Physics: Post your doubts here!

[princess Anu]

but how can I draw a man in the diagram ?? :/

http://www.thestudentroom.co.uk/showthread.php?t=2580146&page=46

 

[Lyfroker]

View attachment 52796 View attachment 52797

Thanks a lot

 

[Lyfroker]

http://onlineexamhelp.com/wp-content/uploads/2012/05/9702_s04_qp_4.pdf

Q#3(c)(ii), how is the ratio 3/4?

 

[Lyfroker]

http://onlineexamhelp.com/wp-content/uploads/2012/05/9702_s04_qp_4.pdf

Q#5(d) how is the r.m.s. value same as the peak value?

 

[Jennifer4678]

Guys can someone please help me with this question. How do you draw the diagram for this experiment

 

[nehaoscar]

http://www.thestudentroom.co.uk/showthread.php?t=2580146&page=46

Thanks! But do i need to always do a third differential if the second differential is zero?
I thought zero may just show that it's positive therefore minimum? or since zero can be positive or negative, you do a third differentiation?

 

[The Sarcastic Retard]

Thanks! But do i need to always do a third differential if the second differential is zero?
I thought zero may just show that it's positive therefore minimum? or since zero can be positive or negative, you do a third differentiation?

You find minimum/maximum of something by differentiating it, and equating the derived formula to zero. Thus to find the minimum of the GRADIENT you need to differentiate dy/dx. This would be 1-8/x^3 and this would be equal to zero when x = 2 . So how do we decide if it's a minimum or a maximum?

There are various ways, one of them would be substituting in the derived equation with a value smaller than 2 and a value greater than 2. If you do so with x =1 and 3 for example, you'd find that the derived formula changes from a negative(decreasing gradient) to a positive (increasing gradient) i.e. a minimum.

Another method would be simple substitution in the gradient equation. Since it's either a minimum or a maximum at x =2, compare the gradient with any value of x and that of 2 ( The second part of the question ), if the gradient of 2 is smaller then it must be a minimum, and that's the case given.

 

[nehaoscar]

You find minimum/maximum of something by differentiating it, and equating the derived formula to zero. Thus to find the minimum of the GRADIENT you need to differentiate dy/dx. This would be 1-8/x^3 and this would be equal to zero when x = 2 . So how do we decide if it's a minimum or a maximum?

There are various ways, one of them would be substituting in the derived equation with a value smaller than 2 and a value greater than 2. If you do so with x =1 and 3 for example, you'd find that the derived formula changes from a negative(decreasing gradient) to a positive (increasing gradient) i.e. a minimum.

Another method would be simple substitution in the gradient equation. Since it's either a minimum or a maximum at x =2, compare the gradient with any value of x and that of 2 ( The second part of the question ), if the gradient of 2 is smaller then it must be a minimum, and that's the case given.

Thanks!
But then you've differentiated d^2y/dx^2 = 1-8/x^3 again to give 24x^-4 to show that this gives a positive value, therefore minimum.
So third differentiation is also a method to find minimum/maximum right?

 

[The Sarcastic Retard]

Thanks!
But then you've differentiated d^2y/dx^2 = 1-8/x^3 again to give 24x^-4 to show that this gives a positive value, therefore minimum.
So third differentiation is also a method to find minimum/maximum right?

When u get 0 try 3rd diff... for no error.

 

[Nowrin Yasmin]

Assalamualaikum,
I have 2 doubts which are from phy p4. Please help me with them.
Here are the questions.

Please solve them asap.

Walaikumassalam,
For the first question:
compression of ideal gas at a constant temp: the gas if being compressed so work is being done on the gas molecules to bring them closer right? So w is +. Now it is said that temp is constant, but it is obvious that the temp will increase when the gas molecules are brought closer due to more frequent collisions, so to keep the temp constant, we have to cool it down i.e. we're removing thermal energy so q is -. Now, acc to the law of thermodynamic, (-)+(+) = 0. Thus, U is 0.

the heating of a solid with no expansion: When you're heating the solid, obviously you're supplying it with thermal energy so q is +. But it is said that it's done with no expansion i.e. no work is done on the solid particles, so w is 0. Overall, internal energy is increasing due to the particles vibrating faster due to greater thermal energy thus U is +.

the melting of ice at 0 degree C to give water at 0 degree C: for melting, the hydrogen bonds between ice has to be broken thus thermal energy is supplied to do so thus q is +. Now, no work is done in compressing the water molecules of ice. Thus w is 0. Overall, U is +.

