Physics: Post your doubts here!

[princess Anu]

The radius of the orbit of a plant X round the sun is 4 times the orbital radius of Earth. If a year on the planet is assumed to be period of its revolution round the sun, what is an equivalent age of an eighty year old in terms of the year on the planet X.



^ Plz help

 

[princess Anu]

Two stars, masses 10^20kg and 2*10^20 kg rotate about their common centres of mass with an angular speed of 'w'. Assuming that the only force on a star is the mutual gravitational force between them, calculate the angular frequency's' assuming the distance b/w the stars is 10^6km.

Help...

 

[Xtreamer]

Why can't the angle ever exceed 90 in vertical circular motion and can in horizontal?

 

[Umna.Asim]

Need topic wise past papers please

 

[My Name]

Need topic wise past papers please

Check here:
https://www.xtremepapers.com/commun...nd-marking-scheme-all-in-one-2001-2014.36824/

 

[ahmedish]

guys, in practical, whats more important in a column??? Significant figures or decimal places? like if I get my results 0.58, 0.96, 1.23. should I make them all 3 s.f? or leave them because they're all the same decimal places?

 

[Copy Cat]

 

[Umna.Asim]

Check here:
https://www.xtremepapers.com/commun...nd-marking-scheme-all-in-one-2001-2014.36824/

Thankyou so much!
jazak Allah

 

[My Name]

Thankyou so much!
jazak Allah

Welcome!
Waeeyaki

 

[qwertypoiu]

guys, in practical, whats more important in a column??? Significant figures or decimal places? like if I get my results 0.58, 0.96, 1.23. should I make them all 3 s.f? or leave them because they're all the same decimal places?

I think you can do either way, and they'd accept it. I used to get confused about this too. I can't remember clearly now but I think what I did was, for time measurements, I would stick to 3 significant figures, so my measurements would be like: 0.568, 0.789, 1.25
However for my measurements of angles, I'd do: 34°, 89°, 125°, etc. For rulers: 12.6cm, 10.9cm, 9.5cm, 8.4cm, etc.
Why?
For time measurements, I measured many oscillations (like 10 or 20) then took the reading, so my stopwatch COULD display 5.68s, so I'd divide by 10 and get 0.568
For angle measurements, I have no way of making 34° into three sig figures! My measuring instrument (protractor) simply doesn't have the ability! On the other hand 123° could be measured, and I wasn't gonna round that down to 120! That'd reduce my accuracy, and why should I when my instrument CAN manage this?
For length measurements, I had a ruler. If something was shorter than 10.0cm, my measurements could only be 2 sig figures, like 9.5cm. So I was limited by my measuring instrument. I cannot make 9.5 into three sig, and I don't wanna round up 12.6cm to 13cm just to stay consistent with the significant figures.

So those are just my thoughts and how I handled the problem, it may or may not be right (although Alhamdulillah I got an A in practical). If anyone wants to add their thoughts or challenge these points I'll be more than happy to read on what they think about this.

 

[manya]

can someone tell me how to find the errors for d^2

 

[qwertypoiu]

can someone tell me how to find the errors for d^2View attachment 57052

Let me do the first example for you:
(2.1±0.1)^2
First, find the percentage error:
0.1/2.1*100% = 4.76%
So our problem now becomes:
(2.1±4.76%)^2

Remember that when we add values, we add their uncertainties. When we subtract values, we also add their uncertainties. When we multiply values, we add their PERCENTAGE uncertainties, and when we divide values, we add their percentage uncertainties as well.
When a value is raised to a power (Eg ^5), the percentage uncertainty is MULTIPLIED by the new power (ie 5)

So here: (2.1±4.76%)^2 = 2.1^2 ± 4.76*2% = 4.41 ± 9.52%

Convert the percentage back to real uncertainty now:
9.52%* 4.41 = 0.419832

So our number now becomes:
4.4±0.4

Note that our number should be to two significant figures since the original number was also to two sig and our uncertainty should only have one significant figure

 

[manya]

Let me do the first example for you:
(2.1±0.1)^2
First, find the percentage error:
0.1/2.1*100% = 4.76%
So our problem now becomes:
(2.1±4.76%)^2

