Physics: Post your doubts here!

[waleedsmz]

Jazakallah tht helped!!

just one thing why did u add 60+20?

U're welcome dude. ; )

Why did I add the 20? Well, let's think of another circuit with only a 12 V cell, a 60 ohm resistor and a 20 ohm resistor all connected in series. How would you calculate the resistance on the 60 ohm resistor? You'd say that the 12 V is shared with all the resistors in the circuit and thus:

20 + 60 ---> 12
60 ---> ?

In your question, it's exactly the same situation, the only difference is that you're getting the resistance of a part of the line AC ( the 60 ohm resistor ), i.e. the 39 ohms. It's as if you divided the line AC into two separate resistors so now you have a 39 (AB), a 21 (BC) and a 20 ohm resistors in the circuit. How to get the voltage across 39?

39 + 21 + 20 ---> 12
39 ---> ?

Hope that helped. : )

 

[waleedsmz]

Minimum will be zero . i am not sure bout this but maybe when the jockey is move towards the extreme position then the voltage or resistance might bcm zero then current will be zero as well :/ plz someone else clarify. Thanks

I believe when the jockey moves to the left, the lamp would actually be shorted. Meaning that no current will pass through the lamp as there is a negligible resistance in the upper wire of the lower loop ( Indicated by the red straight line ).

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By the way, thanks for explaining why it will have that maximum value because I didn't know it. xD
Hope this helped. : )

 

[anonymous123]

U're welcome dude. ; )

Why did I add the 20? Well, let's think of another circuit with only a 12 V cell, a 60 ohm resistor and a 20 ohm resistor all connected in series. How would you calculate the resistance on the 60 ohm resistor? You'd say that the 12 V is shared with all the resistors in the circuit and thus:

20 + 60 ---> 12
60 ---> ?

In your question, it's exactly the same situation, the only difference is that you're getting the resistance of a part of the line AC ( the 60 ohm resistor ), i.e. the 39 ohms. It's as if you divided the line AC into two separate resistors so now you have a 39 (AB), a 21 (BC) and a 20 ohm resistors in the circuit. How to get the voltage across 39?

39 + 21 + 20 ---> 12
39 ---> ?

Hope that helped. : )

such a clear explanation ofc it helped Jazakallah again..

 

[waleedsmz]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_5.pdf
what is the method of finding when the terminal velocity has reachedd??????????????

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. : )

 

[Hussnain]

View attachment 7420View attachment 7423

THANKS

 

[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. 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. : )

im still nt clear how 2 gt the terminal velocity :/

 

[waleedsmz]

im still nt clear how 2 gt the terminal velocity :/

Ops sorry I thought you meant how to make sure that it was the terminal velocity. =/ Sorry about that.

Continuing my previous reply, you basically measure the time taken for the ball to fall the distance AB by a stopwatch. This will allow you to identify the speed through the formula Distance/Time ---> Distance AB ( 5 cm ) / Time taken to fall through AB. How to know that this is the terminal velocity? You find the speed ( by the same procedure ) of the ball through the distance BC. If the speeds are equal, then there is no acceleration and the value of the speed is the terminal value. Read the previous reply again so that you get the whole image of what I'm trying to explain.

Hope this helps. : )

 

[waleedsmz]

View attachment 7420View attachment 7423

DragonCub I wonder what are those? Are those some model answers and you copied them? If so, can I know from where did you get them? Thanks. : )

 

[omg]

Ops sorry I thought you meant how to make sure that it was the terminal velocity. =/ Sorry about that.

Continuing my previous reply, you basically measure the time taken for the ball to fall the distance AB by a stopwatch. This will allow you to identify the speed through the formula Distance/Time ---> Distance AB ( 5 cm ) / Time taken to fall through AB. How to know that this is the terminal velocity? You find the speed ( by the same procedure ) of the ball through the distance BC. If the speeds are equal, then there is no acceleration and the value of the speed is the terminal value. Read the previous reply again so that you get the whole image of what I'm trying to explain.

Hope this helps. : )

AB nad BC are the same distance from the bottom of the cylinder??????

 

[waleedsmz]

Two solid substances P and Q have atoms of mass MP and MQ respectively. They have NP and NQ atoms per unit volume such that NP= NQ. It is found by experiment that the density of P is greater than that of Q.

