[Challenge] - Evaever's Physics questions

[Whatlol]

one force is 20N, the other is -15N so the net force is 5N on the 20N end of the rope

but the force along decreases linearly until the other end where it is 0, so at 0.4m along it will be a fraction of the total net force:




?

i think tension is constant throughout a rope/string (at least in this situation)

[Whatlol]

what are the questions marks meant to indicate... order of magnitude???

[man0005]

would it just be the energy required to push it up?

[m@tty]

For this I'd say calculate the difference in gravitational potential between the initial position of the centre of mass and when it is standing on an edge. So figure out half the diagonal, subtract the original centre of gravity (0.3m)



Though I'm probably wrong again, as usual

[m@tty]

Units perhaps.

X -> Y is really tricky.. I count four possible paths xD

[werdna]

Don't be so self-deprecating, nothing is ever noob on this forum.

Except me, that is. Sometimes I feel like I am a total noob NB: last edit ;p

[Whatlol]

well you know the 50 and 125 ohm resistors are in series... also the 80 and 200 ohm resistors..

so you essentially have a 175ohm resistor, 100 ohm resistor and 280 ohm resistor all in parallel

[m@tty]

No werdna! You're not a noob.

[werdna]

Thanks, but I think you are the noob. We can all see who edited my post last. :coolsmiley:

taiga rules

[evaever]

For this I'd say calculate the difference in gravitational potential between the initial position of the centre of mass and when it is standing on an edge. So figure out half the diagonal, subtract the original centre of gravity (0.3m)



Though I'm probably wrong again, as usual

congrats

[evaever]

what are the questions marks meant to indicate... order of magnitude???

ohms

[evaever]

one force is 20N, the other is -15N so the net force is 5N on the 20N end of the rope

but the force along decreases linearly until the other end where it is 0, so at 0.4m along it will be a fraction of the total net force:




?

i think tension is constant throughout a rope/string (at least in this situation)

gone cold

[Whatlol]

what are the questions marks meant to indicate... order of magnitude???

ohms

ok well i think the answer is just 52 ohms then

[schnappy]

35 N?

I'm fairly confident the tension in the whole string is constant.

Actually if there are forces applied at both ends, is there any tension at all? You've said the string is just sitting there you could just say there's a net force of 5 N (left? Lets say left) then say F=ma, find the acceleration of the entire string... and say 0 tension.

[Shark 774]

From the graph for question 9 in heinemann questions 3.4, if you compare the gravitational potential energy for the 20,000 satellite at 7x10^6m from the earth's centre and 9x10^6, you get:
Ug @ 7x10^6m = mgh = 20,000 x 8 (from graph) x 7 x 10^6 = 1.12 x 10^12
Ug @ 9x10^6m = mgh = 20,000 x 5 (from graph) x 9 x 10^6 = 0.9 x 10^12
From this the gravitational potential energy is higher in the lower orbit, but how can this be right?!?!?!?!

HELP PLEASE!!