Physics: Post your doubts here!

[snowbrood]

u r awe great

hehe thanks buddy

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_ms_1.pdf
Mcq no. 21.
How do we solve it?!

well frequency is going to be same in both cases as frequency depends on the source now just plug in values in the equation v=fw(let w equal to wavelength)
1500=150w1
w1=10
then
300=150w2
w2=2
hope u get it

 

[sandwich147]

Hi, I'm new here. anyone have a good A2 notes on oscillation? I'm having a hard time studying the topic. Thanks in advanced!

 

[Tkp]

can u pls answer this question​

An insect of mass 4.5mg flying with a speed of 0.12m/s-1,encounters a spider's web which brings it to rest in 2.0ms .calculate the average force exerted by the insect on the web.​

​

this is an phy AS level question under momentum topic​

well u have to know the formula.f=(dell p)/dell t
f=mv/t.(just take the mg in kg and ms in s and u will get the ans)

 

[asangietalks]

can someone help me?? im doing may june 2012 paper 1 variant 2, i cant solve no 22. http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
thanks.

 

[Scafalon40]

Guys can someone help me with this?

If you have a raw data value like l= 6.0 +/- o.4 and you're asked to calculate l^2 and write down its absolute uncertainty, how do you know what the no of significant figures for the uncertainty should be?
And what about a number like x=750 +/- 20 and you're asked to do the same thing as above, only this time it's ln(x) which you have to calculate.
All I need to know is how many s.f should the uncertainty be quoted to.
Thanks!

Anyone? ^

 

[Dug]

can someone help me?? im doing may june 2012 paper 1 variant 2, i cant solve no 22. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
thanks.

Density = Mass/Volume
Volume = (3.5 x 10^-25)/(9.2 x 10^3)
Let 'x' be the length of a side of the cube. From the diagram, the shortest distance between 2 atoms is 'x'.
Volume of a cube = x^3
x^3 = (3.5 x 10^-25)/(9.2 x 10^3)
x = 3.4 x 10^-10 m

 

[Ali_00921]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf Q.5 b(ii)?? Kindly.. Thanks in Advance

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf Q.5 b(ii)?? Kindly.. Thanks in Advance

Distance between 2 successive heaps (minimum points) = λ/2

39cm covers 5 times of this distance, therefore:
5(λ/2) = 39 cm
λ = 15.6 cm = 15.6 x 10^-2 m

v = fλ
v = 2.14 x 10^3 x 15.6 x 10^-2 = 334 ms-1

 

[asangietalks]

Density = Mass/Volume
Volume = (3.5 x 10^-25)/(9.2 x 10^3)
Let 'x' be the length of a side of the cube. From the diagram, the shortest distance between 2 atoms is 'x'.
Volume of a cube = x^3
x^3 = (3.5 x 10^-25)/(9.2 x 10^3)
x = 3.4 x 10^-10 m

Thank you so much. 1 more question, why we arent multiply the mass of each atom by the number of atom that shows in the diagram?

 

[1357913579]

Can anyone please help me in
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_1.pdf
number-5

 

[Dug]

Thank you so much. 1 more question, why we arent multiply the mass of each atom by the number of atom that shows in the diagram?

We are finding the shortest distance between the centres of two adjacent atoms. We have the mass of an atom and the density of the crystal is also provided. Volume of the atom can easily be found and then the length 'x' by using cube-root. 'x' is the same as the length measured between the centres of the two adjacent atoms. The keyword here is adjacent. Try to read the statement carefully and understand it.

 

[xxfarhaxx]

asalamu alaikum,
can anyone plss tel me how to calculate uncertinities for P5????

 

[daredevil]

Can anyone please help me in
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
number-5

the answer wud be C (330 m/s) becasue the value is to be written correct to the number of decimal places same as the absolute error that means here it is 3 so we have to remove the decimals from our answer which makes it 328 after rounding off and to bring it in accordance with the absolute error we write 330 m/s. do you get it? if not then ask away any problems u hav

 

[asangietalks]

We are finding the shortest distance between the centres of two adjacent atoms. We have the mass of an atom and the density of the crystal is also provided. Volume of the atom can easily be found and then the length 'x' by using cube-root. 'x' is the same as the length measured between the centres of the two adjacent atoms. The keyword here is adjacent. Try to read the statement carefully and understand it.

oke. thnks anyway.. would you help to explain no 6 in the same question paper?

 

[Dug]

oke. thnks anyway.. would you help to explain no 6 in the same question paper?

The time-base setting is 1 µs/cm. 1 square represents 1 cm.
Therefore,
1 µs = 1 cm
Each pulse occupies 2 squares.
2 cm = 2 µs

 

[Scafalon40]

Guys can someone help me with this?

If you have a raw data value like l= 6.0 +/- o.4 and you're asked to calculate l^2 and write down its absolute uncertainty, how do you know what the no of significant figures for the uncertainty should be?
And what about a number like x=750 +/- 20 and you're asked to do the same thing as above, only this time it's ln(x) which you have to calculate.
All I need to know is how many s.f should the uncertainty be quoted to.
Thanks!

 

[daredevil]

Guys can someone help me with this?

If you have a raw data value like l= 6.0 +/- o.4 and you're asked to calculate l^2 and write down its absolute uncertainty, how do you know what the no of significant figures for the uncertainty should be?
And what about a number like x=750 +/- 20 and you're asked to do the same thing as above, only this time it's ln(x) which you have to calculate.
All I need to know is how many s.f should the uncertainty be quoted to.
Thanks!

your absolute error should be correct to 1s.f. and then the value should be correct to the same no. od d.p. as ur resulting absolute error is.
like if u hav calculated a length to be 20.324 +/- 0.346cm

then ur correct absolute error should be +/- 0.3
and thus the resulting value should be 20.3 as the absolute error comes to 1 d.p.


for another case you have the speed to be 326.55 +/- 1 m/s
then ur answer should be:

327 m/s
because the absolute errror is correct to 0 s.f.

even if the value does not have a number after the decimal but the absolute error is correct to 1 d.p. or so u write abc.0 insead of just writing abc

hope u get it
1357913579 u might wanna check this out ... maybe it will help ^_^

 

[1357913579]

your absolute error should be correct to 1s.f. and then the value should be correct to the same no. od d.p. as ur resulting absolute error is.
like if u hav calculated a length to be 20.324 +/- 0.346cm

then ur correct absolute error should be +/- 0.3
and thus the resulting value should be 20.3 as the absolute error comes to 1 d.p.


for another case you have the speed to be 326.55 +/- 1 m/s
then ur answer should be:

327 m/s
because the absolute errror is correct to 0 s.f.

even if the value does not have a number after the decimal but the absolute error is correct to 1 d.p. or so u write abc.0 insead of just writing abc

hope u get it
1357913579 u might wanna check this out ... maybe it will help ^_^

thanks a lot bro, Jazak Allah khairn

 

[Reh_456]

HI
does anyone know how 2 do this?
9702_s12_qp_12
question 14
thnx in advance!