Physics: Post your doubts here!

[Thought blocker]

Umm..I dont find any.I think its okay.U tel.?

Umm umm I cant make out! it must be correct

Just 3 golden minutes PLEASE ! xD

The time period of the stationary wave is 20ms. After 5 ms, it would be a straight horizontal line, drawn on the dotted line. Take a look at this animation and you will understand the logic

usama321

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_22.pdf

need help!!!
Q2 ii) 2.

 

[Suchal Riaz]

I would start answering physics queries from tomorrow after i am done with my own revision so i don't misguide other people.

 

[Haya Ahmed]

Umm umm I cant make out! it must be correct

The time period of the stationary wave is 20ms. After 5 ms, it would be a straight horizontal line, drawn on the dotted line. Take a look at this animation and you will understand the logic

usama321

okay and now from where and how you got 20ms and what is meant by 5ms is 1/4 of the cycle I didn't get it HOW !

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf

need help!!!
Q2 ii) 2.

friction=6π × 6.6 × 10–4 × 1.5 × 10–3 × 3.7=6.9 × 10-5
resultant force = mass * acceleration = weight-friction
ma=9.81-6.9× 10-5

mass = rho * volume = 1.5 × 10-5

a=(9.81-6.9× 10-5)/1.5 × 10-5
=4.93 m/s²

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf

need help!!!
Q2 ii) 2.

the question says that the radius of the second raindrop is less than the first.
and F=6pi*D*R*v
if the radius for the second is less the frictional force acting on it is also less.
and Density=m/v
both the rain drops have the same density but the mass if the second is approx one-cube of that of the first bcoz the radius is less.

the difference in mass is much greater than the difference in frictional force. so, when if the equation
mg-f=ma
the acceleration of the second will be less than the first. so the slope of the second graph will be below that of the first. and as the mass of the second drop is less the weight (downward force) will b less and it will reach terminal velocity earlier than the first. so the graph will reach a lower constant value earlier.

 

[Suchal Riaz]

okay and now from where and how you got 20ms and what is meant by 5ms is 1/4 of the cycle I didn't get it HOW !

in 1/4 cycle antinode gets from highest point to zero. then in 1/4 it gets to lowest point then in 1/4 it gets to zero. then in 1/4 it gets back to max. so in one T it completes on cycle.

 

[M.Shahzaib Shoaib]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_2.pdf

1 (B)


I know how to calculate the resultant but don't know how to draw a vector diagram..

Please, help needed !!!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf

Question 1 part c(ii).. the markscheme ( http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_21.pdf ) says that for area's uncertaininty, we have to multiply 2 as well.. Why do we multiply 2?

Supposing we have two numbers defined as

a = 5.0 ± 0.2

b = 3.0 ± 0.6

and we have to multiply them together, thus giving us a result of

a * b = (5.0 ± 0.2)(3.0 ± 0.6)

In this case, we convert their absolute uncertainties into percentage uncertainties before multiplying, and this gives us

a * b = (5.0 ± 4.0%)(3.0 ± 20%)

and so we write our answer as 15.0 ± 24%, because whenever we multiply two numbers, we add their percentage uncertainties together and write that percentage uncertainty for the final product.

If we now extend this to the case where we square a particular number, say we square a, then we get

a^2 = a * a = (5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)
= (25 ± 8.0%)

This 8.0% value we have obtained by adding up the percentage uncertainties in the products, and writing that value as the percentage uncertainty for the final product. Note, however, we could have obtained the same value by multiplying the percentage uncertainty in a by the power we were raising a to, i.e. 2.

Suppose we cube a, we get a similar argument:

a^3 = a * a * a= (5.0 ± 0.2)(5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)(5.0 ± 4.0%)
= (125 ± 12%)

Again, we could have gotten this percentage uncertainty by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=3), which for accuracy's sake we can see is 12%, the same as above.

Suppose we raise a to a power of 0.5, i.e. we take the square-root of a. The same principle applies here:

a^0.5 = (5.0 ± 0.2)^0.5
= (5.0 ± 4.0%)^0.5
= (2.23 ± 2.o%)
Where 2.23 is the square root of 5 to 3 significant figures, and the percentage uncertainty has been obtained by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=0.5) to give us 4.0% * 0.5 = 2.0%.

In the question, you are given the diameter and asked to calculate the area. The formula for doing so is given by

Area = π(d^2)/4

where d is the diameter. In this case, there is no uncertainty in π and no uncertainty in 4, so the uncertainty in the Area is equal to the uncertainty caused by squaring the value of d. So, according to what we've seen above, the % uncertainty in the final value is equal to the %uncertainty in d * the power it has been raised to. So, since it is squared in the formula, we get

% uncertainty in Area = 2 * percentage uncertainty in d.

