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Physics: Post your doubts here!

SIstudy

SIstudy

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TimBluesWin said:
1b. Use formula R=(p*l)/A (p is resistivity, l is length, A is cross-sectional area of wire, use formula A=pi*(0.5*d)^2 R is resistance)
So you just input the original measurement into the equation.
5a (ii) Okay so you've found 0.667 for the minimum current. So you can draw a curve with decreasing gradient as R increases. (As V=IR) (I gave a picture of it)
5b. To find the minimum/maximum current, you need to find the minimum/maximum voltage first.
Maximum voltage: 12/12*12=12V. Hence Max current= 12V/6=2A
Minimum voltage: 0/12*12=0V. Hence min current: 0A

Hope I am right, if I'm wrong, just tell me.
Click to expand...
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_21.pdf
fr Q5 a (ii)
how did yu draw the graph, I mean what readings yu took?
 
TimBluesWin

TimBluesWin

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SIstudy said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf
Q1 b(ii)
please solve, showing the calculations
I tried it several times but my answer repeatedly came the same incorrect value
Click to expand...
Have you changed the units to m for the diameters and length?
Btw its my way:
R=p*l/A
7.5=p*(1750/1000)/(phi*(0.5*0.38*10^-3)^2)
7.5*phi*(0.5*0.38*10^-3)^2=1.75p
p=7.5*phi*(0.5*0.38*10^-3)^2/1.75
p=4.8604*10^-7 ohm m.-->4.86*10^-7 ohm m

Hope this helps! :D
 
B

Biocalife

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Did these definitions allright for applications (paper 4) ?

Tuning circuit : It regenerates the signal before trasnmiting it to the R.F amplifier.
Switch : It turn on and off the aerial
 
S

Saad Amjad.

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Biocalife said:
Did these definitions allright for applications (paper 4) ?

Tuning circuit : It regenerates the signal before trasnmiting it to the R.F amplifier.
Switch : It turn on and off the aerial
Click to expand...
switch-it enables the aerial for transmission and receipt of signal
tuning circuit-it fixes the carrier frequency
 
SIstudy

SIstudy

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TimBluesWin said:
Have you changed the units to m for the diameters and length?
Btw its my way:
R=p*l/A
7.5=p*(1750/1000)/(phi*(0.5*0.38*10^-3)^2)
7.5*phi*(0.5*0.38*10^-3)^2=1.75p
p=7.5*phi*(0.5*0.38*10^-3)^2/1.75
p=4.8604*10^-7 ohm m.-->4.86*10^-7 ohm m

Hope this helps! :D
Click to expand...
sir, the next part of the question
what yu solved is b (i)
i am askin fr b (ii)
thanks in advance :)
 
X

XRDXRD

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Need help on paper 1 S06 Q33. find the pd between X and Y

I dont really understand where is "between P and Q". it could be anywhere on the circuit or connect a new wire between P and Q.

Please help.
 
B

Bba321

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Last edited:
A

adarsha

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force and motion

one horse pulls,with a force of X N , a cart of a mass 800 kg along a horizontal road at constant speed. three horses pulling each with a force of X N , give the cart an acceleration of 0.8m/s^2 . find the time it will take two horses to increase the speed of the cart from 2m/s to 5 m/s , given that each horse pulls with a force of X N, and that the resistance to the motion has the same constant value at all time.
 
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