Physics: Post your doubts here!

[Thought blocker]

Can someone plz plz plz explain this question to me... I would really be grateful

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

 

[Asad Moosvi]

May someone please explain questions 13, 14 and 20?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf

 

[Batguy]

/

 

[Thought blocker]

Thanks and please do tell me about the practical also i have it as well

21)


28)
Simple, The flow of charge in electric field is from -ve ( tail) to +ve(head)
So Electron will repel, so would move towards the tail, i.e is downwards.
And As there wiuld be repulsion, it would repel with great force in electric field.

31) and 32) are again theory based, I won't reply to it, read chapters, and get answer yourself And still if you have doubts, don't hesitate to ask

 

[unkidd]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf
q9 16 18 19 23 38

 

[Thought blocker]

/

We know estimate of athlete :
K.E of athlete = 4000 J

LEARN THE ESTIMATES
I have uploaded the file

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf
q9 16 18 19 23 38

9)
Force α Acceleration
So when Force = zero, no change in speed, When force is constant, acceleration is also constant, so speed will increase linearly.

16)
The vertical components of both H and W are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.

18)
P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3

Substitute values, Answer is B.

19) sagar65265

23)
F = kx so x = F/k
For A : x = 4/k
For B : x = 3/k
For C : x = 3k
For D : x = 8/3k
So here constant k is either being multiplied or divide, hence we take k as 1 if we would have addition or subtraction included, we take k as zero
Now put 1 instead k .. A = 4, B = 3, C = 3, D = 2.5 so A is the answer

38) sagar65265

 

[Thought blocker]

May someone please explain questions 13, 14 and 20?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

13)
Basic concept of dynamics, forces.
B is the answer, no explanations

14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

20)
Take the leftmost column of liquid.

There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."

What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)

So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.

Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.

When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write

P + ρgh1 = 16,000.

Repeating that calculation on the right side, we get

P + ρgh2 = 8,000

Eliminating P,

16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06

So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.

 

[Asad Moosvi]

13)
Basic concept of dynamics, forces.
B is the answer, no explanations

14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm

20)
Take the leftmost column of liquid.

There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."

What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)

So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.

Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.

When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write

P + ρgh1 = 16,000.

Repeating that calculation on the right side, we get

P + ρgh2 = 8,000

Eliminating P,

16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06

So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.

Hey, thanks. How about question 25 from the same paper? I don't get that.

 

[papajohn]

Q 17 Thought Blocker plzzzzzzzzzzzzzzzzzzz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

 

[Thought blocker]

Hey, thanks. How about question 25 from the same paper? I don't get that.

25)
The Y is half the amplitude of X = 8/2 = 4
And In onw oscillation of Y there is the frequency of 100Hz, so what will be the frequency of Y which oscillates 3 times when X does for 1, that would be 3 * 100 = 300Hz

 

[Thought blocker]

Q 17 Thought Blocker plzzzzzzzzzzzzzzzzzzz
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf

mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

 

[papajohn]

mgsinteta*speed=40% of total power mgsinteta=force,,,,,,force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
Hope it helps.Please tell if the answer u get by this way is differnt from mark scheme...

Tusi great ho yar!

 

[Hijab]

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

Thankyou so much

 

[Hijab]

Taking a look at the image and the details we are given, it seems to be too cumbersome to count the waves at each point and check if they are 1/8th out of phase. In fact, determining what 1/8th out of phase is would be a big headache. So, let's turn to mathematics and try to form a "phase/wave equation".
(Note - there is a real wave equation that, interestingly, applies to any wave in nature. It is a differential equation, and you do not need it for your exams, but in case you want to put this term in Google, you might not get any of the answers you are looking for)..

Okay, back to the question.

We can see for both waves that each one has a constant frequency. This is very important, because otherwise we wouldn't be able to write down a simple equation to relate the wave phases. Keeping this in mind, let's proceed.

For waveform P, we note that there are 4 wavelengths in 18 seconds. This means that 1 wavelength passes by in 18/4 = 9/2 = 4.5 seconds.
So, taking this line of reasoning further, suppose we have a time "t" in mind. How do we find out how many wavelengths have occurred in this time t?

The answer is rather simple - the number of wavelengths that have passed in time t = t/4.5 . To test this, let us put t = 18 seconds. We get, as expected, the number of wavelengths that have passed in 18 seconds = 18/4.5 = 4. Good!

Repeating this procedure for the other wave, we see that the number of wavelengths of waveform Q that pass in the same "t" seconds are t/4 wavelengths.
So when the difference between these two is 1/8 of a wavelength, then we have our answer!

We can say that the second waveform completes it's waves faster than the first waveform (it has a higher frequency) so we write

t/4 = t/4.5 + (1/8)

Rearranging, t(0.25 - 0.222) = 0.125
0.0277 * t = 0.125
t = 4.5 seconds = B.

