Oxbridge Physics Prelims Revision

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Reply 20

Bezza

If anyone wants shiny, he'll be at the library :p:

Reply 21

shiny

only if cookies were on offer :wink:

Reply 22

Willa

HELP! I know this is statistical physics but i've broken it down into something any physicist can me with:

Given that the escape velocity for Mars is 4700m/s, and that the average velocities for the following molecules is:

H2: 1900m/s
O2: 470m/s
CO2: 270m/s

Discuss whether or not these gases would be present as a Mars atmosphere.

How would I go about deciding what is the upper limit for a velocity to be considered as part of the atmosphere of a planet? Cos surely these gases arent part of the atmosphere on mars?

Reply 23

Paddy-Boi

Depends how exact you need it...
Found this:
http://www.astronomynotes.com/solarsys/s3.htm
As it's statistical physics, thought it may be of use

Reply 24

Willa

ok cheers, i think that helped give a bit of a qualitative explanation

also another question. I have a container that has a small hole in the top, and I fill the container partially with mercury, and hold the container at 273K. How can I calculate the density of the vapour in the container?

Reply 25

Willa

Chloe have you got answer to any of those questions I couldnt answer? I'm curious to find out how to do them!

Reply 26

LennonMcCartney

Hey Willa, sorry for no response on your thermodynamics stuff, it was the one (albeit rather large) bit of advanced physics I took no interest in and , and therefore have no knowledge about prior to lectures on the topic.

Reply 27

Hoofbeat

Willa

Chloe have you got answer to any of those questions I couldnt answer? I'm curious to find out how to do them!

Yup I have indeed. I have the answer to the EM question (according to my tutor) and I *think* I have the answer to the relativity question (I'll know for certain after my tute @ 4:15). I'll try and remember to post up the answers this evening...I should be working right now!

Reply 28

Hoofbeat

Hey, could someone who was in Devenish's Mech Revision Lesson on Monday explain why in his worked example of S02/1 we know that

V_{1}^{'2} = U_{1}^{'2}

Reply 29

Hoofbeat

EDIT: I notice that Devenish uses later the fact that U1' = 4u/5, but I thought it was 3u/5 as surely it's equal to 4u/5 - u/5 (ie. V1' - Vcm)

PS. I would have edited my above post but TSR's gone funny and won't let me! It doesn't have an edit or reply button at the bottom of my post and instead on the left says "Find all posts by Hoofbeat" and underneath that "Add Hoofbeat to your Buddy List" yet all the other posts on the page are normal! eek!

Reply 30

shiny

I informed pig about your funny post. He'll sort it out.

Reply 31

Hoofbeat

shiny

I informed pig about your funny post. He'll sort it out.

thanks...i don't understand why it happened :confused:

Reply 32

shiny

why has it got that funny thing instead of the sig? :confused:

Reply 33

Hoofbeat

pig

What do you want it changing to?

The layout's funny...I've attached a screen print of that post:

EDIT: Coming in a minute...

Reply 34

shiny

It's fixed! :biggrin:

Reply 35

Hoofbeat

shiny

Wheres the screen print? :confused:

Never mind, it's fixed now! I uploaded a screen print, but it wasn't clear, so tried again, then the file was too big blah blah blah!

Thanks Pig neways!

Reply 36

shiny

pig, we salute you! *salutes* :cool:

Reply 37

Bezza

hoofbeat, I just went through the em question with my tutor - basically you have to realise the current goes around the tube, and be careful when you work out the resistance as l is the circumference and the cross sectional area is l*dr. For the rod you integrate for all dr from r=0 to a

pig - as well as hoofbeats weird post, i have most of the tsr background as grey instead of white on the ukl theme

edit: it's back to white now as well as that post being normal - what went wrong?