Physics: Post your doubts here!

[Abdulaziz T]

Any Idea about Paper 51 today

what was the scope

My paper 51 exam is after 4 hours Plz if you know how to ans my doubt , plz share

 

[sagar65265]

Thanks but in q22 How to cancel values ? ( I am very very weak )
In q24 why you took 600 nm ? Y not 400nm or any other value ?

Thanks again, And please post the rest answers.

Let's restate that equation:

(Tension in P) * l / AY = 4 * (Tension in Q) * l/ AY
The 4 comes about because we are multiplying by 2 (2l) and dividing by 1/2 (from A/2) so together they give us multiplied by 4.
We can now divide both sides by "l". This gives us

(Tension in P) / AY = 4 * (Tension in Q) / AY

We can further multiply both sides by Y, since both wires have the same Young's Modulus value:

(Tension in P) / A = 4 * (Tension in Q) / A

Lastly, we multiply by A on both sides, since we have already taken into account the coefficients ( the 1/2 for the cross sectional area of Q is part of the 4 multiplier).

(Tension in P) = 4 * (Tension in Q)
So, (Tension in P)/(Tension in Q) = 4/1

For the next one, there's no problem in using any value; suppose we use 400 nm for our wavelength, we apply the same procedure:

1 / (400 * 10^-9) = 2, 500, 000 = 2.5 * 10^ and this power of 10 is still closest to B.

Hope this helped!

Good Luck for all your exams!

 

[Thought blocker]

Check this site guys! Really helpful for finding uncertainties.
http://spiff.rit.edu/classes/phys273/uncert/uncert.html
Credits: sagar65265

♥♥♥♥

 

[sagar65265]

Guyssss plz heeeelpp , how to find v and the uncertainty of v^2 in Q2 ????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_53.pdf

The marking scheme is really confusing; the two tables shown in the marking scheme for Part 2 (b) show the ratio m/(m+M) in the first column and v^2 in the second column, so don't check your values of v in there, you have to check your values of v^2 there.

In the question body, it is written that the speed v was calculated by

"measuring the time t for the card to pass fully through a light gate connected to a timer."

So, since the length of the card (also given in the question body) is 0.200 meters, we can say that the speed is given by

(Length of Card)/(Time taken) = 0.200/t

So, for the first row, the value of v is approximately 0.200/0.174 = 1.15 ms^-1, and squaring that gives us 1.32 m^2 * s^-2.
For the second row, the value of v is approximately 0.200/ 0.132 = 1.52 ms^-1, and squaring that gives us 2.30 m^2 * s^-2.
For the third row, the value of v is approximately 0.200/0.112 = 1.79 ms^-1, and squaring that gives us 3.19 m^2 * s^-2.

And so on.

To calculate the uncertainty in v, we can note that there are only two terms in the formula we have used to obtain the velocity, which are 0.200 m and t seconds.
There is no uncertainty apparent in 0.200 meters; the uncertainty we can see is in t seconds, as given in the table.
Therefore, since we are dividing the values concerned, we take the (percentage uncertainty in 0.200 meters) + (percentage uncertainty in t seconds) as the final percentage uncertainty in the value of v.

For example, the percentage uncertainty in t for the first row is (0.002/0.174 * 100) = 1.149 %
Therefore, the percentage uncertainty for v in the first row is also 1.149 %.
The value of this uncertainty is (1.149/100) * 1.15 = 0.0132.

As a second example, the percentage uncertainty in t for the second row is (0.002/0.132 * 100) = 1.515 %
Therefore, the percentage uncertainty for v in the second row is also 1.515 %.
The value of this uncertainty is (1.515/100) * 1.52 = 0.0229.

So, now that that part is over, let's move on to v^2.

When we are multiplying two numbers, we add their percentage uncertainties; that sum is equal to the percentage uncertainty in the multiplied product.
So, when we are multiplying v into v to give us v^2, we add together the (percentage uncertainty in v) and the (percentage uncertainty in v) to give us
2 * (percentage uncertainty in v).

As an example, suppose we take the second row; v here is 1.52 and v^2 = 2.30.
So, the percentage uncertainty in v = 1.515 % and accordingly, the percentage uncertainty in v^2 = 2 * 1.515 = 3.030 %.
So, to get the final absolute uncertainty in v^2, we ind out what 3.030% of v^2 is:

Absolute Uncertainty in v^2 = (3.030/100) * v^2 = 0.03030 * 2.3 = 0.0695

That is the final absolute uncertainty in v^2.

Hope this helped!
Good Luck for all your exams!

 

[Abdulaziz T]

The marking scheme is really confusing; the two tables shown in the marking scheme for Part 2 (b) show the ratio m/(m+M) in the first column and v^2 in the second column, so don't check your values of v in there, you have to check your values of v^2 there.

In the question body, it is written that the speed v was calculated by

"measuring the time t for the card to pass fully through a light gate connected to a timer."

