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Physics: Post your doubts here!

littlecloud11

littlecloud11

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hela said:
can you plaese explain in details
c)i and c)ii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_22.pdf
View attachment 24641
Click to expand...

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.
 
daredevil

daredevil

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littlecloud11 said:
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.
Click to expand...
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 :D
 
daredevil

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asd said:
Okaaay, I will. :)
Did you get the answer for your problem?
Click to expand...
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 :)
 
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H

hela

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littlecloud11 said:
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.
Click to expand...
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 QUESTIONUntitled.png
 
littlecloud11

littlecloud11

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hela said:
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
Click to expand...

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
 
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hela

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littlecloud11 said:
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
Click to expand...
your explanation are very good thank you very much you are very helpful
 
gary221

gary221

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Oliveme said:
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.
Click to expand...

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

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gary221 said:
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
Click to expand...
Thank you so much. gary221
I get it now. That was brilliant. (y)
 
daredevil

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Warrior66 said:
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! :)
Click to expand...
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 :)
 
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