Physics: Post your doubts here!

[ashcull14]

Ok so.... firstly, you need to look at the directions the forces are acting. BOTH forces are acting inwards onto the particle, as a result of this, the resultant force will also act inwards.

The next step is figuring out the direction in which it acts, you can find this out by attaching the tail of one force to the head of the other.
View attachment 51828
Then you use your knowledge of vector addition and you would end up with something like this...
View attachment 51829
and then you move towards something like this,
View attachment 51830

Now for the second part, to prove that the resultant force is 6.6 N. You use the diagram in which you applied the nose to tail method.
View attachment 51831
From this you can use the cosine rule, since you have a) two known sides and b) one unknown side with a known angle

R^2 = (2.5)^2 + (7.5)^2 - ( 2 x 7.5 x 2.5 x Cos 60 )
R^2 = 62.5 - 18.75
R^2 = 43.75
R= 6.6143 N

Which rounds up to 6.6 N, since both 2.5N and 7.5N are given to 2 significant figures.

Hope that helped!

very well xplained thnk u so much

 

[DeViL gURl B)]

may june 2007 paper 4 question 3
PLEASE SOMEONE HELP. DETAIL OF HOW TO FIND THE AREA

 

[Straw-Hat]

9702 physics MJ 04
Question no. 9 b
I didn't understand the question? Can any one please explain the questtion and how should I solve it?

 

[smhfifa]

Can anyone tell me why is the papers webpage not working, it is saying that the webpage is not available, this is the site papers.xtremepapers.com

 

[immie.rose]

may june 2007 paper 4 question 3
PLEASE SOMEONE HELP. DETAIL OF HOW TO FIND THE AREA

hope this helps!

 

[DeViL gURl B)]

hope this helps!

thank you soo much.
so basically the formula used is of the area .. right?

 

[ashcull14]

how to do the part (2) of (b) ?Cant we subtract the distance 100 -79=21cm and then find the interval between then ...whereas in markscheme distance of 90cm is used and interval of time found in part 1 is subtracted from it .....can somebody explain WHY?? wid calculations ???

 

[ashcull14]

Can anyone tell me why is the papers webpage not working, it is saying that the webpage is not available, this is the site papers.xtremepapers.com

no idea u can use http://maxpapers.com/

 

[ashcull14]

I just want to know the pattern produced in the tube and C PART???

 

[Xylferion]

View attachment 51884 View attachment 51885
I just want to know the pattern produced in the tube and C PART???

The wavelength is 60 cm, but the tube length is 45 cm.
45 cm is 3/4th of 60 cm. Which means that it is 3/4th the wavelength.

Where I put "AN", you're supposed to write A.
-------------
As the frequency is decreased, the wavelength will increase, because v = f(lambda). Right now the wavelength is 60 cm. So as we decrease our frequency, we need to end up with a wavelength that is >60 cm. This is simply to check whether our final answer makes sense.

For a sound to be produced in an open tube, there must be an antinode at the open end, and a node at the closed end. The only two wavelengths that support this are:


1/4th lambda and 3/4th lambda are both the lengths of the tube. Since 1/4th lambda is the smallest length required for a sound to be produced we can form an equation relating the length and lambda.

Length of tube = lambda / 4
Lambda = Length of tube * 4

They mention in part c that the length is going to remain 45 cm,
So lambda is 45 cm * 4 = 180 cm.

Plug this value of the wavelength into the wave equation, v = f(lambda)
330 = f * (180/100), since the speed is in "metres" per second.
f = 330/1.8
f = 183.333.... Hz which rounds down to 180 Hz. However, you're most welcome to leave your answer as 183.

Hope that made sense, if you're still confused, let me know

 

[ashcull14]

The wavelength is 60 cm, but the tube length is 45 cm.
45 cm is 3/4th of 60 cm. Which means that it is 3/4th the wavelength.
View attachment 51887
Where I put "AN", you're supposed to write A.
-------------
As the frequency is decreased, the wavelength will increase, because v = f(lambda). Right now the wavelength is 60 cm. So as we decrease our frequency, we need to end up with a wavelength that is >60 cm. This is simply to check whether our final answer makes sense.

For a sound to be produced in an open tube, there must be an antinode at the open end, and a node at the closed end. The only two wavelengths that support this are:
View attachment 51890 View attachment 51889

1/4th lambda and 3/4th lambda are both the lengths of the tube. Since 1/4th lambda is the smallest length required for a sound to be produced we can form an equation relating the length and lambda.

Length of tube = lambda / 4
Lambda = Length of tube * 4

They mention in part c that the length is going to remain 45 cm,
So lambda is 45 cm * 4 = 180 cm.

