Physics: Post your doubts here!

[littlecloud11]

Need help with Q2.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf

Attempted this, looked at the mark scheme and I'd done it all wrong.

In the first part I did v2 = 2as. a = 9.81. s = 29.3m. The mark scheme says a = 0.85 and s = 12.8m.
They're asking for the speed BEFORE the brakes are applied so why do we need to use the values for when the car is skidding/decelerating?

Hope I'm making sense.

well, before the car was decelerating it was travelling at a constant speed, so you can't use the formula v^2 = u^2 + 2as for the distance 29.3m, because there was no accleration or change in speed during that period.
The constant speed of the car, which the question asks to find out, is the same as the initial speed of the car while decelerating, and the final speed is 0 m/s as the car comes to rest after deceleration. For this period the deceleration was .85g (given in question) and the car traveled a distance of 12.8 m before coming to rest. so you can use the v^2 =u^2 +2as formula here, where v= o , u= unknown, a= -.85g and s=12.8m

hope you got it!

 

[SuperGirl123]

well, before the car was decelerating it was travelling at a constant speed, so you can't use the formula v^2 = u^2 + 2as for the distance 29.3m, because there was no accleration or change in speed during that period.
The constant speed of the car, which the question asks to find out, is the same as the initial speed of the car while decelerating, and the final speed is 0 m/s as the car comes to rest after deceleration. For this period the deceleration was .85g (given in question) and the car traveled a distance of 12.8 m before coming to rest. so you can use the v^2 =u^2 +2as formula here, where v= o , u= unknown, a= -.85g and s=12.8m

hope you got it!

A little complicated to wrap my head around. But I think I got the gist of it! Thank you so much

 

[littlecloud11]

Can someone please help me with qn 7 and please provide drawings if possible...
qn paper: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf
ms: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_21.pdf
thanx soo much....

The question says circular waves, so you have to use spherical dippers, and to observe interference the sources have to be coherent so the dippers have to be connected to the same vibrating source, the motor.
as for observing the pattern, you need a screen below the tank and a lamp above the tank.
here-
Hope this helps!

 

[Soldier313]

The question says circular waves, so you have to use spherical dippers, and to observe interference the sources have to be coherent so the dippers have to be connected to the same vibrating source, the motor.
as for observing the pattern, you need a screen below the tank and a lamp above the tank.
here-
Hope this helps!

yup it sure did help tx a lot

 

[leosco1995]

when the voltmeter has a resistance of 7800 ohm, the combined resistance in the parallel combination becomes (7800+7800)/ 2 = 3900 ohm ; as resistor R also has a resistance of 7800 ohm.
The thermistor has a resistance of 3900 ohm same as the total parallel combination , so the emf is divided equally between resistor R+ voltmeter and the thermistor. i.e. the emf is 1.5/2= .75 across each component.

Oh, I see. So since both the thermistor and resistor + voltmeter have the same resistance, they get equal V.

BTW, another question if you don't mind.

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_2.pdf

Question 7 (c), I don't get this. Why can't the lower value of V give a lower power dissipation, since P = IV?

 

[littlecloud11]

Oh, I see. So since both the thermistor and resistor + voltmeter have the same resistance, they get equal V.

BTW, another question if you don't mind.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf

Question 7 (c), I don't get this. Why can't the lower value of V give a lower power dissipation, since P = IV?

well, here, you have to answer in reference to the graph, not by interpreting from the equation. you can see that in the graph, as V increases beyond 1.6 the power dissipation decreases.

 

[Mattman]

how do decide how many SFs to use when writing down the awnser and its uncertainty?

 

[leosco1995]

well, here, you have to answer in reference to the graph, not by interpreting from the equation. you can see that in the graph, as V increases beyond 1.6 the power dissipation decreases.

Yeah, didn't read the question properly. :\ Thanks.

 

[Most_UniQue]

Any help would be really appreciated Part B, C(i)(iv)

 

[Most_UniQue]

Btw all the answers must be in 3 SF?

 

[Hassi123]

CAN SOMEONE PLEASE HELP ME WITH PHASE DIFFERENCE? WHAT IS IT??? (SORRY FOR CAPITAL LETTERS IM WORRIED AS SHIT)

 

[TheMan123]

When I was doing this paper as revision:http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
I noticed question 5.b)ii) does not have an answer on the mark scheme, so can anyone tell me the answer?

 

[littlecloud11]

Any help would be really appreciated Part B, C(i)(iv)



View attachment 10897
View attachment 10899

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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.

 

[alim]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf
Q4a(ii)
why the maximum brightness decreases and seperaation increases in 1 n 2???

 

[00tanveer]

Anyone got any notes on potentiometers?? Please it would really help if anyone posted any link!! Thank you!

 

[Tabi Sheikh]

When I was doing this paper as revision:http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
I noticed question 5.b)ii) does not have an answer on the mark scheme, so can anyone tell me the answer?

 

[alim]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf
Q4a(ii)
why the maximum brightness decreases and seperaation increases in 1 n 2???

any1???....
also have problem in http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w03_qp_2.pdf
Q4b ....how to find the spacing of the lines of the grating???...

 

[Tabi Sheikh]

please answer the question with explanation

 

[Tabi Sheikh]

any1???....
also have problem in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_2.pdf
Q4b ....how to find the spacing of the lines of the grating???...

you will just divde 1m/5.8*10^5 and the answer is 1.72*10^-6 because there are 5.8*10^5 lines in 1 meter

 

[Tabi Sheikh]

Actually, D can't be labeled to the right if the metal plate, because the field lines are getting closer, so the strength of the field is increasing.

yes you r right the field lines are getting toward the sphere
thanks for correction