Physics: Post your doubts here!

[Sriyog Sharma]

Hey guys,
I was solving a vector & moments question as a part of my revision and got stuck somewhere. I really need your help.. some sub-questions are solved (a & b) , please correct me if I'm wrong.
Here's the question (Its kinda long) :

The free-body diagram shows three forces that act on a stone hanging at rest from two strings.


Yes the weight you calculated should be correct.
I think for the other question you just have to draw the diagram and consider that the arrows must follow each other . (head to head rule )
For the last part, I am not sure , but try to use the Resultant formula. Not sure though.

(a) Calculate the horizontal component of the tension in each string. Why should these two components be equal in magnitude?

String 1: =F cosθ = 1 x cos60 =0.5 N
String 2: =F cosθ = 0.58 x cos30 = 0.5 N

The components are equal in magnitude because the stone is at rest and there is no horizontal movement.

(b) Calculate the vertical component of the tension in each string

String 1: F Sin θ = 1 x sin60 = 0.87N
String 2: F Sinθ = 0.58 x sin30 = 0.29N

(c) Use your answer to (b) to calculate the weight of the stone

= 0.87 + 0.29 N
= 1.16N
(Not sure about this answer)

(d) Draw a vector diagram of the forces on the stone. This should be a triangle of forces

I don’t know how to solve this one – please help!

(e) Use your diagram in (d) to calculate the weight of the stone

This one too.. help!!


Waiting for your replies..
Thanks!

 

[InnocentAngel]

Could someone please explain to me why for the graphs velocity against time, when we find change in velocity we take initial velocity - final velocity and not the other way round ? for example http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf variant 2
Question 2 part c (i)
answer :http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_ms_2.pdf
Why do we take 9--4.2 and not -4.2-9 ??? normally to find change we do : final vel - initial vel nopes??

 

[geek101]

the more the force the lesser the velocity..??? whhat?? If a car is moving with a higher velocity you need more force to stop unlike the one that is travelling with lesser velocity.Think of the brakes.If you want to keep the stopping distance same..you have to brake harder for the one that is travelling with a higher velocity.

if a car is moving with a velocity lets say 10 ms-1, just one system...and you want to bring the velocity to zero. You apply x newtons in the opp direction, now its moving with 8 ms-1, then you apply lets say 2x newtons and the car stops.
The larger the force you applied, the more the velocity decreased. yah looking at the braking system, for one system. The harder the force you apply on the brakes the faster the velocity decreases and the car stops. your right with ur comparisons before though, but i dunno if you can give a proportionality judgement out of it??

anyway look here...
Force = change in momentum / time
force = m x change in velocity / time
so isnt F directly proportional to change in velocity.

but power = force x velocity
and here they are inversely proportional

 

[axetreme.O]

if a car is moving with a velocity lets say 10 ms-1, just one system...and you want to bring the velocity to zero. You apply x newtons in the opp direction, now its moving with 8 ms-1, then you apply lets say 2x newtons and the car stops.
The larger the force you applied, the more the velocity decreased. yah looking at the braking system, for one system. The harder the force you apply on the brakes the faster the velocity decreases and the car stops. your right with ur comparisons before though, but i dunno if you can give a proportionality judgement out of it??

anyway look here...
Force = change in momentum / time
force = m x change in velocity / time
so isnt F directly proportional to change in velocity.

but power = force x velocity
and here they are inversely proportional

 

[DumbIdiot]

what is mean separation and how do i calculate it?
that question is here: http://xtremepapers.com/papers/CIE/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q3 c(ii)

 

[aniroula]

what is mean separation and how do i calculate it?
that question is here: http://xtremepapers.com/papers/CIE/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf
Q3 c(ii)

The seperation is 4/3 pi r cube..

 

[aniroula]

Plz help with this
May/june 07 q no 5 b(ii)
How to calculate the speed,,

 

[DumbIdiot]

The seperation is 4/3 pi r cube..

Isn't that the volume of a sphere? So, here mean separation is the distance between 2 adjacent molecules right?

 

[DumbIdiot]

Plz help with this
May/june 07 q no 5 b(ii)
How to calculate the speed,,

there are 2.5 waves within 39 cm(count heaps from 1 to 8 so node is at heap 2 then heap 3 and then heap 4. 1 wave has 2 nodes so 2.5 waves)
hence lambda is 0.39/2.5=0.156
v=f(lambda)
so v=2.14 kHz*0.156=333.8 ms^-1

 

[Khunkar]

Need help`` Question no. (5 a ii) and (5 a iii)

Question paper: ( http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf )
Marking scheme: ( http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_ms_22.pdf )

Thanks!!

