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Physics: Post your doubts here!

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Textbook for physics 9702
ASAL Physics Executive preview_Print_Digital.png



The attachment doesn't want to upload properly so I have provided a free a and a premium file host:
When link dies I try again. Feel free to email me at [email protected] for Fresh links or mirrors.
I personally prefer Rapidgator or 1fichier but I know many don't have a Premium leech like alldebrid or real debrid.

Also Since the material is copyrighted its best that its not hosted by the site.
Code:
Free File Host dropapk (Click slow download, Copy in the number ,Click download , Profit!)
https://drop.download/fml2u3harpx9

Debrid(Premium file host for those who have a debrid)
https://rapidgator.net/file/39b23da0bb48453c1eed68d3dd6e41e0/
Best of Luck With Your A levels
 

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How is this Answer 3x?

View attachment 66940
Hooke's law is F=k*x.
And in parallel we add the spring constant of springs.
So, initially: F=3k*x
Then, one spring removed and force doubled: 2F=2k*3x. Must simplify to F=3k*x, as it does.

Another way to look at it is:
Hooke's Law is f=k*x
By changes, new K is made 2/3 times original k
New F is made 2 times original f
As x is directly proportional to f and inversely proportional to k,
new X is 3/2 times * 2 times original x. So, 3 times original x.

I hope this helps.
 
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Hi everyone, AsSalamoAlaikum Wr Wb...

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all. :D ;)

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites! :)
Thanks!
Jazak Allah Khair!

Physics Notes:

Some links & Notes - by destined007

As physics p1 MCQS YEARLY Solved [explaination]
Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery

Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>

Here are the compiled A level topical Physics questions in PDF form...

Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd

Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c

(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)
Can anybody give me please The Answer of Cambridge International AS A Level Physics Coursebook 3rd Edition EXAM-STYLE QUESTIONS?
 
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Can anybody attach the answers for self assesment and exam style questions for

Cambridge International AS & A Level Physics: Coursebook, Third Edition​

Thank you
1667809602176.png
 

PlanetMaster

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View attachment 70291

CAN ANYONE HELP ME SOLVE THIS QUESTION?
Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
Now, the cube floats 42% above the surface and therefore the immersed part i.e. the part that displaces mercury is 100-42% = 58% or 0.58.

Now we can have p1/v1 = p2/v2 where
p1 is the density of mercury,
p2 is the density of iron,
and therefore p1 = p2(v1/v2)
This gives us p1 = 7900/0.58 = 13,621 kg/m^3
 
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Since this is an MCQ, lets first do a quick sanity check. The cube is floating in equilibrium in mercury so the density of mercury should be heaver than the cube itself i.e the answer is either C or D.
Now, the cube floats 42% above the surface and therefore the immersed part i.e. the part that displaces mercury is 100-42% = 58% or 0.58.

Now we can have p1/v1 = p2/v2 where
p1 is the density of mercury,
p2 is the density of iron,
and therefore p1 = p2(v1/v2)
This gives us p1 = 7900/0.58 = 13,621 kg/m^3
thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?
 

PlanetMaster

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thankyou but can you please tell which formula is p1v1=p2v2 and what do v1 and v2 signify here?
Since this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.

To elaborate the steps, lets first consider the immersed part where weight of solid equals the weight of liquid displaced.
[imath]W_{s}=W_{ld}[/imath]
And since w=mg, we get
[imath]m_{s}g=m_{ld}*g[/imath]\
And since density=mass/vol, we get
[imath]P_{s}V_{s}=P_{ld}V_{ld}[/imath] as g cancels out
Now volume of liquid displaced is equal to the volume of the immersed portion, so lets write it as
[imath]P_{s}V_{s}=P_{ld}V_{imm}[/imath]
Now we rearrange as
[imath]\frac{P_{s}}{P_{l}}=\frac{V_{imm}}{V_{s}}=0.58[/imath]
And so
[imath]P_{l}=\frac{P_{s}}{0.58}=\frac{7900}{0.58}=13621\:kg\:m\:^{-3}[/imath]

Hope that helps!
 
