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Physics: Post your doubts here!

H

hela

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syed1995 said:
10 for D and 33 for C?
Click to expand...
an anyone solve
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf Q 10 AND Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q29
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf Q11
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q14

THANKS FOR YOUR HELP
 
H

hela

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hela said:
an anyone solve
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf Q 10 AND Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q29
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf Q11
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q14

THANKS FOR YOUR HELP
Click to expand...
it was solved thanks anyway
 
megamind

megamind

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A star said:
linear scales have advantage
they are stndardised so can be used for all readings
disadvantage
cannot be used accurately for those readings ehich do not vary linearly with time
non linear advantage
sensitivity can be changed to support the reading
can be used suitably for all measurements
Click to expand...
similar question came for 42 thanks man
 
megamind

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gary221 said:
ADC = samples the signal at regular intervals. for each sample, produces an 8- bit binary no. So DAC = will convert the binary no to a decimal no ie to an analogue signal.
Parallel to serial = takes the 8 bits from 1 sample n places them after the 8 bits from the previous sample to produce a continuous series of binary digits.
So, serial to parallel must do the opp. ie assemble the 8 bits, n pass them to the DAC for conversion into analogue. then, pass the next 8 bits to produce the next signal, n so on.
Did I say tht they both r the same??
I guess not : "The serial to parallel converter n the DAC produce an analogue signal" does not imply they r same.
And if I did say they r same, sorry, m bad!!
Click to expand...

DUDE this came for the paper
THANK you SOOO much
 
syed1995

syed1995

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hela said:
thanks I have the MS
I need explanations
Click to expand...

I was asking.. If those are correct then.

D in 10 because.. The acceleration of horizontal component will be zero and constant velocity. But their will be non-constant acceleration there for vertical component.. so the speed will first decrease when going upwards as negative acceleration and then will keep increasing as it comes downwards in the vertical component.

The answer is not C because since there is horizontal velocity even when it reaches the top.. the initial velocity is not zero at top.

Hence the Answer is D.
 
syed1995

syed1995

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hela said:
thanks I have the MS
I need explanations
Click to expand...


33 C because the Charge on the helium nucleus is positive so it will deflect towards the same side as the proton did. But will deflect less since the mass of a Helium nucleus is approximately 4 times the mass of a proton.

More mass means less deviation.
 
SararaIH

SararaIH

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Oliveme

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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf Q2 Q17 Q31 Q35
Click to expand...

2) 3000 revolutions per minute means 50 revolutions in one second. And therefor this the frequency.
Frequency = 50Hz
Now, using the frequency, we find the time taken for one revolution -----> T = 1/f = 1/50 -----> 0.02 seconds.
Time is o.o2 seconds and so, in other words, 20 milliseconds (0.02 x 1000).
So you see, out of all the options given, 10 ms cm^-1 is the closed. So, the answer is B.
If you use 1 s cm^-1 as time base, there would be too many oscillations on the screen and wouldn't give you a good display.
Similarly, 100 microseconds or 1 microsecond cm^-1 will give you a very extended display, you wouldn't be able to see proper oscillations.

31) The total resistance of the cable is 8 ohms. The cable is 800 meters long since that's the distance between the power supply and the relay and there are 2 wires in the cable. 0.005 ohms per m means in one wire the resistance is 0.005 ohms x 800 = 4 ohms. So total resistance is 2 x 4 = 8 ohms.
Add the resistance of the wires to the resistance of the relay. Relay's resistance is R = V/I = 16/0.6 = 26.7 ohms. (26.7 + 8)
To find the output voltage = IR = 0.6 x (26.7 + 8) = 20.82 and the answer is C.
We are treating the relay and wires as one unit. The current through wires does not change and so we can plug the 0.6 amps of the relay into the V=IR formula.
It just seems like it's complicated but it really isn't. Try looking at it in a simplified way.

I'll get back to you on the rest of the questions as soon as possible. :)
 
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