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Physics: Post your doubts here!

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Hey people i have an important question.. Does number of significant figures matter in phy paper 2? If answer scheme gives 2sf for final answer but i gave in 3sf, do i lose any mark?
yeah ih does when they specifically ask for 3 S.F and u lose one mark ,but normally its best to give your answers to 3 s.f and also look at the figures in the question if they are upto 3 s.f then go for 3s.f.

Welcome :)
 
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reposting my question....can someone please help?
and also i need to know the speed of alpha and beta particles
thanx a lot.....God bless
 
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how many significant figures do we have to give when calculating uncertainty?
depends on the qn.....if u refer to mj/11 qp 21 , the sgf of the uncertainty in 1c.) is 1 sgf for the resistivity just lyk the resistance from the qn........so basically i am not sure but i think it's diff fr diff qns
 
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Please someone solve Q.4 d (i) and (ii)

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

With explanation and diagram of-course. Thank you! :)
First draw a rough right angled triangle(drawn in blue) and label the the hypotenuse base and height with their according speed, Now make a scale lets say 5m/s - 5boxes in the graph( I have used cms in my example). Now convert your velocities into box units,(which is basically 18m/s =18boxes) Now draw the right angled triangle again(drawn in red), this time on the graph and according to the box units. Then find the angle using a protractor( you can use trigonometry to make sure your angle measured is indeed correct)
 

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Is there any regulation to how many significant figures you can leave your final answer to? For instance finding out uncertainties in question 1 for http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf, the answer sheet denies a mark if I do not write the answer to the significant figure they want, while in part a. they accept 3 s.f, in part b,I get a mark only if if i write the answer to two s.f. so how the hell do i know what they want? (http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_22.pdf)
 
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please explain me how to make two waves in a same graph paper which have same phase angle eg 30 degree phase angle.how to make?
one complete cycle is 360 deg. so half a wave will be 180 deg. if you start second wave at half point of the first one, you create a phse difference of 180. a quarter you get 90 deg. a thisrd of the quarter and you get 30 deg phase difference. If you still don't get it ask me and I will explain with figures.
 
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A ship is travelling north with a wind speed of 12 kmph relative to the water.
the water is flowing at 4 kmph in an easterly direction relative to the shore.
Determine the velocity of the ship relative to the shore.
 
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one complete cycle is 360 deg. so half a wave will be 180 deg. if you start second wave at half point of the first one, you create a phse difference of 180. a quarter you get 90 deg. a thisrd of the quarter and you get 30 deg phase difference. If you still don't get it ask me and I will explain with figures.
kindly explain with figures.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
6 ii) how to calculate the phase angle difference?
6 ii b) how is the ration of intensities 32?

Q.4 c) how to calculate acceleration from the graph?

7 b iii
 
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Need help with Q2.

http://www.xtremepapers.com/papers/...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? :unsure:

Hope I'm making sense.
 
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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.
 
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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? :unsure:

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!
 
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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 (y)
 
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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!
 

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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:)
 
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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?
 
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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.
 
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