Physics Olympiad Watch

ThunderBeard
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I have gcse physics olympiad thing on Friday, so as a way of revision I tried a past paper, the most recent one. I didnt understand some of the questions so was wondering if someone would be able to help me? Its question 13 part c and f and all of 14 on this paper:

https://www.bpho.org.uk/user/pages/0...Paper_2016.pdf

and the answers:

https://www.bpho.org.uk/user/pages/0...tions_2016.pdf
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Plagioclase
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(Original post by ThunderBeard)
I have gcse physics olympiad thing on Friday, so as a way of revision I tried a past paper, the most recent one. I didnt understand some of the questions so was wondering if someone would be able to help me? Its question 7, 13 part f and all of 14 on this paper:

https://www.bpho.org.uk/user/pages/0...Paper_2016.pdf

and the answers:

https://www.bpho.org.uk/user/pages/0...tions_2016.pdf
Could you explain what you've thought about them so far? We can help you understand, but you need to tell us your thought process.
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ThunderBeard
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(Original post by Plagioclase)
Could you explain what you've thought about them so far? We can help you understand, but you need to tell us your thought process.
Thanks for replying and yeah, course.
Number 7 : Actually I just got this one, thanks so much. You equal gpe and kinetic energy cancelling the mass, getting v^2= 600 (I think)
13c): I got 400 million = h p g
so 400 000 000 = h * 80 000 *10
so h = 500?
but the answers say 5000?
13f) I worked out the cross sectional area 1.2*10^-3
and did force/cross sectional area for 3*10^7 as pressure
then I did P= hpg so h= 422m?
14) I dont know how to do the first part nor what it is asking for?
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Plagioclase
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(Original post by ThunderBeard)
Thanks for replying and yeah, course.
Number 7 : Actually I just got this one, thanks so much. You equal gpe and kinetic energy cancelling the mass, getting v^2= 600 (I think)
13c): I got 400 million = h p g
so 400 000 000 = h * 80 000 *10
so h = 500?
but the answers say 5000?
13f) I worked out the cross sectional area 1.2*10^-3
and did force/cross sectional area for 3*10^7 as pressure
then I did P= hpg so h= 422m?
14) I dont know how to do the first part nor what it is asking for?
13c)
All correct, except you've misread the density! Reread the question

13f)
I'm a bit confused as to what you're doing, the basic equation is  \sigma_{max} = \frac{F_{cable} +F_{elevator}}{CSA}. You know all values apart from F_{cable}, so solve for that, and then solve for the length.

14)
The first part of the question is simply asking you to verify that this equation for the wavespeed is correct. For example, if c=\sqrt{gd} then for a depth of 0.5cm, we end up with a wavespeed of ~0.22 m/s, which works well with the average time recorded to traverse the ripple tank length of 0.7m.
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ThunderBeard
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(Original post by Plagioclase)
13c)
All correct, except you've misread the density! Reread the question

13f)
I'm a bit confused as to what you're doing, the basic equation is  \sigma_{max} = \frac{F_{cable} +F_{elevator}}{CSA}. You know all values apart from F_{cable}, so solve for that, and then solve for the length.

14)
The first part of the question is simply asking you to verify that this equation for the wavespeed is correct. For example, if c=\sqrt{gd} then for a depth of 0.5cm, we end up with a wavespeed of ~0.22 m/s, which works well with the average time recorded to traverse the ripple tank length of 0.7m.
Thanks so much, i fell asleep and then didn’t check. Sorry.
I have understood it all, thank you.
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