Plug this value of the wavelength into the wave equation, v = f(lambda)
330 = f * (180/100), since the speed is in "metres" per second.

f = 330/1.8
f = 183.333.... Hz which rounds down to 180 Hz. However, you're most welcome to leave your answer as 183.

Hope that made sense, if you're still confused, let me know

thnk u so much............MA ure a very competent teacher ....
id like u to explain the above kinematic problem as well ,please?

 

[immie.rose]

thank you soo much.
so basically the formula used is of the area .. right?

yeah Since the area under the graph for that question doesn't have a geometric shape,
we find the area of a single square on the graph,
and multiply the answer by the no. of squares under the graph.

 

[Xylferion]

View attachment 51882 View attachment 51883
how to do the part (2) of (b) ?Cant we subtract the distance 100 -79=21cm and then find the interval between then ...whereas in markscheme distance of 90cm is used and interval of time found in part 1 is subtracted from it .....can somebody explain WHY?? wid calculations ???

No, you cannot just subtract the distance like that. Look carefully, you took 100 cm which is the bottom of the ball, and subtracted 79 which is the top of the ball. It does not work like that. The interval for which the photograph is taken, is either from the bottom of the first ball to the bottom of the other, or vise versa with the top.

The top of the initial position of the ball is at 79 cm. The top of the final position of the ball is at 90 cm.
90 - 79 = 11 cm. Convert it to metres, since the equations of motions deal with metres. You get 0.11m.

Now the reason why what you're trying to do does not work, is because when the ball is at 79 cm or 90 cm or whatever, its initial velocity is no longer zero. It will have some other velocity at which it approaches those points on the scale, since it has gravity acting on it. So in order to solve it your way, you'd need both the velocity of the ball as it approaches the 79 cm mark, and the velocity of the ball as it passes the 90cm mark.

I'd advise using the other method since it is far more straight forward and convenient, but feel free to use the following method.

Alright so.. you have the distance travelled during the interval as 0.11 m.
You have an acceleration of 9.8 ms^-2. ( I know it's 9.81, but it does not make that much of a difference )

In the first part, you found out that it took 0.4 seconds before the ball was captured. That is the time taken to fall 79 cm. The final velocity at the end of those 0.4 seconds, will be your initial velocity during the capture.

So.. v = 0 + (9.8)(0.4), v = 3.9 ms^-1.

Now find the final velocity at 90 cm, you have u = 3.9 ms^-1.
To reach 90 cm, you cover a total distance of 11 cm from the 79 cm mark.

v^2 = (3.9)^2 + 2(9.8)(0.11)
v^2 = 15.21 + 2,156
v^2 = 17.366
v= 4.16 ----> 4.2 ms^-1

Now to find the time for the interval, use: s = (u+v/2)*t
Which is basically average velocity * time.

0.11 = [(3.9 +4.2)/2]*t
0.11 =4.05*t
t = 0.027 s
Which rounds up to 0.03 seconds.

Now tell me, which method would you rather use? Both work but it's all about which one you have more practice with. Ideally you'd want to go with the one that takes less time, but it's all your choice.

Hope that helped

 

[amal sharkawi]

 

[amal sharkawi]

 

[Xylferion]

View attachment 51907

In this question, you have two unknowns. The slit separation and the distance between the slits and the screen.

This allows you to set up two equations, so that when you find one of the unknowns, you may find the other.

The main thing to focus on, is considering the initial and final values of each since the distance is changing.

Here's how I did it..


Hope it helped

 

[princess Anu]

Can somebody please explain part iii :]

 

[<><> Ice <><>]

Hello guys, could someone please help me in this one, I tried reading the marking scheme multiple times but it didn't help at all.
I am unable to solve 9b
I would appreciate a clear explanation ^_^ THANKS!
Paper: 9702/s09/qp4 Question 9b, page 16

 

[ZaqZainab]

Hello guys, could someone please help me in this one, I tried reading the marking scheme multiple times but it didn't help at all.
I am unable to solve 9b
I would appreciate a clear explanation ^_^ THANKS!
Paper: 9702/s09/qp4 Question 9b, page 16

View attachment 51920

possible error is 10% and uncertainity is 2% so it can be 10+ or 2 which could be 8 and 12 but maximum error is 10 so value cant exceed 10 which means it has to be 8 and not 12. 8% is error so we can measure (100-8)=92% accurately.
now find lamda by the formula lamda=ln2/t = ln2/5.27 x 365 x 24 x 3600 = 4.17 x 10^-9
now use the formula A=Aoe^-lamda x t where A=92 and Ao=100 lamda we found out
and now find t by substituting the rest of the values.

 

[Sarosh Jameel]

Can someone plz help me .. how to derive the formula pressure = density*gravity*height