Physics: Post your doubts here!

Snow Angel

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Browny

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How many significant figures should we leave our answer to in the physics theory paper? I would be very thankful to any one who can give me a detailed explanation.:)
 
Thought blocker

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Will come at night to solve everyone's doubt...
The people who asked me for Projectile motion formulas, here you go :

P R O J E C T I L E M O T I O N !​

Motion in two dimensions,
v and angle θ with
horizontal, upward is +ve

x-component → no acceleration
Vx = V cos Ø
x = Vxt = Vt cos Ø

y-component → acceleration is g
Vy = V sin Ø
y = Vyt − ½ gt^2
horizontal range → R =(V^2/g) sin 2Ø
max range at θ=45 deegree.



>>>>>>>MORE DETAILED VIDEO ON YOUTUBE<<<<<<<<​
 
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princeali97

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princeali97

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Can someone please send me some good labelled diagrams for the following questions.
Q:A loudspeaker produces a sound wave of constant frequency.Outline how a cathode-ray oscilloscope (c.r.o.) may be used to determine this frequency.(o/n 10 p23)
Q:The wave nature of light may be demonstrated using the phenomena of diffraction and interference.Outline how diffraction and interference may be demonstrated using light.In each case,draw a fully labelled diagram of the apparatus that is used and describe what is observed.(o/n 09 21)
Q:Outline briefly an experiment that may be used to demonstrate (i) diffraction of a transverse wave. (ii)Suggest how your experiment in (i) may be changed to demonstrate the diffraction of a longitudinal wave.(W08)
Q:The wave in (b) was produced in a ripple tank. Describe briefly, with the aid of a sketch.(w13 qp 23)
diagram, how the wave may be observed
QDescribe apparatus that demonstrates Brownian motion. Include a diagram.(s13 qp23)
Q:A ripple tank is used to demonstrate interference between water waves.
Describe(i) the apparatus used to produce two sources of coherent waves that have circular
wavefronts
Q:Experiment for determining Young Modulus.
Q-Demonstrating interfernce in i)microwaves ii)Water waves.
Q:Observing Stationary waves i)using a stretched string ii)Microwaves iii)Air column.
All these questions carry more than 4 marks and require some explanation too.I have done these questions in the past but i want to make sure what i will draw is correct.I will be very thankful to your response and I hope it helps others too.
Thank you.
Best of luck. :)
 
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omaaaar said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf
Q5 b part pls
Click to expand...
Here you go :) Credits papajohn
Distance S2M =128 (from Pythagoras thereom)
Path difference S2M - S1M
128-100 = 28 (If you do not know this concept then take help from Pacific Physics Vol 2 )

Wavelenght at 1 Khz of sound wave is λ=v/f 330/1^103= 33cm
λ of 4kHz 330/4^103= 8.25 cm

Path difference for minima = odd number of the half the λ
Or simply (2n+1) * λ/2
Put n = 0 in the formula (2*0 +1) λ/2 = 56
Put N= 1 =2(1)+1) * λ/2 =18.7
Put N = 2 2(2) =1) * λ/2 = 11.2
Put N= 4 2(3) +1* λ/2 = 8
Sice λ 56 cm and 8 cm are not within the range of 8.25 to 33 cm, so minima is obtained for λ 11.2 and 18.7cm Therefore two minma are detected.
 
omaaaar

omaaaar

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Thought blocker said:
Here you go :)
Distance S2M =128 (from Pythagoras thereom)
Path difference S2M - S1M
128-100 = 28 (If you do not know this concept then take help from Pacific Physics Vol 2 )

Wavelenght at 1 Khz of sound wave is λ=v/f 330/1^103= 33cm
λ of 4kHz 330/4^103= 8.25 cm

Path difference for minima = odd number of the half the λ
Or simply (2n+1) * λ/2
Put n = 0 in the formula (2*0 +1) λ/2 = 56
Put N= 1 =2(1)+1) * λ/2 =18.7
Put N = 2 2(2) =1) * λ/2 = 11.2
Put N= 4 2(3) +1* λ/2 = 8
Sice λ 56 cm and 8 cm are not within the range of 8.25 to 33 cm, so minima is obtained for λ 11.2 and 18.7cm Therefore two minma are detected.
Click to expand...
Thanks man got i had not understood this concept yet but got it now
 
MiniSacBall

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Thought blocker said:
Here you go :) Credits papajohn
Distance S2M =128 (from Pythagoras thereom)
Path difference S2M - S1M
128-100 = 28 (If you do not know this concept then take help from Pacific Physics Vol 2 )

Wavelenght at 1 Khz of sound wave is λ=v/f 330/1^103= 33cm
λ of 4kHz 330/4^103= 8.25 cm

Path difference for minima = odd number of the half the λ
Or simply (2n+1) * λ/2
Put n = 0 in the formula (2*0 +1) λ/2 = 56
Put N= 1 =2(1)+1) * λ/2 =18.7
Put N = 2 2(2) =1) * λ/2 = 11.2
Put N= 4 2(3) +1* λ/2 = 8
Sice λ 56 cm and 8 cm are not within the range of 8.25 to 33 cm, so minima is obtained for λ 11.2 and 18.7cm Therefore two minma are detected.
Click to expand...

Why do we need to calculate path difference i didn't see any use.:p

And did you calculate those, i mean when n = 0 then it will be λ/2 which is 4.125, λ=8.25
I don't know how did you get 56
 
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Thought blocker

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MiniSacBall said:
Why do we need to calculate path difference i didn't see any use.:p

And did you calculate those, i mean when n = 0 then it will be λ/2 which is 4.125, λ=8.25
I don't know how did you get 56
Click to expand...
o_O o_O
It's like this :¬
ODD NUMBER OF HAVE λ = PHASE DIFFERENCE FOR MINIMA

This implies (2n + 1)*λ/2 = 28

(2(o)+1)* λ/2 = (28)---> Phase difference.
Hence λ = 56..
Same with
(2(1)+1)*λ/2 = 28 ; λ = 18.7
and so on..
 
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