Physics: Post your doubts here!

[omg]

What is your apparatus first? For instance, mine was that you drop a ball into a measuring cylinder ( or a large tube ) full of oil. I believe you can't be 100 % sure that a certain ball will reach a terminal velocity, however you can check it. You can mark two equal distances from the bottom of the tube, like 5 cm each. Why the bottom? To make sure that the terminal velocity has been reached and that there is no more acceleration. Then you measure the time taken to fall through the two distances separately. Let's call them AB and BC. If the time taken for AB is the same as BC then there is no acceleration and the speed is constant ( terminal ). I honestly don't know what would you do if the time is different. Maybe you'll have to choose a larger tube/ a more dense liquid/ smaller balls and repeat the experiment.

Hope that helped. : )

it sure did ^_^
thankssssss

 

[Silent Hunter]

Asalamoalikum people

Wanted help in http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q14,15 JazakAllah

*need help now cuz my mock is tomorrow . Thanks again

 

[omg]

i was having slight confusion with the uncertanity :/
we always write uncertanity to the same sig figures as our value?

 

[confused123]

i was having slight confusion with the uncertanity :/
we always write uncertanity to the same sig figures as our value?

yeah same as the measured quantities.

 

[Hussnain]

You are welcome. And the workings for the rest:
6(b)
View attachment 7432
(b)(i) Magnetic flux in phase with current/voltage in the primary coil
(b)(ii) EMF in the secondary coil is proportional to the rate of change of magnetic flux: E= -dΦ/dt
(b)(iii) phase difference = π/2 rad (or 90 deg)

7(a)
The curve is negative exponential, starting from zero with a positive gradient. The curve converges to a certain height (say h, which is up to you) and be sure to make the curve pass through the points (2.6, h/2) (5.2, 3h/4) (7.8, 7h/8).
7(b)(i)
Divide the mass of manganese by its molar mass, to get the mole number of manganese at the beginning. Then multiply the mole number with Avogadro's constant.
1.4 μg = 1.4 × 10^(-6) g.
1.4 × 10^(-6) g ÷ 56 g mol^(-1) = 2.5 × 10^(-8) mol
2.5 × 10^(-8) × 6.02 × 10^23 = 1.5 × 10^16
7(b)(ii)
A = λN
λ = (ln 2)/halflife
halflife = 2.6 hrs = 9360 s
λ = (ln 2)/9360 = 7.41 × 10^(-5) s^(-1)
A=7.41 × 10^(-5) × 1.5 × 10^16 = 1.11 × 10^12 Bq

THANKS ALOT

 

[Silent Hunter]

Asalamoalikum people

Wanted help in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q14,15 JazakAllah

*need help now cuz my mock is tomorrow . Thanks again

ANY BODY ?

 

[waleedsmz]

Asalamoalikum people

Wanted help in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q14,15 JazakAllah

*need help now cuz my mock is tomorrow . Thanks again

Sorry for the late reply bro. ; )

14) This is actually a very tricky question. However, if we call the initial speed of the projectile V1 and the speed at the highest point V2, we'll find that V2 = V1 cos45. Why? In a projectile motion there is only a vertical acceleration, meaning that the horizontal component of the velocity doesn't change. At the highest point in motion, the vertical component of velocity is zero but the horizontal component is constant. This would mean that the speed at the highest point is only the speed of the horizontal component which is Vcos45.

The initial K.E. is E, let's call it E1 and let's call the final kinetic energy E2. Keep in mind that E is proportional to V^2 ( square of the speed ).

Now please check my awesomely drawn solution. : p

​

Thus the answer is A.

I will answer number 15 in the next post.

 

[waleedsmz]

ANY BODY ?

15) The idea of this question is simply energy conservation. When compressed, the spring had a certain elastic potential energy that changes to the kinetic energies of the two cars once released. So you just need to find the kinetic energies of the two cars. However, you only got the speed of 1 car. To find the other one's speed, we resort to momentum conservation rules.

Since the two cars were at rest, therefore their momentum was 0. From there, m1v1 + m2v2 = 0
m1v1 = -m2v2
2 x 2 = -1 x v2 ---> v2 = -4

Now to find the kinetic energy which is 0.5 x mass x velocity squared:

0.5 x 2 x 4 + 0.5 x 1 x 16 = 12

So the answer is D.

