Physics: Post your doubts here!

[daredevil]

Okaaay, I will.
Did you get the answer for your problem?

yeah i did thanks a bunch I got it now... i was doing with the drawing the waves method and reading the answer i realized i was going about it all wrong and caught my mistake

 

[hela]

ci)Two other forces would be-
1) the normal reaction force, which acts upwards from the rod.
2) Friction between the rod and the card as the card swings.

cii) the card comes to rest when it's center of gravity is directly along the vertical axis through the rod. This is because at this position the weight does not create any moment about the rod. No resultant moment so the card rests.

THANKS
GOOD EXPLANATION
WOULD YOU PLAESE EXPLAIN THIS
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf CAN YOU PLAESE EXPLAIN WITH IN DETAILS GTHIS QUESTION

 

[daredevil]

THANKS
GOOD EXPLANATION
WOULD YOU PLAESE EXPLAIN THIS
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf CAN YOU PLAESE EXPLAIN WITH IN DETAILS GTHIS QUESTIONView attachment 24643

woaaahhh the first thing i noticed was the number of tabs on the browser in ur pic!! O__O
ummm sorry i can't explain it tht well that's y i wasn't answering :/

 

[littlecloud11]

THANKS
GOOD EXPLANATION
WOULD YOU PLAESE EXPLAIN THIS
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf CAN YOU PLAESE EXPLAIN WITH IN DETAILS GTHIS QUESTIONView attachment 24643

First you connect the output of the loudspeaker to the c.r.o. This is done by connecting the microphone detecting the output to the y-plates of the c.r.o
You can notice a dot move across the screen. Next you can adjust the time-base setting of the c.r.o until you can see a few cycles of the wave. The time base allows you to measure the time period for the wave. Frequency = 1/time period

 

[littlecloud11]

woww i explained that question too earlier today but my exlanation seeems like layman language and kindergarten stuff when compared to urs ... hats off man #RESPECT!! O__O

Haha, thanks.

 

[hela]

First you connect the output of the loudspeaker to the c.r.o. This is done by connecting the microphone detecting the output to the y-plates of the c.r.o
You can notice a dot move across the screen. Next you can adjust the time-base setting of the c.r.o until you can see a few cycles of the wave. The time base allows you to measure the time period for the wave. Frequency = 1/time period

your explanation are very good thank you very much you are very helpful

 

[Oliveme]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_42.pdf
Question 9b (ii) and (iii)
Can someone please explain this to me? I don't understand this at all.
Thank you very much.

 

[raamish]

s05: http://www.freeexampapers.com/index... Level/Physics/CIE/2005 Jun/9702_s05_qp_4.pdf

i need help in qs 3b(ii) dont know why the mc(Delta theta) of ice turned into water was taken. Help me somebdy?

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_42.pdf
Question 9b (ii) and (iii)
Can someone please explain this to me? I don't understand this at all.
Thank you very much.

9 b ii) Since the Vout is the difference btw the two input voltages... Vout will only be zero when both V+ n V- are the same ie V+ = V- ---> V2 = V1
n to calculate the value of V2 (which is in a potential divider circuit), the formula ---> V2 = [R1/(R1 + R2)] * Vtotal.
So, V2 = [1000/(1000 + 125)] * 4.5
V2 = V1 = 4.0 V

iii) V1 = 4.0 V
V2 = [R1/(R1 + R2)] * Vtotal
= [1000/1128] * 4.5
= 3.99 V
Vin = V2 - V1 = 3.99 - 4 = -0.01 V
So, gain = 12 = Vout/Vin
12 = Vout/(-0.01)
Vout = 12 * 0.01
Vout = -0.12 V

 

[Oliveme]

9 b ii) Since the Vout is the difference btw the two input voltages... Vout will only be zero when both V+ n V- are the same ie V+ = V- ---> V2 = V1
n to calculate the value of V2 (which is in a potential divider circuit), the formula ---> V2 = [R1/(R1 + R2)] * Vtotal.
So, V2 = [1000/(1000 + 125)] * 4.5
V2 = V1 = 4.0 V

iii) V1 = 4.0 V
V2 = [R1/(R1 + R2)] * Vtotal
= [1000/1128] * 4.5
= 3.99 V
Vin = V2 - V1 = 3.99 - 4 = -0.01 V
So, gain = 12 = Vout/Vin
12 = Vout/(-0.01)
Vout = 12 * 0.01
Vout = -0.12 V

Thank you so much. gary221
I get it now. That was brilliant.

 

[gary221]

Thank you so much. gary221
I get it now. That was brilliant.

np..

 

[Warrior66]

Asalam-o-Alaikum!
can anyone please
explain http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_22.pdf question 2, part b & ci?
gary221
thanks a bunch!

 

[Warrior66]

Asalam-o-Alaikum once again..!
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
Question 3ci.. when they ask us to calculate change in momentum...is it like mv - mu or mu - mv ?
thanks a bunch..!

 

[Ashique]

9 b ii) Since the Vout is the difference btw the two input voltages... Vout will only be zero when both V+ n V- are the same ie V+ = V- ---> V2 = V1
n to calculate the value of V2 (which is in a potential divider circuit), the formula ---> V2 = [R1/(R1 + R2)] * Vtotal.
So, V2 = [1000/(1000 + 125)] * 4.5
V2 = V1 = 4.0 V

iii) V1 = 4.0 V
V2 = [R1/(R1 + R2)] * Vtotal
= [1000/1128] * 4.5
= 3.99 V
Vin = V2 - V1 = 3.99 - 4 = -0.01 V
So, gain = 12 = Vout/Vin
12 = Vout/(-0.01)
Vout = 12 * 0.01
Vout = -0.12 V

Hey, just one question- in the potential divider why did you take R1 to be 1000 ohms? Shouldn't 125 be taken as R1, since the voltage that passed through the strain gauge is the voltage is passes through the op- amp?

 

[daredevil]

Asalam-o-Alaikum!
can anyone please
explain http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_22.pdf question 2, part b & ci?
gary221
thanks a bunch!

I'm sorry i can't make the graph here but wat u need to do is make a line with a positive gradient first and after it reches the maximum velocity the line bends with a very negative gradient (remember this is rebound of the ball and the time for that is 20ms only) then the graph bends again and another straight line is made again with a positive gradient but it is in the negative area of the graph i.e. below the x-axis. Remember this second area under graph (distance) should be less than the one before because the ball attains a lesser height this time.

in ci u need to calculate the two kinetic energies and find their difference...
u have the velocity of the ball at A so calculate the kinetic energy at that point with Ek=(1/2)mv^2
use the velocity of the ball at B to calculate the Ek at B and find the difference of the two values...
if i am wrong or u don't get what i did then plz do tell

 

[daredevil]

Asalam-o-Alaikum once again..!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Question 3ci.. when they ask us to calculate change in momentum...is it like mv - mu or mu - mv ?
thanks a bunch..!

I think it should be mv-mu

 

[Tkp]

I think it should be mv-mu

yup mv-mu

 

[syed1995]

Asalam-o-Alaikum once again..!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Question 3ci.. when they ask us to calculate change in momentum...is it like mv - mu or mu - mv ?
thanks a bunch..!

final momentum - initial momentum ..
mv - mu
m(v-u)

 

[SalmanslK]

Physics A2..June 05 P4 Q1 .

' State The Origin Of Centripetal force? '

the Earth is orbiting the sun in this question...what is the origin anyone?

 

[daredevil]

may june 2012 p22 Q3 and Q7