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Your overall question about internal ranks was very similar to mine when I asked D94 about it, and he put it very nicely; (Just in case you get confused the assessment mark percentage determines our ranks) D94: As for your question, the 'moderated' assessment marks are definitely based on...

Please evaluate; %5Cint_%7B%5Cfrac%7B-3%282%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B4%7D%7D%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cfrac%7B1%7D%7B%28%5Cfrac%7B9%7D%7B4%7D-x%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%7D%20dx

Wow very nice trebla Pls tell us the origins of that neat idea

Use the formula S=n/2(a+l) for the bottom question answer should be 10 trips, 1km apart

Yeah it was in my 2U exam too

Were you trying to question her?

I can't see the Pythagoras thereom in the question Also Ty for reply

A man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is a further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr, how far from A should he land?

What year were you in when you did this exam anomalous?

Re: HSC Physics Marathon 2014 Worth 2 marks

Yes please!

Re: HSC Physics Marathon 2014 YAY finally got it, scroll up for the rest of the working out Centripetal acceleration outwards: a=[(2pi(69911000)/35760)^2]/(69911000) =2.158292148ms^-2 Outwards Therefore net acceleration on the surface of Jupiter, if we take the outward direction positive...

Re: HSC Physics Marathon 2014 ffs im so bad pls share answers 2014ers

So for b and d, you have to use the log rule logabn=nlogab

Re: HSC Physics Marathon 2014 Oh so "r" in the centripetal acceleration calculation needs to be the equatorial radius of Jupiter? EDIT: Yes, r needs to be the equatorial radius of Jupiter So that means the centripetal acceleration = [(69911000/35760)^2]/778x10^9...

You have to use the log rule: alogab=b Proof should be in the theory section of the Cambridge book So for a) turn the negative log into positive by making whatever is in a) into a fraction For c) you can add the two logs in the power because they have the same base, then go on from there by...

Re: HSC Physics Marathon 2014 Length of day on Jupiter:9 hrs and 56 mins (35760s) Mass of Jupiter: 317.8x Mass of Earth (317.8x6.0x1024) Equatorial radius: 69911km (69911000m) Jupiters orbital radius :778 million km (778x109m) The NET acceleration due to gravity along Jupiter's equator...

Re: HSC Physics Marathon 2014 Oh wow, we won't be given the gravity formula? Where can I find online a list of formulas that will be given to us in the hsc exam? Also, that would mean I would have to derive the formula by equating the law of gravitational attraction and F=mg? But I feel like...

Re: HSC Physics Marathon 2014 The distance between the Moon and the Earth from their centres is 3.84x10^8 m. The moon circles the Earth once every 27.3 days and the mass of the Moon is 7.475x10^22kg. Calculate the Moon's linear velocity

Re: HSC Physics Marathon 2014 Using the g=(Gm)/r^2 formula and assuming that Earth's approximate mass is 6x10^24kg; g=(6.67x10^-11x317.8x6x10^24)/(69911000)^2 =30ms^-2(1s.f) Did I get the sig fig right?(did I even get the answer right lol) Btw, thanks for posting the thread