OCR Physics A G484 Jan 2012 - The Newtonian World

I said so that the radius of the orbit is the right value to give an orbital period of 24 hours.

I thought have an equatorial orbit ensured that the satellite was geosynchronous? A geostationary satellite is merely a special case of a geosynchronous satellite.

Reply 841

Stevo F

18

Original post by sweetascandy

Guys you're all gunna get 10 warning points (like I got yesterday :cool:

)

why would we? u can discuss ocr exams as soon as they finish

Reply 842

livvyyy

For the points plotted on the graph. Did anyone put all 3 on equilibrium :O

Reply 843

Stevo F

18

Original post by livvyyy

For the points plotted on the graph. Did anyone put all 3 on equilibrium :O

almost everyone did yes, its right

Reply 844

Shack93

6

Original post by sweetascandy

Guys you're all gunna get 10 warning points (like I got yesterday :cool:

)

I got 10 warning points for discussing c2 exam 20 mins after it finished. :cool:

But this ones fine I think :smile:

Reply 845

Stevo F

18

c2=edexcel "For Edexcel: morning exams may not be discussed until 12:00 midnight (BST) and afternoon exams until 4:30 AM (BST)." ocr u can discuss straight after

Reply 846

wig44

1

Original post by godkid308

You cannot be penalised for TOO MANY significant figures. Only too little. I asked my teacher this many times, and he said they test your precision in the practical assessments, not the exam. So even if you gave the answer as an exact answer (fraction) you would still get the mark. Only if you put it to something stupid like 1 sf then you lose marks.

Thank you. I hope that's true, if I can be bothered I might look for some examples of the mark scheme seemingly penalising answers with a higher degree of accuracy than expected (or perhaps I'm interpreting them wrong/recalling them incorrectly).

As for the power dissipated to offset the 0.25N damping force, I got 2.16/2.2N like others did. My original reasoning was that (using E = W*D and P = E/T) the distance was 2* the amplitude as you would not need to provide energy when the object was moving towards the equil. point, as it would now be using potential energy instead of kinetic energy. Now that I think about it though, 4* seems attractive too - both values requie you to assume something and we can only wait for someone like teachercol (please? :colondollar:

) to post up solutions.

Reply 847

Stevo F

18

Original post by wig44

Thank you. I hope that's true, if I can be bothered I might look for some examples of the mark scheme seemingly penalising answers with a higher degree of accuracy than expected (or perhaps I'm interpreting them wrong/recalling them incorrectly).

As for the power dissipated to offset the 0.25N damping force, I got 2.16/2.2N like others did. My original reasoning was that (using E = W*D and P = E/T) the distance was 2* the amplitude as you would not need to provide energy when the object was moving towards the equil. point, as it would now be using potential energy instead of kinetic energy. Now that I think about it though, 4* seems attractive too - both values requie you to assume something and we can only wait for someone like teachercol (please? :colondollar:

) to post up solutions.

he said he's going to do it this evening, give him a break :P

Reply 848

H_Rattan

Original post by Moiraclaire

they wew all on the line x=0 :smile:

i don't think they were to be honest - the first two were...but the last one was when potential energy is a minimum...isn't potential energy a minimum when the oscillation is at its lowest ie. at -0.4m?

Reply 849

wig44

1

Original post by Kingz

Referring to the power question..it said the mean force was 0.25N, and I've forgotten what the frequency was, but what I did was work out the period (1/frequency) and then divided 0.25N over the period which have me the answer 600N (sounds right because it's a a full number) but I probably got that one wrong!!

Dividing the force by the period would not give you the correct unit of power, so we know that's wrong. Sorry! :frown:

Reply 850

sweetascandy

OP

18

Oh okay, my bad! :colondollar:

Reply 851

teachercol

9

Original post by Stevo F

he said he's going to do it this evening, give him a break :P

I have a job teaching A level Physics too :wink:

Reply 852

sweetascandy

OP

18

Just ONE more left now before it's all OVER!

Reply 853

t01

1

Original post by H_Rattan

i don't think they were to be honest - the first two were...but the last one was when potential energy is a minimum...isn't potential energy a minimum when the oscillation is at its lowest ie. at -0.4m?

For SHM I thought p.e is lowest when k.e is at max which is when x=0.

Reply 854

just george

Original post by Stevo F

almost everyone did yes, its right

I think the max velocity had to be onthe equilibrium point after negative displacement going into positive displacement though? otherwise it would be minimum velocity rather than max velocity?

Original post by Sheheardvoices

I thought have an equatorial orbit ensured that the satellite was geosynchronous? A geostationary satellite is merely a special case of a geosynchronous satellite.

Isnt it because to the only way for it to stay above the same point on earth is for it to be orbiting above the equator?

Reply 855

Megatron94

Original post by Stevo F

almost everyone did yes, its right

Why is that right I thought that it said oscillating mass i thought P.E would be minimum below the equilibrum?

Reply 856

Stevo F

18

Original post by H_Rattan

i don't think they were to be honest - the first two were...but the last one was when potential energy is a minimum...isn't potential energy a minimum when the oscillation is at its lowest ie. at -0.4m?