So, since the length of the card (also given in the question body) is 0.200 meters, we can say that the speed is given by

(Length of Card)/(Time taken) = 0.200/t

So, for the first row, the value of v is approximately 0.200/0.174 = 1.15 ms^-1, and squaring that gives us 1.32 m^2 * s^-2.
For the second row, the value of v is approximately 0.200/ 0.132 = 1.52 ms^-1, and squaring that gives us 2.30 m^2 * s^-2.
For the third row, the value of v is approximately 0.200/0.112 = 1.79 ms^-1, and squaring that gives us 3.19 m^2 * s^-2.

And so on.

To calculate the uncertainty in v, we can note that there are only two terms in the formula we have used to obtain the velocity, which are 0.200 m and t seconds.
There is no uncertainty apparent in 0.200 meters; the uncertainty we can see is in t seconds, as given in the table.
Therefore, since we are dividing the values concerned, we take the (percentage uncertainty in 0.200 meters) + (percentage uncertainty in t seconds) as the final percentage uncertainty in the value of v.

For example, the percentage uncertainty in t for the first row is (0.002/0.174 * 100) = 1.149 %
Therefore, the percentage uncertainty for v in the first row is also 1.149 %.
The value of this uncertainty is (1.149/100) * 1.15 = 0.0132.

As a second example, the percentage uncertainty in t for the second row is (0.002/0.132 * 100) = 1.515 %
Therefore, the percentage uncertainty for v in the second row is also 1.515 %.
The value of this uncertainty is (1.515/100) * 1.52 = 0.0229.

So, now that that part is over, let's move on to v^2.

When we are multiplying two numbers, we add their percentage uncertainties; that sum is equal to the percentage uncertainty in the multiplied product.
So, when we are multiplying v into v to give us v^2, we add together the (percentage uncertainty in v) and the (percentage uncertainty in v) to give us
2 * (percentage uncertainty in v).

As an example, suppose we take the second row; v here is 1.52 and v^2 = 2.30.
So, the percentage uncertainty in v = 1.515 % and accordingly, the percentage uncertainty in v^2 = 2 * 1.515 = 3.030 %.
So, to get the final absolute uncertainty in v^2, we ind out what 3.030% of v^2 is:

Absolute Uncertainty in v^2 = (3.030/100) * v^2 = 0.03030 * 2.3 = 0.0695

That is the final absolute uncertainty in v^2.

Hope this helped!
Good Luck for all your exams!

Thaaaaaaaaaannkkkkk youuuu very much <3 it helped a lot .. Thanks and good luck to you too

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

The relationship needed for this question is the vector relationship F(net) = ma.

This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.

Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.

 

[sagar65265]

Thanks but in q22 How to cancel values ? ( I am very very weak )
In q24 why you took 600 nm ? Y not 400nm or any other value ?

Thanks again, And please post the rest answers.

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

 

[Thought blocker]

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

How do you get symbols like, lambda, rho.. etc etc ?

 

[DeViL gURl B)]

Quest

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 7 I don't understand it!!
Please explain!
Thank you

Question 4 as well please..

 

[psychiatrist]

Let's restate that equation:

(Tension in P) * l / AY = 4 * (Tension in Q) * l/ AY
The 4 comes about because we are multiplying by 2 (2l) and dividing by 1/2 (from A/2) so together they give us multiplied by 4.
We can now divide both sides by "l". This gives us

(Tension in P) / AY = 4 * (Tension in Q) / AY

We can further multiply both sides by Y, since both wires have the same Young's Modulus value:

(Tension in P) / A = 4 * (Tension in Q) / A

Lastly, we multiply by A on both sides, since we have already taken into account the coefficients ( the 1/2 for the cross sectional area of Q is part of the 4 multiplier).

(Tension in P) = 4 * (Tension in Q)
So, (Tension in P)/(Tension in Q) = 4/1

For the next one, there's no problem in using any value; suppose we use 400 nm for our wavelength, we apply the same procedure:

1 / (400 * 10^-9) = 2, 500, 000 = 2.5 * 10^ and this power of 10 is still closest to B.

Hope this helped!

Good Luck for all your exams!

ty

 

[psychiatrist]

25) There is an equation that states that "the intensity of a wave is proportional to the square of its amplitude". Therefore, doubling the amplitude increases the intensity by a factor of 4; tripling the amplitude increases the intensity by a factor of 9.

So, for a particular point, we can write

I = k * A²

using the variables in this situation.
To find the answer we are being asked for, we rewrite the equation, assuming that the new intensity is 2I, the constant k is still the same (it always is for a particular wave) and the amplitude A has increased by a factor of C (i.e. the amplitude at the relevant point is CA):

2I = k * (CA)² and dividing both sides by 2, we get
I = [k * (CA)²]/2

Since the magnitude of I is the same in both equations, we can equate the first equation and this most recent equation to get

[k * (CA)²]/2 = k * A²

Dividing both sides by k, we get

[(CA)²]/2 = A²

Multiplying both sides by 2 and opening out the terms in the brackets. we get

C² * A² = 2 * A²

Dividing both sides by A², we get

C² = 2

Taking the square root on both sides, we get C = √2.

Therefore, the amplitude in the new, relevant position is CA = √2 * A = √2A = B.