Hope this helped!

Good Luck for all your exams!

 

[Thought blocker]

I AM GONE, BYE.

 

[Haya Ahmed]

in 1/4 cycle antinode gets from highest point to zero. then in 1/4 it gets to lowest point then in 1/4 it gets to zero. then in 1/4 it gets back to max. so in one T it completes on cycle.

what about the horizontal line how did we come up with it ?

 

[Suchal Riaz]

what about the horizontal line how did we come up with it ?

i think you need some pictures to help you learn it.

does that make anything clear?

 

[Haya Ahmed]

i think you need some pictures to help you learn it.
View attachment 42439
does that make anything clear?

hmmm okay I got the idea now ! .. so in the question it is asking for , draw the stationary wave at time (t + 5.0 ms) so it means we have to add 1/4 to the 1 cycle right ??.. then why did we draw the straight line it isn't 2/4 or 4/4 it is 5/4 !!

 

[AhsanAfzal]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_23.pdf
Q6 (c) (ii) explain this
shouldnt pd remain same in parallel and series arrangement across R2when R1 is zero

 

[Suchal Riaz]

hmmm okay I got the idea now ! .. so in the question it is asking for , draw the stationary wave at time (t + 5.0 ms) so it means we have to add 1/4 to the 1 cycle right ??.. then why did we draw the straight line it isn't 2/4 or 4/4 it is 5/4 !!

it was 5 ms which is =1/4T
it was initially at the 1/4th from my diagram so add 1/4 we get 2/4th stage.

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_23.pdf
Q6 (c) (ii) explain this
shouldnt pd remain same in parallel and series arrangement across R2when R1 is zero

you attached the wrong paper.

 

[AhsanAfzal]

you attached the wrong paper.

sorry...now it is right

 

[AhsanAfzal]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_23.pdf
Q6 (c) (ii) explain this
shouldnt pd remain same in parallel and series arrangement across R2when R1 is zero

 

[sma786]

K

Supposing we have two numbers defined as

a = 5.0 ± 0.2

b = 3.0 ± 0.6

and we have to multiply them together, thus giving us a result of

a * b = (5.0 ± 0.2)(3.0 ± 0.6)

In this case, we convert their absolute uncertainties into percentage uncertainties before multiplying, and this gives us

a * b = (5.0 ± 4.0%)(3.0 ± 20%)

and so we write our answer as 15.0 ± 24%, because whenever we multiply two numbers, we add their percentage uncertainties together and write that percentage uncertainty for the final product.

If we now extend this to the case where we square a particular number, say we square a, then we get

a^2 = a * a = (5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)
= (25 ± 8.0%)

This 8.0% value we have obtained by adding up the percentage uncertainties in the products, and writing that value as the percentage uncertainty for the final product. Note, however, we could have obtained the same value by multiplying the percentage uncertainty in a by the power we were raising a to, i.e. 2.

Suppose we cube a, we get a similar argument:

a^3 = a * a * a= (5.0 ± 0.2)(5.0 ± 0.2)(5.0 ± 0.2)
= (5.0 ± 4.0%)(5.0 ± 4.0%)(5.0 ± 4.0%)
= (125 ± 12%)

Again, we could have gotten this percentage uncertainty by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=3), which for accuracy's sake we can see is 12%, the same as above.

Suppose we raise a to a power of 0.5, i.e. we take the square-root of a. The same principle applies here:

a^0.5 = (5.0 ± 0.2)^0.5
= (5.0 ± 4.0%)^0.5
= (2.23 ± 2.o%)
Where 2.23 is the square root of 5 to 3 significant figures, and the percentage uncertainty has been obtained by multiplying the percentage uncertainty in a (=4.0%) by the power we were raising a to (=0.5) to give us 4.0% * 0.5 = 2.0%.

In the question, you are given the diameter and asked to calculate the area. The formula for doing so is given by

Area = π(d^2)/4

where d is the diameter. In this case, there is no uncertainty in π and no uncertainty in 4, so the uncertainty in the Area is equal to the uncertainty caused by squaring the value of d. So, according to what we've seen above, the % uncertainty in the final value is equal to the %uncertainty in d * the power it has been raised to. So, since it is squared in the formula, we get

% uncertainty in Area = 2 * percentage uncertainty in d.

Hope this helped!

Good Luck for all your exams!

Thabkyou for the fantastic reply <3

 

[Suchal Riaz]

no. when in parallel to LDR the p.d = 20
when in series with ldr the p.d <20 as p.d of both LDR+R2 = 20