Hope this helped. : )

 

[omg]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_5.pdf
PLEASE cn smbdy upload the diagram fr this experiment :/

 

[waleedsmz]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_5.pdf
PLEASE cn smbdy upload the diagram fr this experiment :/

Well I can give you some details that my teacher said about this.

The idea is that you want to measure the sound beyond the glass when the pressure is varied. To do this, you need to cover the air outlet with a vacuum pump ( to vary the pressure ) and a pressure gauge ( to measure the pressure ). Then you need a sound source that shouldn't change through the experiment. Moreover, you put a microphone beyond the glass connected to a C.R.O. to detect and measure the amplitude of the sound. There you have it, the pressure and the amplitude. Now you can just plot a graph of the amplitude against the root of the pressure and the gradient would be K.

This is basically my teacher's answers. I'd love if someone shared a better answer. If you need more details just ask. Hope that helped. : )

 

[waleedsmz]

hw will we connect the glass windw wid a CRO??

The microphone is what is connected to the C.R.O. and not the glass itself. xD

 

[unique840]

The microphone is what is connected to the C.R.O. and not the glass itself. xD

we can also connect a piezo electric crystal instead of microphone.

 

[omg]

The microphone is what is connected to the C.R.O. and not the glass itself. xD

argh sorry my bad :/
whts the method of ensurin that the output frm the speakr is constantttttttttttttt

 

[waleedsmz]

we can also connect a piezo electric crystal instead of microphone.

Well the microphone is actually made of piezo electric stuff as far as I know.

argh sorry my bad :/
whts the method of ensurin that the output frm the speakr is constantttttttttttttt

Well first of all, use the same sound source (e.g. speaker) for all measurements with the same sound level. Furthermore, keep the distance between the sound source and the glass constant.

 

[omg]

Well the microphone is actually made of piezo electric stuff as far as I know.



Well first of all, use the same sound source (e.g. speaker) for all measurements with the same sound level. Furthermore, keep the distance between the sound source and the glass constant.

THANKS A LOTTTTTTTTTT

 

[waleedsmz]

Oct/Nov 2011/23 As level
Q1/ d
Q4/ (iv)
Help please :|

I'll answer both of your questions tomorrow inshaAllah since it's pretty late in here. Sorry about that. : /

 

[Hussnain]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_qp_2.pdf
AS PHYSICS
Q5 (b)(i)
Phase angle.
Please answer as soon as possible.

 

[DragonCub]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_2.pdf
AS PHYSICS
Q5 (b)(i)
Phase angle.
Please answer as soon as possible.

One complete cycle has phase angle 360 deg. 60 deg means one-sixth cycle.
T2 is one-sixth cycle behind T1, or 1/6 period behind.
According to the graph, the period is 3 sec.
T2 is then 0.5 sec behind T1.
Graph:

 

[Tabi Sheikh]

One complete cycle has phase angle 360 deg. 60 deg means one-sixth cycle.
T2 is one-sixth cycle behind T1, or 1/6 period behind.
According to the graph, the period is 3 sec.
T2 is then 0.5 sec behind T1.
Graph:
View attachment 7512

still couldnt get the point

 

[waleedsmz]

Oct/Nov 2011/23 As level
Q1/ d
Q4/ (iv)
Help please :|

Alright sorry for the late reply. ; )

Q1: I honestly don't know how to solve it by using the triangle of forces... But I do know how to do so with resolution:

Since the object is in equilibrium, therefore sum of horizontal forces equals zero and some of vertical forces equals zero as well.

From this:
T1 cos50 = T2 cos40 ( Horizontally ) ---> T1 = T2 cos40 / cos50
T1 sin50 + T2 sin40 = 7.5 ( Vertically )

By substituting in T1, T2 cos40 sin50 / cos50 + T2 sin40 = 7.5
T2 = 4.8 and by further substitution in the T1 = T2 cos40/cos50 ---> T1 = 5.7

I'm not really sure about Q4 so we should just wait for anyone to answer that. : /