You are Here: Home >< Physics

# Unit 5 Physics Edexcel A2 and Edexcel IAL

Announcements Posted on
Last day to win £100 of Amazon vouchers - don't miss out! Take our quick survey to enter 24-10-2016
1. i know u cant really do this for physics but any predictions??
2. (Original post by SuruthiG)
i know u cant really do this for physics but any predictions??
No clue ! Predicting Edexcel is hard as hell.
3. How do you do this question?
4. (Original post by target21859)
How do you do this question?
It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
5. (Original post by Moe21)
It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
ah thanks makes perfect sense
6. Can someone please explain question 3 to me I just don't get it at all?
https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf
7. (Original post by Moe21)
Can someone please explain question 3 to me I just don't get it at all?
https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf
Well you know that the change would be a decrease, so you can immediately discount C and D.
I did it by saying g mars = GM/r^2. Since the height of Olympus Mons is 0.6% of r, then the height is 0.006r, hence g mountain = GM/(r+0.006r)^2 which is GM/(503r/500)^2
Since GM and r are constants, you can simply approximate g Mars to be 1 and g Mountain to be 1/(503/500)^2 which is approximately 0.9881
0.9881-1 is approximately - 0.0119 which is equivalent to - 1.19% so the answer is B
8. (Original post by Moe21)
It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
But if the question was about a pendulum then the answer would be 1Hz because
?
9. ?
10. (Original post by PhysicsIP2016)
Well you know that the change would be a decrease, so you can immediately discount C and D.
I did it by saying g mars = GM/r^2. Since the height of Olympus Mons is 0.6% of r, then the height is 0.006r, hence g mountain = GM/(r+0.006r)^2 which is GM/(503r/500)^2
Since GM and r are constants, you can simply approximate g Mars to be 1 and g Mountain to be 1/(503/500)^2 which is approximately 0.9881
0.9881-1 is approximately - 0.0119 which is equivalent to - 1.19% so the answer is B
I get it now thank you!

But if the question was about a pendulum then the answer would be 1Hz because
?
Yea that's right
11. (Original post by candycake)
Sorry, there isn't one - unit 5 was first examined in June 2010.
Thats really strange because there is a unit 4 jan 2010 :')
12. (Original post by CasioGamer98)
?
Flux= Luminosity/ 4pid^2 so we know that Luminosity has the unit W and d has the unit m^2 so

F= W/m^2. Try simplifying the unit W as this is a derived unit into J/s and then J into Nm. You will get the answer as A.
13. Not sure what you're meant to do here.
14. (Original post by target21859)
Not sure what you're meant to do here.
is 6.67x10^24 molecules correct?
15. (Original post by CasioGamer98)
is 6.67x10^24 molecules correct?
Yes thanks I think I understand now.
17. (Original post by apocolyptic)
The answer is B. The density of the universe is not accurately known to us due to dark matter being undiscovered till now. If Dark Matter is present, then the actual density would be less than the critical density so this means that the universe will keep on expanding i.e it is an open universe because there is continuous expansion taking place.
18. (Original post by sabahshahed294)
The answer is B. The density of the universe is not accurately known to us due to dark matter being undiscovered till now. If Dark Matter is present, then the actual density would be less than the critical density so this means that the universe will keep on expanding i.e it is an open universe because there is continuous expansion taking place.
How will the actual density be less if dark matter is present?
19. (Original post by apocolyptic)
How will the actual density be less if dark matter is present?
I'm not exactly sure myself but I recall reading somewhere this.(Idk whether my logic is right though :/ )
20. (Original post by sabahshahed294)
I'm not exactly sure myself but I recall reading somewhere this.(Idk whether my logic is right though :/ )
but dark matter increases the mass of the universe? Surely it would increase the density by that logic?

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 29, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources