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Physics: Post your doubts here!

VelaneDeBeaute

VelaneDeBeaute

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A*(a*) said:
Can you please help me with this? https://www.xtremepapers.com/community/attachments/23499/
Click to expand...

Velocity is the gradient of the acceleration-time graph. Following that, the graph C seems the viable option. If you look closely, toh the gradient of a-t graph increases from a horizontal slope, decreases and reaches back a 0 gradient at its maximum point. That is what the graph C shows! :)
Concentrate a little and you'll surely understand! :)
 
haha101

haha101

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VelaneDeBeaute said:
Velocity is the gradient of the acceleration-time graph. Following that, the graph C seems the viable option. If you look closely, toh the gradient of a-t graph increases from a horizontal slope, decreases and reaches back a 0 gradient at its maximum point. That is what the graph C shows! :)
Concentrate a little and you'll surely understand! :)
Click to expand...
The answer is A my friend :)
 
A*(a*)

A*(a*)

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VelaneDeBeaute said:
Velocity is the gradient of the acceleration-time graph. Following that, the graph C seems the viable option. If you look closely, toh the gradient of a-t graph increases from a horizontal slope, decreases and reaches back a 0 gradient at its maximum point. That is what the graph C shows! :)
Concentrate a little and you'll surely understand! :)
Click to expand...

haha101 said:
The answer is A my friend :)
Click to expand...

haha101 said:
A way to look at this question is by looking at the end of the graph . when acceleration is zero it doesnt mean velocity is zero so B & C arent correct . and D isnt correct as it only shows acceleration increasing n no decrease . So A has to be the ans
Click to expand...

yes guys the answer is A, I also thought B was correct but, acceleration never goes negative, so velocity would be ever increasing here :D
thanks haha101 :)
 
salvatore

salvatore

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snowbrood said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf q no 7 VelaneDeBeaute gary221 salvatore
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The uncertainty in g is simply obtained by adding the uncertainties of s and t.
In this case, the ucertainty of 2s is 2% (The constant '2' is ignored).
The uncertainty in t^2 is 3% + 3% = 6%. This is because t^2 is the same as t x t, so the uncertainties are supposed to be added.
Total uncertainty in g = 2% + 6% = 8%

Hope that helped :)
 
VelaneDeBeaute

VelaneDeBeaute

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haha101 said:
A way to look at this question is by looking at the end of the graph . when acceleration is zero it doesnt mean velocity is zero so B & C arent correct . and D isnt correct as it only shows acceleration increasing n no decrease . So A has to be the ans
Click to expand...

Apologies, I missed that thing. Thank you for the correction! :)
 
S

sagar65265

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Scafalon40 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_42.pdf
Q11 part b (ii)
why is the frequency 10 kHz?
Click to expand...

Since the wave is amplitude modulated (the frequency of the carrier wave appears constant and thus unmodulated), the carrier wave's amplitude follows the envelope of the signal wave; in other words, if you connect the peaks of every wave to the next one, you should get something like a sine wave. In this case, the maximum amplitude occurs first at

t = approximately 2 microseconds - you can also estimate this to be at t = 0 microseconds since you can shift the graph to the left along the x axis without changing the general shape of the graph.

The wavelength is from one peak till the direct next peak, so this next peak (of both the carrier and the signal) occurs at

t = approximately 102 microseconds (or 100 microseconds if you have approximated 2 microseconds to 0)

so the difference is either 100 - 0 or 102 - 2, both of which are equal to 100 microseconds.
So the time period of the wave is 100 microseconds and as

frequency = 1/Time Period,

time period = 1 / (100 x 10^-6) = 10,000 Hz = 10 kHz.

Hope this helped!

(A neat little thing is that if you start with the beginning of the graph at t = 0 at the top of your webpage and scroll up till the top, you can actually see the carrier wave as it is!)

Good Luck for your exams!
 
nodyed

nodyed

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hassankhan said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_2.pdf

Plz need help with Q5 part c
Click to expand...

First of all data you are provided with
Frequency = 125 hz
tension = 4n
wavelength = 17.8/100 = 0.178m (Note : this length is given for anti node seperation, and wave length is between 2 alternative anti nodes or nodes)
therefore : 0.178 * 2 = 0.356 m (this length is of wavelength )

V= f*wavelength
v = 125 * 0.356
v = 44.5 m/s

now rooting on both sides
m = 4/44.5^2
m = 2.1 x 10^-3
 
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