Remember that when we add values, we add their uncertainties. When we subtract values, we also add their uncertainties. When we multiply values, we add their PERCENTAGE uncertainties, and when we divide values, we add their percentage uncertainties as well.
When a value is raised to a power (Eg ^5), the percentage uncertainty is MULTIPLIED by the new power (ie 5)

So here: (2.1±4.76%)^2 = 2.1^2 ± 4.76*2% = 4.41 ± 9.52%

Convert the percentage back to real uncertainty now:
9.52%* 4.41 = 0.419832

So our number now becomes:
4.4±0.4

Note that our number should be to two significant figures since the original number was also to two sig and our uncertainty should only have one significant figure

oh thankyou so much

 

[nehaoscar]

I need help with finding the gradient of the lines in P5
These are the two papers
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_51.pdf
Please look at question 2 of both papers.

In the first link (s12 paper) to find the gradient I did,
(0.82-0.36)/(1.95-1.67)x10ˉ²
This gives 164
But the answer is 1.64 so my teacher said don't multiply by 10ˉ² since the graph says I/10ˉ²

In the second paper (s10) to find the gradient I did what my teacher said above
(334-152)/(4.5-2)
This gives 68.8
Now in part d, the answer should be 10^-6 whereas I get 10^-3
So in this case I had to multiply the
(334-152)/(4.5-2)x10^-3
to get the correct answer

So I am confused when to multiple by the power given on the axis and when not to? :/
Please help!

 

[nehaoscar]

Why is q negative for the first row?
no thermal energy is supplied to the system since it's constant temperature right so then why zero??

 

[OneOfAKind]

Can someone please solve this

any help is much appreciated.

 

[qwertypoiu]

Can someone please solve this

any help is much appreciated.

When it's on the floor horizontally:
Fr = uR
u = Fr/R = 10N/12N = 0.833
When it's titled:
R = 12cos@
12sin@ > Fr
12sin@ > uR
12sin@ > 0.833*12cos@
tan@ > 0.833
@ > 39.8°

 

[nehaoscar]

I need help with finding the gradient of the lines in P5
These are the two papers
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_51.pdf
Please look at question 2 of both papers.

In the first link (s12 paper) to find the gradient I did,
(0.82-0.36)/(1.95-1.67)x10ˉ²
This gives 164
But the answer is 1.64 so my teacher said don't multiply by 10ˉ² since the graph says I/10ˉ²

In the second paper (s10) to find the gradient I did what my teacher said above
(334-152)/(4.5-2)
This gives 68.8
Now in part d, the answer should be 10^-6 whereas I get 10^-3
So in this case I had to multiply the
(334-152)/(4.5-2)x10^-3
to get the correct answer

So I am confused when to multiple by the power given on the axis and when not to? :/
Please help!

 

[qwertypoiu]

I need help with finding the gradient of the lines in P5
These are the two papers
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_51.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_51.pdf
Please look at question 2 of both papers.

In the first link (s12 paper) to find the gradient I did,
(0.82-0.36)/(1.95-1.67)x10ˉ²
This gives 164
But the answer is 1.64 so my teacher said don't multiply by 10ˉ² since the graph says I/10ˉ²

In the second paper (s10) to find the gradient I did what my teacher said above
(334-152)/(4.5-2)
This gives 68.8
Now in part d, the answer should be 10^-6 whereas I get 10^-3
So in this case I had to multiply the
(334-152)/(4.5-2)x10^-3
to get the correct answer

So I am confused when to multiple by the power given on the axis and when not to? :/
Please help!

When making calculations based on values of gradients, make sure you use the REAL values (ie the one with powers of ten multiplied)
When asked to calculate the gradient of a graph, I'd prefer to use the real value still, but it seems they want without power of ten adjustments.

 

[nehaoscar]

When making calculations based on values of gradients, make sure you use the REAL values (ie the one with powers of ten multiplied)
When asked to calculate the gradient of a graph, I'd prefer to use the real value still, but it seems they want without power of ten adjustments.

That's what i thought as well like always take the powers
but then in the first paper they're not using the power... is it something to do with logs?