The GPe questiona dn the mass and distanceH (1 Viewer)

[suchet_i]

guys i just think they whole thing was a trick paper .. thats y i didnt bother doing any of it.. my god told me to slaughter 5 goats and 2 pigs and meditate for 6 months and i will instantly get a 100 uai

 

[tauren]

i said that it is literally not possible as spacecraft cannot reach such distance and i drew a diagram saying how far spacecraft can travel into space

 

[tojaco]

ok
E = -GMm/r
1x10^6 x (10 000 - 20 000) x 10^3/-6.67x10^-11 = Mm
Mm = 1.499 x 10^23 kg^2
now Mm is constant BUT it asks for the next change of work ie from 20 000 to 80 000 not 10 000 to 80 000
thus E = -GMm/r
E = -6.67 x 10^-11 x 1.499 x 10^23/ (20000 - 80000)x10^3

E = 1.66666.67

E = 0.16 MJ

for the other question
F=kIIl/d

1/d = m/I x a/KIL
1/d = (gradient) x a/kIl
1/d = (0.5488 - .5465)/(0- 20) x (-9.8/2x10^-7x 50 x 2.6)
1/d = 43.346
d=0.02307m
d=23mm

i think those are correct.... the second one i thought i had wrong and i was checking it for half an hour at the end trying to work it out and i wasted half of my checking over time

 

[P_Dilemma]

Uh, there seems to be two questions being discussed here, so i'll put my answers for both.

The distance between the two 2.6m long wires, I got 0.23m (23cm)

The work needed to push the spacecraft that extra 60,000m, i used the equation on the data sheet: W=Fs (Work equals Force times Displacement)

Obviously then, F can be calculated as 100 with the given values, then you do the calculation:

W = 100 x 60,000
= 6,000,000
= 6MJ

Seemed simple enough. But maybe i'm wrong, it was, i duno, too simple?

-P_D

 

[dartm2]

i said that it is literally not possible as spacecraft cannot reach such distance and i drew a diagram saying how far spacecraft can travel into space

Ha ha.. lol yeah cause they will hiot the ceiling will all the stars painted on....

 

[Rax]

Yeah I got 6MJ but I am pretty damn sure thats wrong. I just went linear scale and wrote some crap.

I ended up with around 0.023..... for the distance, it was somewhere around there. Initially I had 0.0023 and went....na that is way too close, and realised that I used 5 A rather than 50.
Not too sure about that paper though

 

[HazamataX]

For the GPE question this is what I put:
KL says:
you know how it said that 1 MJ of work had to be done to get from 10000km to 20000 km?
KL says:
I just said that GPE at 20000 km - GPE at 10000 km was equal to 1 MJ
KL says:
so then -Gmm/r-(-Gmm/2r)=1 MJ, where r = 10000 km
KL says:
so -Gmm/r=2 MJ
KL says:
and at 80000, it's -Gmm/8r
KL says:
which = -Gmm/r * 1/8 = 2 * 1/8 = .25 MJ
KL says:
and so the work required to get from 20000 km to 80000 km was 1 MJ - 0.25 MJ
KL says:
= 0.75 MJ
KL says:
anything wrong with what I've done?

and for the rods question it said in the question that the upper rod had a current of 50A running through it. So I just took a point from the line of best fit and substituted it in. Force = weight of the rod - the force experienced then just rearrange the equation.

You're absolutely right theres no other way. I even tried the year 11 formula with FD. Btw, if u supply 2 answers, is that zero? even if one is correct?

 

[dereck]

for the GPE question i got 750000J. I first thought it was 6MJ but it seemed too simple.

so i used the values given to work out the constant mm in E=-Gmm/r--Gmm/r.


