Physics: Post your doubts here!

[ashcull14]

If the answer is B then this is how.
We know that PE= 80 J just before release, so use efficiency equation and PE=mgh

Efficiency is 28%
We know efficiency = output/input
28/100= output/ 80
Output = 22.4 J
Now use PE= mgh
Mass of arrow= 120 g= 0.12 Kg since arrow is shot alone
22.4= 0.12(10)(h)
h= 18.7 m ~= 19 m
I'm not sure though I might be wrong

youre abslutly right thnks

 

[manya]

can someone help me with this question

 

[Mayarzawaydeh]

can someone help me with this question

first off you need to know the timing between the emitted and the reflected pulse, which when you start counting from the centre of each pulse, gives you 4 blocks in separation, (between the emitted and the reflected pulse.) but you're going to consider that as 2 blocks between the source of the emitted pulse and the reflector.
The time base setting is 0.20 μs cm–1. which makes the time --->2 x 0.20 μs cm–1 = 0.0000004s
and since a radio active pulse travels in the speed of 3.0 x 10^8 ms-1,
using the good old formula of distance =speed x time gives you ---> 3.0x10^8 x 0.0000004 = 120m.

hope it helps.

 

[The Sarcastic Retard]

first off you need to know the timing between the emitted and the reflected pulse, which when you start counting from the centre of each pulse, gives you 4 blocks in separation, (between the emitted and the reflected pulse.) but you're going to consider that as 2 blocks between the source of the emitted pulse and the reflector.
The time base setting is 0.20 μs cm–1. which makes the time --->2 x 0.20 μs cm–1 = 0.0000004s
and since a radio active pulse travels in the speed of 3.0 x 10^8 ms-1,
using the good old formula of distance =speed x time gives you ---> 3.0x10^8 x 0.0000004 = 120m.

hope it helps.

why only 2? :/

 

[manya]

first off you need to know the timing between the emitted and the reflected pulse, which when you start counting from the centre of each pulse, gives you 4 blocks in separation, (between the emitted and the reflected pulse.) but you're going to consider that as 2 blocks between the source of the emitted pulse and the reflector.
The time base setting is 0.20 μs cm–1. which makes the time --->2 x 0.20 μs cm–1 = 0.0000004s
and since a radio active pulse travels in the speed of 3.0 x 10^8 ms-1,
using the good old formula of distance =speed x time gives you ---> 3.0x10^8 x 0.0000004 = 120m.

hope it helps.

thank you so much but why is the timing not taken from the end of the waves that makes it 3 blocks

 

[Mayarzawaydeh]

thank you so much but why is the timing not taken from the end of the waves that makes it 3 blocks

Because if you chose to start counting from the right end of the emitted pulse, you should stop your count also at the right end of the reflected pulse, which again makes it 4 blocks. I think you must have got confused counting from the end, so to avoid this problem I suggest you start your count from the centre

 

[princess Anu]

The question was
Outline how a c.r.o may be used to determine the frequency of a loudspeaker.


I don't get why have they mentioned frequency= 1/lamda * time base setting?? :/
Why multiplying by LAMDA

 

[manya]

Because if you chose to start counting from the right end of the emitted pulse, you should stop your count also at the right end of the reflected pulse, which again makes it 4 blocks. I think you must have got confused counting from the end, so to avoid this problem I suggest you start your count from the centre

oh okay thankyou so much

 

[crazytaylorfanXD]

can someone help me out?
http://maxpapers.com/wp-content/uploads/2012/11/9702_s14_ms_42.pdf ( question 10 biii)


http://onlineexamhelp.com/wp-content/uploads/2014/02/9702_w13_ms_43.pdf ( question 9 biii)


http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_ms_42.pdf ( question 9b)


http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_43.pdf (question 9cii and 9dii)

 

[Asif1223]

Question asks about Apparent weight loss but mark scheme talks about pressure differences between liquids.. Does the contents relate each other? If then, How?

 

[Llamas]

http://www.sheir.org/a-level-physics-43-nov2012.pdf

Could someone please solve question number 8 for me?

Also, I tried the blog http://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-42-43-worked.html#uds-search-results and it does not have the solved version of this paper, so I'd love it if someone could please walk me through the solution?

 

[amal sharkawi]

Can anyone plz explain , why the answer id c

 

[amal sharkawi]

Can anyone answer plz ???

 

[Llamas]

http://www.sheir.org/a-level-physics-43-nov2012.pdf

Can someone please walk me through question eight, part b and c please? This seems pretty basic but I think there's either something wrong with my notes, or there's something I'm missing?


I've already tried the website http://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-42-43-worked.html#uds-search-results, it does not have the answers I'm looking for.

 

[_Ahmad]

can someone help me out?
http://maxpapers.com/wp-content/uploads/2012/11/9702_s14_ms_42.pdf ( question 10 biii)


http://onlineexamhelp.com/wp-content/uploads/2014/02/9702_w13_ms_43.pdf ( question 9 biii)


http://maxpapers.com/wp-content/uploads/2012/11/9702_s13_ms_42.pdf ( question 9b)


http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_43.pdf (question 9cii and 9dii)

For the first one

Vin = Votage at the non inverting input - voltage at the inverting input

gain=Vout/Vin

open loop gain is usually very large almost around (x 10^5) if not not specified
Votage at the non inverting input is constant at 1.0V (given) and the voltage at the inverting input is varying.

for the first few seconds voltage at the inverting input is more than that at non inverting input, therefore Vin is neagtive.
Vout will be a large neagtive number which is greater than the supply voltage hence op-amp will undergo negative saturation giving max Vout -5v and therfore diode which is reverse biased (R) will emit light
vice versa

 

[hammadullah]

Hello everyone,

Here are Physics Notes for A Level (A2) chapter wise.

Syllabus Code: 9702

Along with these notes you would also need to refer to the Physics A level Cambridge book for the last three chapters that are:

1. Direct Sensing
2. Medical Imaging
3. Communication Systems

Hope that these notes are helpful

The link for the Notes is given below

Link: http://cieoandalevelnotes.blogspot.com/2015/04/physics-cambridge-level-2-notes-9702.html?m=1

Thank You

 

[eliyeap]

Hi guys, I need help for question 5 (b) iii. Cant quite wrap my head around the answer. Any help is much appreciated. Oh, if its possible, please explain in detail. Thanks.
http://www.acethem.com/pastpapers/a...uestion-paper-2013-summer-paper-2-15506.html/

 

[eliyeap]

Question asks about Apparent weight loss but mark scheme talks about pressure differences between liquids.. Does the contents relate each other? If then, How? View attachment 52345 View attachment 52346

The question is mainly talking about Archimedes Principle. In order for an object to float in water when it is partially submerged, the pressure that is acting upwards from the bottom of the object needs to be equal or more then the weight of the object that is acting downwards. The difference in pressure between this 2 forces gives rise to the buoyant force.

 

[ashcull14]

just want a sketch of the ball when it rebounds horizontally as referred in markscheme

 

[The Sarcastic Retard]

View attachment 52374 View attachment 52375
just want a sketch of the ball when it rebounds horizontally as referred in markscheme

It should be like this.