Physics: Post your doubts here!

[Bishnu Dev]

View attachment 62048
View attachment 62049View attachment 62052
explain?

b) i)
The phase shift between the waves (T1 and T2) is 60 degrees. Draw wave T2 5 small squares ahead of T1. Remember, both waves have same Period and Amplitude. To draw this, move the wave T2 to the right by 5 small squares. Check attachment 1.

ii) Now at time T=1 second, count the number of small squares down the X-axis until you reach the T1 wave. This is the displacement of the wave T1. Recall that 10 squares = A. Since it is 5 small squares below the time axis, it is negative. Displacement of T1 = -A/2 (Don't ignore the '-' Sign) Check attachment 2.

For wave T2, the displacement is above the time axis, so that's positive. Repeat the same counting process and you will get 5 small squares for wave t2 as well. Hence displacement = A/2

Resultant is sum of individual displacements, so that's zero.

 

[shahzaib ihsan]

b) i)
The phase shift between the waves (T1 and T2) is 60 degrees. Draw wave T2 5 small squares ahead of T1. Remember, both waves have same Period and Amplitude. To draw this, move the wave T2 to the right by 5 small squares. Check attachment 1.

ii) Now at time T=1 second, count the number of small squares down the X-axis until you reach the T1 wave. This is the displacement of the wave T1. Recall that 10 squares = A. Since it is 5 small squares below the time axis, it is negative. Displacement of T1 = -A/2 (Don't ignore the '-' Sign) Check attachment 2.

For wave T2, the displacement is above the time axis, so that's positive. Repeat the same counting process and you will get 5 small squares for wave t2 as well. Hence displacement = A/2

Resultant is sum of individual displacements, so that's zero.

in b1, how to know how many small squares ahead we have to plot? i seem to have forgotten the formula/method...

 

[shahzaib ihsan]

in b1, how to know how many small squares ahead we have to plot? i seem to have forgotten the formula/method...

np i get it now... so time period of the wave is 30 small boxes
phase difference = (360/timeperiod)x timeinterval difference, so: 60=(360/30) x time interval difference , so (60x30)/360 = time interval difference = 5, which means 5 small boxes ahead of it.

 

[ka2017]

Hi!

O/N 2009 P22, Q.1 I am having difficulty deciding how many significant figures or decimal places to use... For L = 92±1 cm, the percentage uncertainty is 1.09%. Shouldn't we write the final answer as 1%, because the smallest no. of sig fig in the given data L = 92±1 is one?

Thanks!

 

[Bishnu Dev]

Hi!

O/N 2009 P22, Q.1 I am having difficulty deciding how many significant figures or decimal places to use... For L = 92±1 cm, the percentage uncertainty is 1.09%. Shouldn't we write the final answer as 1%, because the smallest no. of sig fig in the given data L = 92±1 is one?

Thanks!

NEVER write the final answer in one significant figure or as an integer.

92 is two significant figure and the uncertainty is 1 which is usually quoted in one sf. You are writing the percentage uncertainty so you shouldn't keep it in 1 sf.

 

[Jamil Zaitouni]

Is it okay to write part 1 from P5 as bullet points?

 

[student2]

Pleasseee sm1 help

 

[Myaaa_20]

Hey, does anyone have worked solutions for paper 5, the latest ones ( from 2014 onward)?

 

[student2]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_43.pdf

12 (b) i) Physicist cn u help pls?

 

[i_try9621]

View attachment 62066


Pleasseee sm1 help

 

[i_try9621]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w15_qp_43.pdf

12 (b) i) Physicist cn u help pls?

They are asking for the minimum bits needed to encode the sample (to convert from analogue to digital). You can see the maximum voltage here is 15mv which requires at least 4 bits, above 15mv such as 16mv, you will need at least 5 bits. So the minimum bits needed here is 4
Ans is 4

 

[i_try9621]

Should an oscilloscope be connected in parallel in a circuit or in series?

 

[sj0007]

Should an oscilloscope be connected in parallel in a circuit or in series?

It's always connected in parallel

 

[student2]

They are asking for the minimum bits needed to encode the sample (to convert from analogue to digital). You can see the maximum voltage here is 15mv which requires at least 4 bits, above 15mv such as 16mv, you will need at least 5 bits. So the minimum bits needed here is 4
Ans is 4

thnx soooo much.

 

[student2]

View attachment 62076

thnx again. i understood ths

 

[Sharukh khan]

can some one explain this please?

 

[Sharukh khan]

caca some one explain b part ii

 

[sj0007]

can some one explain this please?

They have already provided you with the electric field strength and the charge and are asking you what the additional charge should be to have the electric field strength as 2x10^6
We know that E=kq/(r^2) so we can use the given information to get a value for k/(r^2) since these values will remain constant throughout the question (there is nothing about a change in distance and k is a constant).
So put in the given values and you will find q/(r^2) to be equal to (E/q =) 2.5 x 10^12
Next, use this to find q, plugging in E as 2 x 10^6 (from the same equation E = kq/(r^2)).......... you will get the answer as 8 x 10^-7
Subtract 6 x 10^-7 from this to find the additional charge as 2 x 10^-7

 

[sj0007]

caView attachment 62084ca some one explain b part ii

Use the fact that <c^2> is directly proportional to the kelvin temperature
Cross multiply: 1.9 x 10^6 ------------> (32+273)
? -------------> (80+273)
Once you get the mean square speed at 80 (that is, '?'), square root it to get root mean square speed

 

[Sharukh khan]

Use the fact that <c^2> is directly proportional to the kelvin temperature
Cross multiply: 1.9 x 10^6 ------------> (32+273)
? -------------> (80+273)
Once you get the mean square speed at 80 (that is, '?'), square root it to get root mean square speed

thank youu sooo sooo much bro. thanks a lot . i just got it