 kΩ. So total resistance: 1/1200 + 1/600=1/R
R is 400 Ω.

 

[IGCSE13]

For maximum light intensity, the resistance will be lowest i.e. 1.2 kΩ. So total resistance: 1/1200 + 1/600=1/R
R is 400 Ω.

Thanks a lot I was a little bit confused

 

[TheJDOG]

Can you explain how did you get the wavelength as 2.5 lambda

It's a pattern, Compression Rarefaction Compression Rarefaction....
Each heap is in the pattern of C R C R C R
A wavelength is from C to C or from R to R. In this case, the 0.39 m distance of 6 heaps is 2.5 lambda.
Try it on your figure

 

[Ahmed Aqdam]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf

Question 6 complete PLEASE!!!
Especially the b part!!! :$

(i)Total p.d.= 9V.
p.d. across 1.2 kΩ resistor is 5V as 4V is the voltmeter reading.
The total resistance will be the ratio of p.d.'s as current is same.
4/5 × 1200
=960 Ω

(ii) for parallel combination: 1/R=1/960-1/1600
R= 2400 Ω
According to graph: temperature at R=2.4kΩ = 11 °C

 

[Mohammed salik]

I am on!

Some One Plz DRAW THEM Plz..!
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_23.pdf Q6 C )iv)


http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_21.pdf Q5)b)

Thnx In ADVANCE..!

 

[Ahmed Aqdam]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Can you please explain Q5 a ii) and iii)

(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1
As I2 and I1 both pass half through the metal wire their p.d. will be divided by two.
(iii) E2 and I3 is in negative direction. p.d. across R2 will be half as the other half is not included in the loop. I1 is common across R1, r1 and R2.
E1 – E2 = –І3r2 + І1 (R1 + r1 + R2 / 2)

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Can you please explain Q5 a ii) and iii)

The resistance of the first half of the metal wire is R2/2 and so is that of the second part but the current in the first part of the metal wire is I2 because both it gets voltage from E1 and E2 this is not the same as the current in the second half of the metalwire which gets voltage only from E1 so E1= I1r1+I2R2/2+I1R2/2+I1R1

 

[IGCSE13]

(ii)According to Kirchhoff's second law: E1 = І2R2/2 + І1R2/2 + І1R1+ І1r1
As I2 and I1 both pass half through the metal wire their p.d. will be divided by two.
(iii) E2 and I3 is in negative direction. p.d. across R2 will be half as the other half is not included in the loop. I1 is common across R1, r1 and R2.
E1 – E2 = –І3r2 + І1 (R1 + r1 + R2 / 2)

I don't get why are we not including I2R2/2 in iii

 

[omaaaar]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf
Q1 b part 2 uncertainty never understood it can u explain clearly how to do

 

[Ahmed Aqdam]

I don't get why are we not including I2R2/2 in iii

The loop does not pass through BJ so I2 isn't included.

 

[_Ahmad]

I don't get why are we not including I2R2/2 in iii

because the loop HBCDJFGH doesn't include BJ so we will not include I2R2/2