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Since this is a MCQ question, I skipped a lot of steps and used just ratios and the ratio based on simply density=mass/vol. This is where p1v1=p2v2 comes from.

To elaborate the steps, lets first consider the immersed part where weight of solid equals the weight of liquid displaced.
[imath]W_{s}=W_{ld}[/imath]
And since w=mg, we get
[imath]m_{s}g=m_{ld}*g[/imath]\
And since density=mass/vol, we get
[imath]P_{s}V_{s}=P_{ld}V_{ld}[/imath] as g cancels out
Now volume of liquid displaced is equal to the volume of the immersed portion, so lets write it as
[imath]P_{s}V_{s}=P_{ld}V_{imm}[/imath]
Now we rearrange as
[imath]\frac{P_{s}}{P_{l}}=\frac{V_{imm}}{V_{s}}=0.58[/imath]
And so
[imath]P_{l}=\frac{P_{s}}{0.58}=\frac{7900}{0.58}=13621\:kg\:m\:^{-3}[/imath]

Hope that helps!
thankyou so much :)
 

PlanetMaster

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View attachment 70388

CAN SOMEONE PLEASE EXPLAIN THIS QUESTION. THE ANSWER IS D
Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
And cross sectional area [imath]A = \frac{πd^{2}}{4}[/imath]

When the wire is stretched, the diameter is reduced to 94%.
This causes the area of stretched wire to be
[imath]A_{s} = \frac{π(0.94d)^{2}}{4} = 0.8836(\frac{πd^{2}}{4})[/imath] = 0.8836A

Volume of the wire is cross-sectional area x length,
[imath]V = AL[/imath]
And since the volume of the wire is unchanged,
[imath]AL=A_{s}L_{s}[/imath]
and the diameter is reduced to 94%, the length has to be
[imath](\frac{πd^{2}}{4})L=(\frac{π(0.94d^{2})}{4})(L_{s})[/imath]
[imath]L_{s}=\frac{L}{0.94^{2}}[/imath]

So, the length of the stretched wire is [imath]\frac{L}{0.94^{2}}[/imath] or [imath]1.132L[/imath]

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]

So Resistance of stretched wire
[imath]R_{s} = \frac{ρL_{s}}{A_{s}} = \frac{ρ(1.132L)}{0.8836A} = 1.28\frac{ρL}{A} = 1.28R[/imath]

Hope this explains!
 
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Resistance of wire [imath]R = \frac{ρL}{A}[/imath]
And cross sectional area [imath]A = \frac{πd^{2}}{4}[/imath]

When the wire is stretched, the diameter is reduced to 94%.
This causes the area of stretched wire to be
[imath]A_{s} = \frac{π(0.94d)^{2}}{4} = 0.8836(\frac{πd^{2}}{4})[/imath] = 0.8836A

Volume of the wire is cross-sectional area x length,
[imath]V = AL[/imath]
And since the volume of the wire is unchanged,
[imath]AL=A_{s}L_{s}[/imath]
and the diameter is reduced to 94%, the length has to be
[imath](\frac{πd^{2}}{4})L=(\frac{π(0.94d^{2})}{4})(L_{s})[/imath]
[imath]L_{s}=\frac{L}{0.94^{2}}[/imath]

So, the length of the stretched wire is [imath]\frac{L}{0.94^{2}}[/imath] or [imath]1.132L[/imath]

Resistance of wire [imath]R = \frac{ρL}{A}[/imath]

So Resistance of stretched wire
[imath]R_{s} = \frac{ρL_{s}}{A_{s}} = \frac{ρ(1.132L)}{0.8836A} = 1.28\frac{ρL}{A} = 1.28R[/imath]

Hope this explains!
wont we subtract 0.94 from 1D as it says DECREASED BY 94%
 
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