Physics: Post your doubts here!

[Xylferion]

And isn't absolute uncertainity with
Max - Min / 2 =Absolute uncertainity ?

you've got 3 readings... and those readings are not proper either. Like I've already said twice, have your values differ by 1.00 from each other.

 

[The Sarcastic Retard]

you've got 3 readings... and those readings are not proper either. Like I've already said twice, have your values differ by 1.00 from each other.

I face lot of troubles in practical papers.
I generally do it like 0.2/(reading) * 100 for %UC :/ though idk if its correct or not acc to above post but i get marks in my exams.. Recently I scored 33/40

 

[TheJDOG]

The number of the page we are on right now ain't cool man xD

 

[Midnight dream]

I face lot of troubles in practical papers.
I generally do it like 0.2/(reading) * 100 for %UC :/ though idk if its correct or not acc to above post but i get marks in my exams.. Recently I scored 33/40

and unluckily i am even worse than you..i got 29/40

how do you get the last part done ..Which wants limitations..?

 

[amal sharkawi]

Can anyone answer this question please???

 

[Mayarzawaydeh]

View attachment 52289
Can anyone answer this question please???

Answer is B.
Stress = force per unit area = F / A
Let the force (weight) and the (cross-sectional) area in the full-size crane be F and A respectively.
Stress in cable of full-size crane = F / A
As mentioned in the question, the model is one-tenth full-size in all linear dimensions.
For the model,
The load, which has a cube-shaped load, is in 3-dimension. So, for each of the 3 dimensions, the length of the model is reduced by a factor of 1/10. Therefore, the volume of the load is reduced by a factor of (1/10)3 = 1 / 1000.
Weight = mg and Mass = density x volume. Since the same material is used, the density is the same. So, the mass is proportional to the volume. A reduction in the volume causes the mass to be reduced by the same factor. The weight, which depends on the mass (g is constant), is also reduced by the same factor.
Force (weight) in model = F / 1000
Similarly, the cross-sectional area (which depends on (diameter)2) will be reduced by a factor of (1/10)2 = 1 / 100.
(Cross-sectional) Area in model = A / 100
Stress in cable of model crane = (F/1000) / (A/100) = 0.1 (F/A)
Ratio = (F/A) / 0.1(F/A) = 10

 

[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.