 

[aniroula]

You are correct,, But according to mark scheme,, it is done so... By calculating the difference in volume,,

 

[aniroula]

there are 2.5 waves within 39 cm(count heaps from 1 to 8 so node is at heap 2 then heap 3 and then heap 4. 1 wave has 2 nodes so 2.5 waves
hence lambda is 0.39/2.5=0.156
v=f(lambda)
so v=2.14 kHz*0.156=333.8 ms^-1

Do 2 node make a wavelength...

 

[emkay]

can someone explain b iii

qp: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_2.pdf

ms: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_ms_2.pdf

since you haven't mentioned the question number and as i had a look there's only Q#5 which has a part iii of (b).
So, if that is what you are asking for then here it is.

it says clearly, (for time t=1.0sec) and in terms of A.
1. the displacement due to wave T1 alone = put your finger on t=1.0 sec then drag it down and see where it touches the wave T1 alone that would be -0.5A
2. the displacement due to wave T2 alone = you must have drawn wave T2 before to do this, and if you have drawn it correctly, do the same you did in 1. to find the displacement and you should be able to get 0.5A.
3. Since both the amplitude's of wave's T1 and T2 are same, it cancels out and the resultant displacement is zero.

 

[A_K_D]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf


could someone explain 4b))

i thought that the extension would be 1/2 x 2

because 2 springs halve the extension when in same row and 2 springs double extension when in the same column...

I cant visualise why the 2 springs in the same column would just contribute to another extension

 

[aniroula]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf


could someone explain 4b))

i thought that the extension would be 1/2 x 2

because 2 springs halve the extension when in same row and 2 springs double extension when in the same column...

I cant visualise why the 2 springs in the same column would just contribute to another extension

The extension for first arrangement would be 2e because two springs are attached together and one spring has extension e...

It is same a series and parallel circuits.. Now for the second arrangement two springs are attached parallely with a load so extensoion is 1/e +1/e,,. And it results e/2,,

 

[emkay]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf


could someone explain 4b))

i thought that the extension would be 1/2 x 2

because 2 springs halve the extension when in same row and 2 springs double extension when in the same column...

I cant visualise why the 2 springs in the same column would just contribute to another extension

You should know that a spring constant ‘k’ is a measure of how stiff the spring is. A spring that is very hard to stretch out has a large spring constant. A spring that is easy to stretch has a small spring constant. As the term spring constant implies, the spring constant is always the same for a given spring. It is the force applied per unit extension or there is another way, take this as Electricity (Springs in series and Springs in Parallel) which come's under my category of "Desi Totka's"
Now, here in this question. at first the extension is double (2E) as springs are in series and spring constant is half (1/2K).
In the second part, extension is halved(1/2E) as springs are in parallel and spring constant is doubled(2K).
And the last one, focus on the parallel series first and then the single spring. For extension parallel + single = 3/2E and for Spring constant parallel + single = 2/3K

 

[aniroula]

Can some one help.me please,,

May/june 2009 q no 4 b spring constant arrangement.. I knew how to find extension but not spring constant,,

 

[aniroula]

You should know that a spring constant ‘k’ is a measure of how stiff the spring is. A spring that is very hard to stretch out has a large spring constant. A spring that is easy to stretch has a small spring constant. As the term spring constant implies, the spring constant is always the same for a given spring. It is the force applied per unit extension or there is another way, take this as Electricity (Springs in series and Springs in Parallel) which come's under my category of "Desi Totka's"
Now, here in this question. at first the extension is double (2E) as springs are in series and spring constant is half (1/2K).
In the second part, extension is halved(1/2E) as springs are in parallel and spring constant is doubled(2K).
And the last one, focus on the parallel series first and then the single spring. For extension parallel + single = 3/2E and for Spring constant parallel + single = 2/3K

Could you plz explain the last part,,, for spring constant

 

[emkay]

Could you plz explain the last part,,, for spring constant

for the last part, consider the stretching of parallel springs by that single spring (it would be hard compared to the previous springs). Recall the definition that i previously wrote about spring constant and then relate that with this condition. it's pretty easy, a single spring takes a lot of effort in stretching the parallel springs. when springs are in parallel, spring constant is added Ktotal = (K1+K2)k so 2/3K.
arrangement value of k / K
single spring = 1
two springs in series = ½
three springs in series = ⅓
two springs in parallel = 2
three springs in parallel = 3

 

[aniroula]

for the last part, consider the stretching of parallel springs by that single spring (it would be hard compared to the previous springs). Recall the definition that i previously wrote about spring constant and then relate that with this condition. it's pretty easy, a single spring takes a lot of effort in stretching the parallel springs. when springs are in parallel, spring constant is added Ktotal = (K1+K2)k so 2/3K.
arrangement value of k / K
single spring = 1
two springs in series = ½
three springs in series = ⅓
two springs in parallel = 2
three springs in parallel = 3

Thanks bro..,,