 

[Nowrin Yasmin]

http://onlineexamhelp.com/wp-content/uploads/2012/05/9702_s04_qp_4.pdf

Q#3(c)(ii), how is the ratio 3/4?

Assalamualaikum.
I'm not quite sure about the ratio they used in the mark scheme but this is how I solved it

 

[forest822]

Can anyone explain to me these questions? Thanks a lot.

Paper 1: May/02/40, June/04/39, June/05/39 Why not A?, June/07/40 Why not D?, Nov/08/38

 

[Nowrin Yasmin]

i need help with p2 o/n 14 q1 c, q2 iii

Since you didn't mention the variant, I'm assuming it's variant 2 since that is the only variant which has part 1)c).

 

[amal sharkawi]

Can anyone answer the question plz

 

[nehaoscar]

Hey guys, suppose I'm not sure of the answer for a question and I write 2 varied answers for it.
Suppose one of then is correct and the other one is wrong.
Would i get marks for the correct answer or would they not give me any marks since I also have the wrong answer written?

 

[The Sarcastic Retard]

Hey guys, suppose I'm not sure of the answer for a question and I write 2 varied answers for it.
Suppose one of then is correct and the other one is wrong.
Would i get marks for the correct answer or would they not give me any marks since I also have the wrong answer written?

They will assume that u were confused and give no marks.

 

[anasaziz18]

Walaikumassalam,
For the first question:
compression of ideal gas at a constant temp: the gas if being compressed so work is being done on the gas molecules to bring them closer right? So w is +. Now it is said that temp is constant, but it is obvious that the temp will increase when the gas molecules are brought closer due to more frequent collisions, so to keep the temp constant, we have to cool it down i.e. we're removing thermal energy so q is -. Now, acc to the law of thermodynamic, (-)+(+) = 0. Thus, U is 0.

the heating of a solid with no expansion: When you're heating the solid, obviously you're supplying it with thermal energy so q is +. But it is said that it's done with no expansion i.e. no work is done on the solid particles, so w is 0. Overall, internal energy is increasing due to the particles vibrating faster due to greater thermal energy thus U is +.

the melting of ice at 0 degree C to give water at 0 degree C: for melting, the hydrogen bonds between ice has to be broken thus thermal energy is supplied to do so thus q is +. Now, no work is done in compressing the water molecules of ice. Thus w is 0. Overall, U is +.

Jazak illahu khairan.

 

[nehaoscar]

They will assume that u were confused and give no marks.

Oh man... no luck in guessing then

 

[anasaziz18]

View attachment 52907
Can anyone answer the question plz

I got the ans A. The reason is first calculate the total resistances of 3 and 6 ohm together and separately for 2 and 2 ohm. So, the total resistance for 3 and 6 comes 2 ohms and for the 2 and 2 ohms, the total R is 1 ohm. Due to this, the pd across 2 ohms ( 3 ohm and 6 ohm) will be double of that across 1 ohm (2 and 2 ohm). So, we now know that V1=2V2. As a result, V1>V2. Now just consider that the total current flowing is I, so the current flowing thru the 3 ohm resistor can be written as V1/3 and the current flowing thru the 2 ohm can be written as V2/2.
Therefore,
I1=V1/3= 2V2/3
I2= V2/2.
So, 2V2/3 is obviously greater than V2/2. Hence, I1>I2.

 

[AhsanAfzal]

http://theallpapers.com/papers/CIE/AS_and_ALevel/Physics (9702)/9702_s11_qp_42.pdf
someone explain q4 b (ii)