Which of the following deductions from this experiment must be correct?
A MP > MQ
B NP > NQ
C MPNP > MQ NQ
D MP/NP = MQ/NQ

Ok look, if the first answer is the correct one ( which is the more probable ) then my explanation would make sense. What did the question ask for? It asked for a "deduction". In the question he said that the density of P is greater than that of Q. This means that the answer C is pretty much what he said but in different words. However, I can deduce from the density of P being greater and the numbers of atoms per volume being equal that the mass of the atom P is greater than that of Q. So A would be the "more correct" answer. That's what I think...

 

[omg]

Ops sorry I thought you meant how to make sure that it was the terminal velocity. =/ Sorry about that.

Continuing my previous reply, you basically measure the time taken for the ball to fall the distance AB by a stopwatch. This will allow you to identify the speed through the formula Distance/Time ---> Distance AB ( 5 cm ) / Time taken to fall through AB. How to know that this is the terminal velocity? You find the speed ( by the same procedure ) of the ball through the distance BC. If the speeds are equal, then there is no acceleration and the value of the speed is the terminal value. Read the previous reply again so that you get the whole image of what I'm trying to explain.

Hope this helps. : )[/quote
THese are the points???
:/
we'llmeasure the distance the ball travelled btwn A and B ??

 

[waleedsmz]

AB nad BC are the same distance from the bottom of the cylinder??????

Umm no, check this lil drawing that I've made. : )

​

 

[omg]

Umm no, check this lil drawing that I've made. : )

View attachment 7436​

thank you so much!

so its like we'll measure the speed of the ball traveling from A to B and dn from B to C!
if its the same the ball has reached the terminal velocity!
and if they r nt the same we'll repeat the experiment with different distances to get the terminal velocity????

 

[DragonCub]

THANKS

You are welcome. And the workings for the rest:
6(b)

(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

 

[DragonCub]

DragonCub I wonder what are those? Are those some model answers and you copied them? If so, can I know from where did you get them? Thanks. : )

This is not model answer. I typed these using Microsoft Word 2010 which has an "insert formulae" function to make the text very formal.

 

[waleedsmz]

thank you so much!

so its like we'll measure the speed of the ball traveling from A to B and dn from B to C!
if its the same the ball has reached the terminal velocity!
and if they r nt the same we'll repeat the experiment with different distances to get the terminal velocity????

You're welcome!

Well you got it right. But if they're not the same, I honestly am not sure what to do. The question didn't put any marks for that as far as I can remember. But your solution of repeating with different distances isn't really going to fix it since the problem is not with the distance, it's with the speed ( i.e. distance/time ratio ), and as I mentioned earlier we marked from the bottom of the tube. I believe as a solution to this we can choose a larger tube/ a more dense liquid/ smaller balls and repeat the experiment.

Hope this helped. : )

 

[waleedsmz]

This is not model answer. I typed these using Microsoft Word 2010 which has an "insert formulae" function to make the text very formal.

Thanks for the reply. And damn that text is foxy!
I don't have microsoft word 2010 unfortunately so I'll just keep replying in the "old fashion" way.

 

[DragonCub]

Thanks for the reply. And damn that text is foxy!
I don't have microsoft word 2010 unfortunately so I'll just keep replying in the "old fashion" way.

It's really labourious to type in the conventional way, and the expression can be vague sometimes. So I prefer "natural displays" of formulae.

 

[confused123]

I believe when the jockey moves to the left, the lamp would actually be shorted. Meaning that no current will pass through the lamp as there is a negligible resistance in the upper wire of the lower loop ( Indicated by the red straight line ).

View attachment 7428​

By the way, thanks for explaining why it will have that maximum value because I didn't know it. xD
Hope this helped. : )

where is the lamp in the circuit , by lamp you mean the 6 ohm resistor? how will be the resistance zero at this point as 6 ohm resistor is still present. will this not effect the I2 in any way?

p.s no problem at least now you know how to calculate max current. don't forget now

 

[waleedsmz]

where is the lamp in the circuit , by lamp you mean the 6 ohm resistor? how will be the resistance zero at this point as 6 ohm resistor is still present. will this not effect the I2 in any way?

p.s no problem at least now you know how to calculate max current. don't forget now

Oh shoot I thought it's a lamp. xD

Anyway it's pretty much the same. I didn't say the resistance is zero in the lamp, I said in the wire. You know when you connect a wire and a lamp for instance parallel to each other, the lamp will not light. In our example, if you remove the variable resistor ( but keep the wire in the middle ) and still connect the 6 ohm resistor across that wire, it won't have any current passing through it. Why? It's a phenomenon called short circuiting. The current won't have to go through the resistor because the wire is perfect for it to go through ( it has a 0 resistance ).