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October 03, 2025, 12:51:56 am
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anonymous12

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Re: suggested exam solutions
« Reply #30 on: June 11, 2008, 07:33:23 pm »
For question 17 can you do.

This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)

Use R as 3.83*10^6
      g=2.91

then I got T=2.7*10^10

am I wrong?
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Mao

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Re: suggested exam solutions
« Reply #31 on: June 11, 2008, 07:40:10 pm »
Quote from: chid on June 11, 2008, 06:50:17 pm
are you sure it wasn't XZ. That's what I thought all this time.

that was the first one

the second one was on YZ, where I was wrong [i am pretty sure i'm wrong.... i hope i double-wrong myself tho :P ]
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4xrp

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Re: suggested exam solutions
« Reply #32 on: June 11, 2008, 08:32:36 pm »
Quote from: anonymous12 on June 11, 2008, 07:33:23 pm
For question 17 can you do.

This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)

Use R as 3.83*10^6
      g=2.91

then I got T=2.7*10^10

am I wrong?

How can you do that? You don't have g do you...?

I used T=[(4pi^2R^3)/(GM)]^(1/2)

Thats seems to get the same answer as Keplers law.
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2007
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2008
English 42 ----> 42
Literature 42 ----> 43
Methods (CAS) 32 ----> 39
Physics 41 ----> 43.8
Economics 38 ----> 40

ENTER: 96.00
(Hopefully its very conservative!!)

Mao

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Re: suggested exam solutions
« Reply #33 on: June 11, 2008, 08:37:11 pm »
Quote from: anonymous12 on June 11, 2008, 07:33:23 pm
For question 17 can you do.

This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)

Use R as 3.83*10^6
      g=2.91

then I got T=2.7*10^10

am I wrong?

that is wrong.

your formula is suggesting that:

, where

that'll create , which is really a bad transposing from keplar's law.


what your formula should have read is

[ps btw 2000 posts, woot, lol]
« Last Edit: June 11, 2008, 08:46:32 pm by Mao »
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4xrp

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Re: suggested exam solutions
« Reply #34 on: June 11, 2008, 08:52:57 pm »
Quote from: Mao on June 11, 2008, 08:37:11 pm
Quote from: anonymous12 on June 11, 2008, 07:33:23 pm
For question 17 can you do.

This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)

Use R as 3.83*10^6
      g=2.91

then I got T=2.7*10^10

am I wrong?

that is wrong.

your formula is suggesting that:

, where

that'll create , which is really a bad transposing from keplar's law.


what your formula should have read is

[ps btw 2000 posts, woot, lol]


So was my way right? I'm assuming it was because I got a similar answer...
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2007
Revolutions 39 ----> 41
2008
English 42 ----> 42
Literature 42 ----> 43
Methods (CAS) 32 ----> 39
Physics 41 ----> 43.8
Economics 38 ----> 40

ENTER: 96.00
(Hopefully its very conservative!!)

Mao

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Re: suggested exam solutions
« Reply #35 on: June 11, 2008, 08:58:55 pm »
yep, yours is just keplar's law.

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4xrp

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Re: suggested exam solutions
« Reply #36 on: June 11, 2008, 09:04:30 pm »
Yeah...I knew that alright...lol
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2007
Revolutions 39 ----> 41
2008
English 42 ----> 42
Literature 42 ----> 43
Methods (CAS) 32 ----> 39
Physics 41 ----> 43.8
Economics 38 ----> 40

ENTER: 96.00
(Hopefully its very conservative!!)

anonymous12

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Re: suggested exam solutions
« Reply #37 on: June 11, 2008, 09:21:10 pm »
Fuck TSFX man. That's the formula they gave me. Shit!
« Last Edit: June 11, 2008, 09:28:29 pm by anonymous12 »
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Mao

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Re: suggested exam solutions
« Reply #38 on: June 11, 2008, 09:30:41 pm »
Quote from: anonymous12 on June 11, 2008, 09:21:10 pm
Fuck TSFX man. That's the formula they gave me. Shit!


you will find that they all do the same:

http://vcenotes.com/forum/index.php/topic,2645.0.html
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4xrp

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Re: suggested exam solutions
« Reply #39 on: June 11, 2008, 09:35:40 pm »
Quote from: anonymous12 on June 11, 2008, 09:21:10 pm
Fuck TSFX man. That's the formula they gave me. Shit!


I went to TSFX as well and got that same sheet...That formula isn't on it. You must have it confused with the one I used.
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2007
Revolutions 39 ----> 41
2008
English 42 ----> 42
Literature 42 ----> 43
Methods (CAS) 32 ----> 39
Physics 41 ----> 43.8
Economics 38 ----> 40

ENTER: 96.00
(Hopefully its very conservative!!)

dooopy

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Re: suggested exam solutions
« Reply #40 on: June 11, 2008, 10:10:02 pm »
hi every1, ive got a question abt the resultant force in the motion part, most of my friends ddint get P or R, they got O and soemthing else. Basically they did a vector diagram, and found a "new" force.
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porsca911

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Re: suggested exam solutions
« Reply #41 on: June 11, 2008, 10:17:37 pm »
JEBUS I KNOW ALL THE ONES I GOT WRONG NOW.. I SWEAR I FAILED MOTION
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dcc

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Re: suggested exam solutions
« Reply #42 on: June 11, 2008, 10:24:22 pm »
As the car is undergoing circular motion, the net (or resultant, however you want to term it) force MUST be acting towards the centre of the circle (i.e. the components of the car's acceleration are always perpendicular to the cars direction of velocity, so the speed of the car remains the same, but the direction constantly changes).
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dooopy

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Re: suggested exam solutions
« Reply #43 on: June 11, 2008, 10:35:34 pm »
i hope so as well =D, other wise i am GGed
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Pop

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Re: suggested exam solutions
« Reply #44 on: June 11, 2008, 10:39:53 pm »
I did the tsfx classes, got the same sheet and wrote it out in the right hand corner of my cheat sheet. Used like i said, both the tsfx derived formula nad the conventional one later when i had time and got the same 7.2x10^3s
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