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Physics: Post your doubts here!

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image-jpg.56053


can anyone help plz ???
 
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View attachment 56395

need help plz in this question
i) there will be a point in the field between the two, where Moon's gravitational field will cancel out earth's Gravitational field.

See, the resultant gravitational force on a test particle between the two will be equal to Fe (force by EARTH pulling the object towards its centre)- Fm (force by MOON pulling the object towards its centre). At some point, both will be EQUAL AND OPPOSITE. Hence, there would be 0 resultant force on the object!!

ii. Gravitational field will be zero when field by earth = field by moon

GMe/x2 = GMm/ (6ORe - x)^2 ; where Me and Mm are used to show mass of earth and moon respectively.
 
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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
 
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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...
 
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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?
 
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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.
:)
 
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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
 
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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 :):)
 
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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!
 
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