For the rod question, well, i cant remember what i did but i got 0.02 metres

 

[iamben]

For the GPE question this is what I put:
KL says:
you know how it said that 1 MJ of work had to be done to get from 10000km to 20000 km?
KL says:
I just said that GPE at 20000 km - GPE at 10000 km was equal to 1 MJ
KL says:
so then -Gmm/r-(-Gmm/2r)=1 MJ, where r = 10000 km
KL says:
so -Gmm/r=2 MJ
KL says:
and at 80000, it's -Gmm/8r
KL says:
which = -Gmm/r * 1/8 = 2 * 1/8 = .25 MJ
KL says:
and so the work required to get from 20000 km to 80000 km was 1 MJ - 0.25 MJ
KL says:
= 0.75 MJ
KL says:
anything wrong with what I've done?


i'm not sure but i thought that 0.25Mj is the gpe from 0 to 80000, not from 10000 to 80000, so deducting 0.25 from 1 gives you the gpe difference between 80000 and 20000 PLUS the gpe from 0 to 10000?????

 

[Wackedupwacko]

tried that at first and was like... how come i need less energy to move it 70000 than i do to move it10000.

basically i think for that question u need to use W = Fs (work = force x distance)

and then work out F from the given.... and then work out the 2nd part... ... i THINK not too sure though

for the distance of the rods i worked out the mass (duh took me a minute but then i realised all i needed was the graph). basically i took me a while to get my logic around it but i think i have it.

when the balance reads 0kg it means the force UP is equal to the force DOWN thuse
F=kI1I2l/d = mg

so whats I2 when balance is 0? work it out from the graph and it was about.... 4392A or somfin like that... then plug it all into the equation solve for d and i think i got 0.2m althoguht im not 100% sure how mnay zeros there are atm since ive had 2 exams and cant remember well at this moment

 

[Angry.student]

i stuffed the formulars and i dont know whether what iver done is right, but heres my explaination.

Gravity, like light, disperses inverse to the square.

So if it took 1MJ to move it from 10km 20km theirfor it must take 1MJ to move it from 20k, to 40k, and another one from 40k to 80k.

This gave me the answer that it took 2MJ to move the object from.

 

[yaseng1988]

Yep, I got the answer about x 10^3 too

 

[iamben]

tried that at first and was like... how come i need less energy to move it 70000 than i do to move it10000.

basically i think for that question u need to use W = Fs (work = force x distance)



Can anyone remember the exact wording for this question. I didn't think you needed to use the W=Fd equation, im sure its just a gpe difference.

Because of the minus sign in the gpe equation, -1Mj is actually less than -0.25Mj. Also as you get further away from the planet, the stregth of gravity decreases thus the magnitude of energy required to move something becomes less per unit length.

 

[_ShiFTy_]

Hmmm...i did everything right except i let 1MJ = 1000J, so my answer was 750J which would've been right if i sub'd in teh correct value at first :S

Is it possible to get 2marks here?

 

[suchet_i]

i did everything rite but i forgot to read the "more" bit so i answered energy from 10000km to 80000km instead of from 20000km .. how many marks outa 3 would i get?

 

[bboyelement]

i did everything rite but i forgot to read the "more" bit so i answered energy from 10000km to 80000km instead of from 20000km .. how many marks outa 3 would i get?

yeh my friend did that ... definitely 2 but, i think they might accept it as well so maybe 3

 

[meekzy89]

tried that at first and was like... how come i need less energy to move it 70000 than i do to move it10000.

basically i think for that question u need to use W = Fs (work = force x distance)



Can anyone remember the exact wording for this question. I didn't think you needed to use the W=Fd equation, im sure its just a gpe difference.

Because of the minus sign in the gpe equation, -1Mj is actually less than -0.25Mj. Also as you get further away from the planet, the stregth of gravity decreases thus the magnitude of energy required to move something becomes less per unit length.

it went:

an object is stationary in space and located at a distance 10 000km from the centre of a certain planet. It is found that 1.0MJ of work needs to be done to move the object to a stationary point 20 000km from the centre of the planet.

Calculate how much more work needs to be done to move the object to a stationary point 80 000km from the centre of the planet.



I TOTALLY FORGET WAT I DID... all i kno is that i had NFI wat 2 do. lol

[ps] did anyone else keep a spare exam paper?