I have one more question... Kindly explain its answer as well
Q33
If the image is not clear...
U can check the paper... Its oct/nov 2012 variant 13 q33
Thanks in advance

 

[Thought blocker]

I have one more question... Kindly explain its answer as well
Q33
If the image is not clear...
U can check the paper... Its oct/nov 2012 variant 13 q33
Thanks in advance

Thnks to zaqy
okay so we have a wire and it is being stretched. By stretching a wire you only change its length or area you can't change the volume (to change volume you should either remove a part of it or add more copper to it)
and so Volume of stretched=volume of unstretched

A copper wire had a diameter of 1.0mm and it was stretched to 0.5mm diameter. The word stretched makes it quite clear that the length changes
Now the question is by how much did the length change?
We didn't add extra to it we didn't remove from it so that means we can use ratios

I guess you aren't getting this part
let me use an example. lets say you have got 2 bowls of chocolate each bowl contain 10 so you have a total of 20 it will always stay 20 unless you decide to eat one from it or maybe just add another chocolate to it but that's not happening You decide to pick one chocolate from bowl A and you drop it in bowl B
total number still 20 but there is 1 less in bowl A where did it go? to bowl B!

From our question Bowl A is Diameter and Bowl B is the length and total number is the total volume
You working shows that you did consider the change in Diameter but not length
Well basically if you do that your total will change to 19 and in the length case you are just removing a part of it which is not indicated in the question you are suppose to use the same 'long,stretched' wire .Have you played with play doh? if you have you might have noticed while making a snake or anything long the more you reduce the cross sectional area the longer the snake.

We have to find out the change in area in other words the area which was removed to take shape as a length
I have done a lot of talking lets start with the calculations
First thing first finding the change in area which you did already
Area 1 (pie(0.5)^2 )
Area 2 (pie(0.25)^2 )
you have done it and found the ratio as 4/1
so goes for the length which you ignored
find the initial length using the formula R=resistivity*length/area
but as it is the same material i take resistiviy as 1
0.2=l/pie(1*0.5)^2
l=1/20 pie
this is the initial it should also change by 4/1 as the area did
so final l=(1/20 pie)*(4/1) =(1/5 pie)
now you can use the formula again to get the new R with Area=pie(o.25)^2 and l=(1/5 pie)

The best shot (trick) of the question was, we have to change length also with the ratio of 1 : 4
Other than that your work was correct, as I saw in that pic, I did the same mistake though!
The thing matter is, to get your concepts cleared and hope best for future, and if you keep working hard, A is your's

 

[Asad Moosvi]

9)
Force α Acceleration
So when Force = zero, no change in speed, When force is constant, acceleration is also constant, so speed will increase linearly.

16)
The vertical components of both H and W are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.

18)
P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3

Substitute values, Answer is B.

19) sagar65265

23)
F = kx so x = F/k
For A : x = 4/k
For B : x = 3/k
For C : x = 3k
For D : x = 8/3k
So here constant k is either being multiplied or divide, hence we take k as 1 if we would have addition or subtraction included, we take k as zero
Now put 1 instead k .. A = 4, B = 3, C = 3, D = 2.5 so A is the answer

38) sagar65265

I don't get 18

 

[Thought blocker]

I don't get 18

P = E / t ( E = 0.5 * m * v^2)
P = ( 0.5 * m * v^2 )/ t
Now as you know mass = density (ρ) * volume (V)
P = ( 0.5 * ρ * V * v^2 ) / t
Now you know Volume = Area (A) * Length (l)
P = ( 0.5 * ρ * A * l * v^2 ) / t
Now you know Velocity = l / t
P = 0.5 * ρ * A * v^3
Now substitute the values :
Keep in mind
It converts the power available in the
wind to electrical power with an efficiency of 50%
i.e 50% P
Now substitute the values :¬
(100/50)P = 0.5 * 1.3 * 2000 * 10^3
2P = 1300000
P = 650000 = 650kW = B

 

[Asad Moosvi]

P = E / t ( E = 0.5 * m * v^2)
P = ( 0.5 * m * v^2 )/ t
Now as you know mass = density (ρ) * volume (V)
P = ( 0.5 * ρ * V * v^2 ) / t
Now you know Volume = Area (A) * Length (l)
P = ( 0.5 * ρ * A * l * v^2 ) / t
Now you know Velocity = l / t
P = 0.5 * ρ * A * v^3
Now substitute the values :
Keep in mind
It converts the power available in the
wind to electrical power with an efficiency of 50%
i.e 50% P
Now substitute the values :¬
(100/50)P = 0.5 * 1.3 * 2000 * 10^3
2P = 1300000
P = 650000 = 650kW = B

Wow, couldn't tell you how much I appreciate this! Number 17, please?

 

[Thought blocker]

Wow, couldn't tell you how much I appreciate this! Number 17, please?

v = (2gh)^1/2
HEIGHT IS 0.72 NOT 0.8! Measure from the base of the ball. Diameter is 0.08. We have to subtract that.

Use 1/2mv^2 = initial KE to calculate mass.

u = (2gh)^1/2
Again, height is 0.45-0.08 m.
KE after impact = 1/2mu^2 = 0.39 J

Answer is C