Q27)

You need to know how a standing wave looks; this is a very important topic not only in the MCQ paper but in the theory paper as well, so revise that and this problem will become easy. I'm going to assume you know the concept; if you have any doubts after reading this, be sure to revise and ask in the forums until you've grasped the concept. From what I recall, suchalriaz has done a really good post on this but there's a little drought in my head on where it is.

Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength.

Therefore, here half a wavelength is 15 mm = 0.015 meters.
In a last step, we can calculate that if λ/2 = 0.015, λ = 0.030 meters.

Since the waves concerned are electromagnetic waves, they travel at the speed of light (approximated as 3.0 * 10^8 ms^-1) and follow the equation v = fλ. So, we can write the following equation:

3 * 10^8 = 0.030λ
Therefore, λ = 1 * 10^10 meters = C.

Q33)
The formula that relates the resistance of a sample of material to it's physical properties is what we need to know here:

R = ρL/A

Where R is the resistance of the sample, ρ is a property of the material that sample is made of, L is the length of cross-section through which the current flows, and A is the cross-sectional area through which current flows in that sample.

We can rearrange this to give

ρ = RA/L

Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D.

Hope this helped!
Good Luck for all your exams!

25)the amplitude A has increased by a factor of C <--- Where does this came from Is it written in question and I can't see or its a mistake ?
27) Isn't Lambda / 2 for 1st harmonic, I know what you mean by "Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength." But, How did you know we dont use 2nd or 3rd harmonic i.e :
Lambda or Lambda / 4 ?
33) I din't understood "Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D."
:'( I am so weak in Electronics part.

 

[DeViL gURl B)]

The relationship needed for this question is the vector relationship F(net) = ma.

This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.

Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.

Thank you

 

[sagar65265]

25)the amplitude A has increased by a factor of C <--- Where does this came from Is it written in question and I can't see or its a mistake ?
27) Isn't Lambda / 2 for 1st harmonic, I know what you mean by "Either ways, on a standing wave, there is a distance of half a wavelength between any two maxima; from one maxima/anti-node to another/ from one minima/node to another is a distance equal to half a wavelength." But, How did you know we dont use 2nd or 3rd harmonic i.e :
Lambda or Lambda / 4 ?
33) I din't understood "Therefore, the left side is constant for any material, so we can equate the piece of soft metal before rolling and after rolling out to give us the answer.

Note, that part about the volume remaining constant is very, very important. Supposing that the initial length is L and the initial cross sectional area is A.
The initial volume, therefore, is AL.
Taking a look at the final case, we see that the final length is 2L, and the final cross sectional area is some variable say A(2).
In this case, the volume is 2LA(2).

Equating these two, since the volume is constant in both situations, we see that

AL = 2LA(2)

Dividing both sides by 2L, we get

A/2 = A(2)

Therefore, not only does the length change, but the cross sectional area also changes. So, for the final situation, we get (where R(2) is the final resistance, the variable we want to obtain):

ρ = R(2) * (A/2) / (2L)
ρ = R(2)A/(4L)

Since ρ remains constant in both situations, we equate the first equation and this most recent one to get

RA/L = R(2)A/(4L)

Dividing both sides by A, we get

R/L = R(2)/(4L)

Multiplying both sides by 4L, we get

4R = R(2)

So the final resistance is 4R = D."
:'( I am so weak in Electronics part.

26) C is just a random value I included; we know that the value of A changes by a factor; from A it becomes (something) * A, not (something) + A. So, I just decided to make that (something) C - it isn't part of the question, it is just something I put it. You will get the same answer by replacing (CA) with A(final). It doesn't matter, you can put whatever you want there.

27) The harmonics do not matter here;

I'll try posting the rest, I have to go right now.

 

[Thought blocker]

Thank you

For that q4) I am doubting, do you know how to calc uncertainties not in this but in any other question ?

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 24 and 29 .. Please anyone?
Thank you

 

[DeViL gURl B)]

For that q4) I am doubting, do you know how to calc uncertainties not in this but in any other question ?

I was posting the wrong question ll I know how to do that

 

[Thought blocker]

I was posting the wrong question ll I know how to do that

ok then

 

[Hadi Murtaza]

Quest

Question 4 as well please..

Maximum speed = Upper limit of distance/Lower limit of time = 40.1/2.55 ≈ 16.4
Minimum speed = Lower limit of distance/Upper limit of time = 39.9/2.45 ≈ 15.6
Range of uncertainty = (Maximum - Minimum)/2 = (16.4 - 15.6)/2 = ± 0.4 m/s
Answer : C

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_1.pdf

Question 24 and 29 .. Please anyone?
Thank you

24)
This concerns values you need to keep in memory unless they are on the data sheets, i'm not sure. Either ways, visible light falls in the range of
400 nm< λ < 700 nm. We can take any value in this range to represent out wavelength. Suppose we take 600 nm.
Then, the number of wavelengths in 1 meter = 1 / (600 * 10^-9) = 1.6667 * 10^6. So our answer is closest to B. Therefore, that's our answer.
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