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Physics: Post your doubts here!

F

Faaiz Haque

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usama321 said:
4 Period is the time taken for one complete oscillation. Now, one complete oscillation would be the time taken between two successive crests. Let's take the crests as the peak of the wave on the graph. Travelling horizontally, it takes 4 block till the next peak. So, 2.5 * 4 = 10. Looking at the vertical axis, we can see that the peak voltage is about a bit more than 3 blocks from the horizontal axis. As each block represents 5mV, 17 would be our answer. It cant be 25 because that would require 5 blocks. Answer C

7. Both the other quantities are velocities. Thus X should represent the velocity due to the force F. Answer is thus C, as product of a and t would give us the velocity due to that acceleration for the specified time.

13 Resolve the forces perpendicularly to the beam. 8 * sin60. Now just multiply it by the length of the beam as it is a couple. 8 *sin60 * .6 = 4.15 = B

19 1/2 * force* extension... force = 2N. Extension = 0.9-0.5 = 0.4 Answer A

26 lamdba = ax/d... increasing lambda would increase the fringe spacing according to the formula, and as red light has a higher wavelength than violet light, the answer is D

33 I've explained this type of question a couple of times. Just scroll back a few pages and you will find the solution.

36 Terminal potential difference is the PD of the battery after we have taken account of the internal resistance. First find the current in the circuit with V = IR
I =3/6 = 0.5 A. Now calculate the PD due to internal resistance. V = 0.5 * 2 = 1V. Subtract this from the EMF and we have our terminal voltage = 2V. Output power = IV = 0.5 * 2 =1 answer is C
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Thanks a lot this was very helpful I just couldn't find where you explained question 33
 
usama321

usama321

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The Godfather said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf
1 b,c-i)ii)
3b)
4b,cii)-1,2,3
6c,di)ii)
Please help.............
ZaqZainab
asma tareen
Thought blocker
Suchal Riaz
snowbrood
:)
Click to expand...

Well if you can't do part b of question 1, i'd suggest you read the uncertainty topic first, and gets your concepts right.

For c part, you can only have equal to or one less than the number of significant figures you used for calculating that value. T and L have 3 and 2 sig figures, thus g which was calculated from them can't have 4 sig figures


For 3 b, just note down the scale of the horizontal vector on the paper as it is supposed to be 4m/s. Draw a vector downwards according to that scale of 6.2m/s. Use head to tail rule to find the resultant. Use pythagoras just to confirm your answer.

4b Work done = force * distance. Now, when a spring obeys Hooke's law, its graph of force against extension would act as force and distance. Thus we get W = fx. However, as a spring that obeys Hooke's law always has a graph which starts from origin and is a line of constant gradient, it always gives us a triangle. Thus W = 1/2 fx... now we know that f = kx... So replace it and we get 1/2 kx square

c1 Use mgh, with h as change in height
c2 Use 1/2 kx square... and subtract from it the initial elastic potential energy to get the change
c3 Work done = sum of energies... mgh decreased so it would be negative and elastic potential increased so it would be positive

6c As the lamps are identical and in series, the pd would be divided across them so each would have a pd of 1.5V. Figure out the current they have at 1.5V pd, and use it to calculate R = V/I.. As they are in series there resistance would be added. In parallel, they both will have the pd of 3V across them. Again use the graph for R and then use the formula for resistance in parallel.
di is pretty straightforward?
dii We can see from the graph that as the PD increases, the current in the circuit does not increase lineraly with it. This shows that resistance is increasing with increasing pd. Though current should increase as a rule with PD, due to heating the resistance would also increase leading to a decrease in current in the circuit.
 
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usama321 said:
Well if you can't do part b of question 1, i'd suggest you read the uncertainty topic first, and gets your concepts right.

For c part, you can only have equal to or one less than the number of significant figures you used for calculating that value. T and L have 3 and 2 sig figures, thus g which was calculated from them can't have 4 sig figures


For 3 b, just note down the scale of the horizontal vector on the paper as it is supposed to be 4m/s. Draw a vector downwards according to that scale of 6.2m/s. Use head to tail rule to find the resultant. Use pythagoras just to confirm your answer.

