Physics: Post your doubts here!

[Norton]

Can someone help me to "Design an experiment to show interference of light waves using a laser beam"
for AS level PLZZZZZZZZZZZZZZZZZZZ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[Kumkum]

can someone pls help me with this
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 4 and 12

 

[Dug]

can someone pls help me with this
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
question 4 and 12

4)
T = 1/50 = 0.02s = 20ms
Time-base = 10ms
Therefore, 1 complete wave takes 2 divisions horizontally.

Y-gain = 5V
Amplitude = 5

D is the answer.

12)
Initial Momentum = Final Momentum
1000 x 5 = 10x
x = 500m/s

 

[Kumkum]

4)
T = 1/50 = 0.02s = 20ms
Time-base = 10ms
Therefore, 1 complete wave takes 2 divisions horizontally.

Y-gain = 5V
Amplitude = 5

D is the answer.

12)
Initial Momentum = Final Momentum
1000 x 5 = 10x
x = 500m/s

Jazakallah

 

[Hi_all]

Can anyone tell me about waves and superposition on skype? Please

 

[Yousuf Ykr]

guys please help me to solve this questions 10,19,21,23,26,28,29,31,34,36,37,38

The answer is attached to the MS word. Didn't do the last three coz didn't have time. maybe some other time

 

[Ali_00921]

Kindly, can someone explain the relation between acceleration and velocity? And how to find acceleration from a velocity time graph?? Thank's in advance fella's

 

[salvatore]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_2.pdf

In question no. 3(b) of the paper above, why do we have to multiply the mass of alpha particle (4) by 1.66 x 10^-27?
If you consider qn c(ii), the mass used is just 4 (Without multiplying by 1.66 x 10^-27).

Could anyone please explain this?
Thanks

 

[salvatore]

Kindly, can someone explain the relation between acceleration and velocity? And how to find acceleration from a velocity time graph?? Thank's in advance fella's

In short, acceleration is the rate of change of velocity... and velocity is the speed in a given direction (Speed is a scalar quantity & velocity is a vector quantity).
To calculate acceleration from a velocity-time graph, you are supposed to find the gradient of the linear section of the graph. If the acceleration to be found is at a curved point, draw a tangent at that point and find the gradient.

Hope that helped

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_2.pdf

In question no. 3(b) of the paper above, why do we have to multiply the mass of alpha particle (4) by 1.66 x 10^-27?
If you consider qn c(ii), the mass used is just 4 (Without multiplying by 1.66 x 10^-27).

Could anyone please explain this?
Thanks

In 3(b) you are only considering the Alpha particle which is a helium nucleus

4
2​

He. So while finding the mass you have to multiple it my the atomic mass constant but in c(ii) you have to form an equation 'momentum of francium=
momentum of alpha + momentum of astatine' the atomic mass of all 3 particles has to be multiplied by the atomic mass constant so you can just cancel it out from both sides to simplify the calculation. Even if you don't the answer will be the same, it'll just be a lengthier calculation.

 

[hela]

QUESTION 5 PLAESE URGENT

 

[hela]

HELP PLAESE QUESTION 5

 

[littlecloud11]

HELP PLAESE QUESTION 5

5a) Two condition would be equal amplitude and phase difference of 180 degree so that the vales cancel each other out at M and the intensity is 0.

5b) First you have to find the path difference between S1 and S2. For that calculate the hypotenuse of of the triangle, √(80^2+100^2) = 128. So the path difference is 128-100 = 28cm. Then calculate the range of wavelength of the sound due to the change of frequency.
330=(1000)λ , λ = 33 cm
330=(4000)λ , λ = 8.25 cm

For the minima of sound the equation for the wavelength is (n+1/2)λ, where n= 0, 1, 2..so on. So, the minima occurs at 1/2λ , 3/2λ, 5/2λ ,7/2 λ and so on.
Equate the values of λ with the path difference-
1/2λ=28 , λ=56
3/2 λ = 28, λ=18.7
5/2λ= 28, λ = 11.2
7/2λ =28, λ =8

Now, for these only two values (11.2 and 18.7) lie in the range 8.25cm and 33cm, so there are 2 minimas.

 

[hela]

Thank you very much for your help

 

[Ali_00921]

In short, acceleration is the rate of change of velocity... and velocity is the speed in a given direction (Speed is a scalar quantity & velocity is a vector quantity).
To calculate acceleration from a velocity-time graph, you are supposed to find the gradient of the linear section of the graph. If the acceleration to be found is at a curved point, draw a tangent at that point and find the gradient.

Hope that helped

Aaaahhh, now i get it.. Thank you so muchh!! But what if we have constant velocity increase in a VELOCITY-TIME GRAPH, can we use the formula (a=v-u/t)?

 

[hela]

help q8

 

[snowbrood]

HELP QUESTION 5

for first part for destructive interference waves phase difference between the two waves should be 180 or we can say that they are out of phase by 180 thats how complete destructive interference occurs.
for second condition
these waves must have same amplitude and same frequency so that to form a destructive interference i.e to get a minima

for the second part of the question this is a bit lengthy
step 1 find the distance from s2 to M
using Pythagoras theorem squareroot{(100/100)^2+(80/100)^2} we get 1.28
now lets find the phase difference
1.28-1=0.28
let wavelength=w (as i find it difficult to write lamda)
w=330/f(w=v/f we know from question that v=330)
formula for destructive interference is (n-0.5)w
(n-0.5)w=0.28
now plug in 330/f into w
(n-0.5)330/f=0.28
(n-0.5)330=0.28f
f=(n-0.5)1180(correct to 3 s.f)
plug in values from 1 lets see what we get
when n=1
f=590 and w=56cm
when n=2
f=1770 and w=18.7cm
when n=3
f=2950 and w=11.2
when n=4
f=4130 and w=8cm
so there are only two values that satisfy our range of f values
as we are given the range of 1000hz to 4000hz.
only 2 and 3 are considered.
so there are only 2 minima

 

[snowbrood]

help q8

x=ut+1/2(at^2)
we know that motion started from rest so u=0
x=1/2(at^2)
l=1/2(gt^2) equation 1

if t=0.5
1/2g(0.5t)^2
0.5*g*0.25t^2 equation 2
comparing equation 1 and 2 we get 0.25l
as 0.25 will cut each other at both ends and we will get our original equation 1 hope u understand

 

[hela]

help this question plaese 8 and 11 mcq
thanks

 

[snowbrood]

help this question plaese 8 and 11 mcq
thanks

dear i have answer q8 already i am goin to answer 11 just in a minute