Physics: Post your doubts here!

[Most_UniQue]

ci) For the stationary wave the particles between two consecutive nodes or 2 consecutive antinodes vibrate in the same phase. therefore, they have no phase difference between them. the answer will be 0.

iv) the wave will be a straight horizontal line over the dotted line. For a stationary wave there is no onward motion of disturbance from one particle to the adjoining particle, so beyond a particular instant of time the displacement of all the particles is 0.

a1) the diagram is below. for a wave s = f x lamda, therefore s = (1/T) x lamda so, lamda = s x T
the speed is same for both the first and the second wave, as T changes by .25 the wavelength also shifts by .25, which is 20cm here. thus the second wave will be 20 cm to the right of the first.

Oh okay so if it wasnt a stationary wave then it wud be 180* right?

For a1 I got it but how does it start from 20?

 

[littlecloud11]

View attachment 10909please answer the question with explanation

S= ut + 1/2 a t^2
for t1 sec distance traveled = 0 + 1/2 g t1^2
for t2 sec distance traveled = 0 + 1/2 g t2^2

so the difference = 1/2 g t2^2 - 1/2 g t1^2
that's D

 

[Tabi Sheikh]

thanks

 

[Silent Hunter]

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

need help in question 6 part (a) (ii) Why are we using total emf here? Its only asking about the battery ...... am confused

 

[Most_UniQue]

No, it wouldn't. for the wave to have a phase difference if 180* the difference between them has to be 1/2 the wavelength. here, it if were a moving wave the phase difference would be approx. 45* that is 1/8 the wavelength.

The initial wave starts from zero, and by then the second wave has already traveled 1/4 of the wavelength, that is 20m. So, it starts from 20.

Oh ok gt it Tnx

 

[Silent Hunter]

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

need help in question 6 part (a) (ii) Why are we using total emf here? Its only asking about the battery ...... am confused

 

[xxxtoughxxx]

q(7) b......in the ms and er it says for damage of supply to take place lamp A must be shorted ....why especially lamp A wasnt lamp c da shorted one in the quetion b4 it??? any 1 plzz thx in adv

 

[xxxtoughxxx]

its easy if u understand (a)i.2 to get the total emf of the circuit we would add the emf of battery nd charger but as they have opposite polarity it would be the emf of charger minus that of battery bec normally charger has higher emf so its 14-E now for (ii) the V=IR u will substitute 14-E for V and equate it so that 14-E=42x0.16 and solve for E hope i didnt confuse u

 

[kristf]

Hi all, i was just wondering if alternate currents come up in the aslevel exam?? cause on the syllabus it says its for alevel, but our teacher taught it to us this year, so im confused where to study it or not? help please

 

[xxxtoughxxx]

Hi all, i was just wondering if alternate currents come up in the aslevel exam?? cause on the syllabus it says its for alevel, but our teacher taught it to us this year, so im confused where to study it or not? help please

nop its not in da as level my teacher never explained it nd dere r no questions about in da pastpapers

 

[dayeed]

PLEASE someone help me with the potential divider ,,,,,, please post any examples based on it ....
and please tell how do we solve q5b may june 2010 paper 21 !!!!!!!!!!!!!!!!!!!!!!!!!

 

[NokiaN95638]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf

Guys i need help on Q5) aii
AND Q7. b

 

[Gémeaux]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf

Guys i need help on Q5) aii

for Q.5 aii)

 

[dayeed]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf

Guys i need help on Q5) aii
AND Q7. b

In q5 you just find out different values of current for given vlaues of voltage on x axis and plot a graph.....
in q7b u just take a look at revision note on xtremepapers revision page u have to ratofy it SIMPLY.

 

[iKhaled]

need help in may june 2007..

question 4 d) i. what is the resultant vector? :S

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

 

[dayeed]

need help in may june 2007..

question 4 d) i. what is the resultant vector? :S

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

make a horizontal vector and a vertical vector by working out the vertical velocity and these will be 90 degrees to each other and simply use pythagoras theorem to find the resultant =)

 

[Jaf]

make a horizontal vector and a vertical vector by working out the vertical velocity and these will be 90 degrees to each other and simply use pythagoras theorem to find the resultant =)

Uh... the resultant velocity is 25 (given in part b)).
So you have to take an appropriate scale and draw in the horizontal first. Then draw a vertical from the end of the horizontal (at 90 degrees). Then start the resultant from the other end of the horizontal and join this to the vertical where the resultant is equal to 25 (from your scale). Then rub out the vertical and measure the angle between the horizontal and resultant velocities.

 

[iKhaled]

Uh... the resultant velocity is 25 (given in part b)).
So you have to take an appropriate scale and draw in the horizontal first. Then draw a vertical from the end of the horizontal (at 90 degrees). Then start the resultant from the other end of the horizontal and join this to the vertical where the resultant is equal to 25 (from your scale). Then rub out the vertical and measure the angle between the horizontal and resultant velocities.

isnt the vertical velocity is the one equal to 25ms^-1 ?

 

[iKhaled]

need help in may june 2008

question 5 c) why they used the tensian as 4.00 ?

 

[Jaf]

isnt the vertical velocity is the one equal to 25ms^-1 ?

Nope, doesn't fit in.

If we calculate the time taken for the ball to reach the ground using the vertical velocity:
s=ut+(gt²)/2
Since u=0, this gives us t= 1.81 seconds (s=16, g=9.8)

Now if you put in the values in v=u+at:
v = 9.8 * 1.81 = 17.7 ms⁻¹ ≠ 25 ms⁻¹

But √(17.7² + 18²) ≈ 25 ms⁻¹!