Physics: Post your doubts here!

[Unicorn]

CAN SOMEONE PLEASE ANSWER QUESTIONS 6b and 6c

for question 6b
they tell u that the distance between where sound is heard is 32.4 and if u look closely in fig 6.1 you will see it is a node
and the distance between 2 nodes is 0.5ƛ
and we know s=f * ƛ
so 0.5ƛ=32.4cm so ƛ=64.8cm =0.648

therefore 0.648*512 =331.78 -> 332m/s

for 6c
distance between a node and antinode it is 0.25ƛ which is half distance between 2 nodes
so 32.4/2 =16.2

and the length of air is 15.7 so 16.2-15.7 =0.5 above the end of the tube or simply 16.2 cm above water level

hope it helped

 

[Jaf]

What is that formula for? If it is for area, I am getting a different answer for the area, using that formula.

You shouldn't. It's the same formula.
Cross-section area of the wire = πr² or π(d/2)² ---> π d²/2² ---> πd/4

We're not halving the uncertainty because we shouldn't! Think of how an uncertainty occurs when you're measuring the diameter. It is because of the markings in the ruler. You're halving the value read by the same ruler so the uncertainty will not change.
You will only multiply fractional uncertainties by a number if the value is raised to a power of the same number in the equation you're dealing with (the number could be a fraction too).

 

[Peter Check]

A question on planning an apparatus to find the young modulus
A question part states that some has measured the original length of a wire, its extension, the load at extension, and deduced the cross-area. Now the question is Describe how the measurements taken can be used to determine the Young modulus.

A part of the answer in ms is:
plot a graph of force against extension (B1)
determine gradient of graph for F / e (B1)
calculate E from E = F l / e A or gradient × l / A (B1)

Is it necessary to plot a graph?? I mean I have all the values, can i just plug it into the equation for young modulus??

 

[hussamh10]

What is that formula for? If it is for area, I am getting a different answer for the area, using that formula.

7.5=p x 1750/Pi(0.38x10^-3)^2/4 (put square only on D) Use the same for the Uncertanities

 

[hussamh10]

Plz help Whole question M/J 2006 Q5
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s06_qp_2.pdf

 

[Peter Check]

You shouldn't. It's the same formula.
Cross-section area of the wire = πr² or π(d/2)² ---> π d²/2² ---> πd/4

We're not halving the uncertainty because we shouldn't! Think of how an uncertainty occurs when you're measuring the diameter. It is because of the markings in the ruler. You're halving the value read by the same ruler so the uncertainty will not change.
You will only multiply fractional uncertainties by a number if the value is raised to a power of the same number in the equation you're dealing with (the number could be a fraction too).

okay yeah, i thought the formula was pieD^.5

 

[applepie1996]

for question 6b
they tell u that the distance between where sound is heard is 32.4 and if u look closely in fig 6.1 you will see it is a node
and the distance between 2 nodes is 0.5ƛ
and we know s=f * ƛ
so 0.5ƛ=32.4cm so ƛ=64.8cm =0.648

therefore 0.648*512 =331.78 -> 332m/s

for 6c
distance between a node and antinode it is 0.25ƛ which is half distance between 2 nodes
so 32.4/2 =16.2

and the length of air is 15.7 so 16.2-15.7 =0.5 above the end of the tube or simply 16.2 cm above water level

hope it helped

but there is only one node so y r u taking the distance between 2 nodes and thnx!!!

 

[Unicorn]

but there is only one node so y r u taking the distance between 2 nodes and thnx!!!

in fig 6.1 there is 1 node and that is where sound is heard and in fig 6.2 sound is heard at the other node which is 32.4cm away

 

[Hassi123]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf
Q5 cii) how do we do this and please show working! Thankyou!

 

[applepie1996]

in fig 6.1 there is 1 node and that is where sound is heard and in fig 6.2 sound is heard at the other node which is 32.4cm away

THNX SO MUCH I GET IT NOW

 

[Unicorn]

THNX SO MUCH I GET IT NOW

welcome ^_^

 

[Peter Check]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
question no. 5(b) . from the first variant paper. Why is my answer wrong? I used the formula dsinθ =(n+0.5)λ I got theta has 38.66 degrees, and my d was .8m(80cm). Lambda was found using V=fλ . I got my n value as 5, which is way off. Can anyone make me understand why my method/approach is wrong

 

[00tanveer]

The answer matches and thanks man.

 

[00tanveer]

How do you this range thingy? Explain.

 

[Soldier313]

What is the speed of an alpha particle? is it 0.1(3x10)^8?? and what about the speed of beta particles?

 

[perkypearl]

anyone can help me to do oct/nov 2007 Q#2 and Q#1(b)
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf

 

[Soulgamer]

Please someone solve Q.4 d (i) and (ii)

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf

With explanation and diagram of-course. Thank you!

 

[hussamh10]

anyone can help me to do oct/nov 2007 Q#2 and Q#1(b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

Q1 )b First of all find out the radius which is 0.488
Then Work out the uncertainity
Delta R/R=DeltaV/V +DeltaL/L
Delta R/0.488=0.5/15+0.1/20
We will get Delta R=0.018706 ---->Now this is the uncertainity of R^2 so we will divide it By 2 to get the Uncertainity Of R
Delta R=0.018706/2------>0.00935
Radius =0.488 +/- 0.009

 

[Navi Don]

please explain me how to make two waves in a same graph paper which have same phase angle eg 30 degree phase angle.how to make?

 

[haroon740]

Is current Vector or Scalar ?