4b Work done = force * distance. Now, when a spring obeys Hooke's law, its graph of force against extension would act as force and distance. Thus we get W = fx. However, as a spring that obeys Hooke's law always has a graph which starts from origin and is a line of constant gradient, it always gives us a triangle. Thus W = 1/2 fx... now we know that f = kx... So replace it and we get 1/2 kx square

c1 Use mgh, with h as change in height
c2 Use 1/2 kx square... and subtract from it the initial elastic potential energy to get the change
c3 Work done = sum of energies... mgh decreased so it would be negative and elastic potential increased so it would be positive

6c As the lamps are identical and in series, the pd would be divided across them so each would have a pd of 1.5V. Figure out the current they have at 1.5V pd, and use it to calculate R = V/I.. As they are in series there resistance would be added. In parallel, they both will have the pd of 3V across them. Again use the graph for R and then use the formula for resistance in parallel.
di is pretty straightforward?
dii We can see from the graph that as the PD increases, the current in the circuit does not increase lineraly with it. This shows that resistance is increasing with increasing pd. Though current should increase as a rule with PD, due to heating the resistance would also increase leading to a decrease in current in the circuit.
Click to expand...
I was too not knowing few of the answers so I did not replied, but its awesome. Thanks to share :)
 
The Godfather

The Godfather

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usama321 said:
Well if you can't do part b of question 1, i'd suggest you read the uncertainty topic first, and gets your concepts right.

For c part, you can only have equal to or one less than the number of significant figures you used for calculating that value. T and L have 3 and 2 sig figures, thus g which was calculated from them can't have 4 sig figures


For 3 b, just note down the scale of the horizontal vector on the paper as it is supposed to be 4m/s. Draw a vector downwards according to that scale of 6.2m/s. Use head to tail rule to find the resultant. Use pythagoras just to confirm your answer.

4b Work done = force * distance. Now, when a spring obeys Hooke's law, its graph of force against extension would act as force and distance. Thus we get W = fx. However, as a spring that obeys Hooke's law always has a graph which starts from origin and is a line of constant gradient, it always gives us a triangle. Thus W = 1/2 fx... now we know that f = kx... So replace it and we get 1/2 kx square

c1 Use mgh, with h as change in height
c2 Use 1/2 kx square... and subtract from it the initial elastic potential energy to get the change
c3 Work done = sum of energies... mgh decreased so it would be negative and elastic potential increased so it would be positive

6c As the lamps are identical and in series, the pd would be divided across them so each would have a pd of 1.5V. Figure out the current they have at 1.5V pd, and use it to calculate R = V/I.. As they are in series there resistance would be added. In parallel, they both will have the pd of 3V across them. Again use the graph for R and then use the formula for resistance in parallel.
di is pretty straightforward?
dii We can see from the graph that as the PD increases, the current in the circuit does not increase lineraly with it. This shows that resistance is increasing with increasing pd. Though current should increase as a rule with PD, due to heating the resistance would also increase leading to a decrease in current in the circuit.
Click to expand...
Okay :) thanks a lot :)
 
U

unique111

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Can somebody help me with this? It seems quite easy, cause its in number 5, but i couldn't figure out where to start from. Much help appreciated.
 

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usama321

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sadiaali said:
No one helping me :cry::cry:.I m crying.
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At time 0.25T, the wave should have passed 1/4th of its disance along the string. You can also calculate it to be sure.

T = 1/15
0.25T = 1/60
we have already calculated the speed as 12 m/s. So after 1/60 seconds, it would be 12/60 = 20 cm

We did not need to do that calculation as we can easily see from the graph that 1/4 of the wave would be at 20cm. Now you will have to draw something like this and continue it on
asd.png

For the other part, we know that the points between the nodes in a stationary wave just keep moving up and down. We also know that in the time of one complete ossicillation, the particles will go all the way to their lowest amplitude and then back to their original positions. Now at 0.5T, we know that the particles will be at their lowest point. Calculate half of this, which is 0.25T, and we know that the particles will be halfway between their highest and lowest point, which would be a straight horizontal line through the x axis. all the points will have zero amplitude
 
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usama321 said:
At time 0.25T, the wave should have passed 1/4th of its disance along the string. You can also calculate it to be sure.

T = 1/15
0.25T = 1/60
we have already calculated the speed as 12 m/s. So after 1/60 seconds, it would be 12/60 = 20 cm

We did not need to do that calculation as we can easily see from the graph that 1/4 of the wave would be at 20cm. Now you will have to draw something like this and continue it on
View attachment 38101

For the other part, we know that the points between the nodes in a stationary wave just keep moving up and down. We also know that in the time of one complete ossicillation, the particles will go all the way to their lowest amplitude and then back to their original positions. Now at 0.5T, we know that the particles will be at their lowest point. Calculate half of this, which is 0.25T, and we know that the particles will be halfway between their highest and lowest point, which would be a straight horizontal line through the x axis. all the points will have zero amplitude
Click to expand...
ty :)
 
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