Distance between the rods??? It changes! (1 Viewer)

[yankyfly]

thats complete bullshit. the gradient is absolutely relevant; thats why they got u to draw the graph.

the reason they got you to draw the graph was to get the initial weight of the rod. From there you just had to sub a point into the equation.

 

[mlinger]

i got 2.5cm... (were u suppose to go from the gradient aye??? cos i didnt) i used a point that fitted on my line of best fit... i found the weight force in newtons wen the current was 0 and wen the current was 20 amps in the lower rod... i then subtracted them to give the force of attraction... then used the parallel lines formula..

 

[nittyc]

the reason they got you to draw the graph was to get the initial weight of the rod. From there you just had to sub a point into the equation.

How can you though? ur ratio of mass to amps is completely incorrect! look at the first point 0 and something else?!

 

[nittyc]

ah u did it a different why, u used change in--> which is the same, but i'm not sure about your method: look at the solution i posted a few up and comment.

 

[SL33pY]

arghh... i just used the formula... at 8 amps..
man scrwed up this question hardcore i did everything except minus the weight of the rod.. everything would be wrong after that...
dammit

 

[mlinger]

ah u did it a different why, u used change in--> which is the same, but i'm not sure about your method: look at the solution i posted a few up and comment.

what answer did u get??? because i recon both ways are correct...

 

[Benmc]

i got 2.5cm... (were u suppose to go from the gradient aye??? cos i didnt) i used a point that fitted on my line of best fit... i found the weight force in newtons wen the current was 0 and wen the current was 20 amps in the lower rod... i then subtracted them to give the force of attraction... then used the parallel lines formula..

Yeah thats what i did. Use a current of 8 Amps as my other amp tho but i calculated the weight force in newetons when the current was 0 and 8. Then used that force and substituted all the other values in to get a distance value of 0.026 m or rougly 2.6cm.

To calculate the force i used F=ma with a =9.8.

Hopefully thats around what everyone did.

 

[bboyelement]

i remembered i put down 0.02 metres .... it really depends on what point you chosen to work your calculation

and what gravity you used ...

 

[Rax]

Yeah thats what i did. Use a current of 8 Amps as my other amp tho but i calculated the weight force in newetons when the current was 0 and 8. Then used that force and substituted all the other values in to get a distance value of 0.026 m or rougly 2.6cm.

To calculate the force i used F=ma with a =9.8.

Hopefully thats around what everyone did.

Thats what I did, I used the 8 amps too.
I got around that answer 0.023m I think , I used g as 9.8.

Atleast this question made mild sense, and yes I am sure the only reason for the graph was the mass using your line of best fit, because I dont know what kg/Amps give you at all.......

But damn I stuffed that test

 

[astrobunny]

2.21

 

[gamecw]

2.21

cm? then yea

 

[theodore]

i got 2.3cm. what did every1 else get?

i got around there, 2.3 or 2.5cm can't remember exactly

 

[raikkonen_rulez]

I didn't really get the question... I subbed multiple values of current and the mass, but the distance changes. In the end I subbed in the first value which was 2.8A cause I thought it would be the closest to the right value. But I don't think they're gonna give me marks for that. Maybe 1 for writing the formula.

Btw, I got 2.48* 10^-2 (or -3, I seriously can't remember)

 

[dunno04]

i got 23mm
Which is 2.3 mm

and i'm pretty sure i am right
....
Pretty....

 

[doraemon]

i got 23mm
Which is 2.3 mm

edited.

 

[charismo]

the mass doesnt change... the weight force does due to the increased current flowing throwing the lower rod.. increase in current = increase in attractive force towards the upper rod thereby counteracting the downward force (of gravity) acting on the scale and thus reudcing its apparent weight.. i think i had a a 2 and a 3 somewhere in my answer but i have this nagging feeling i put the decimal place in the wrong spot and wrote 0.23 m instead of 0.023m

oh and also you work out the mass by extended your line of best fit so it touches the y-axis and then by observation i think its 0.54885 or something

 

[Wackedupwacko]

mass doesnt change to work it out... just draw ur graph to touch the y axis as stated. then i think the distance is worked out by

F=mg = (formula the I1I2 which i cant be bothered to write)

this occurs when the balance reads 0kg which you can work out from the graph... mine suggested abotu 4392 A needed for it. chuck that as your I2 in ..... got about 0.2m or 0.02 cant remmeber which atm..... head spinnning

 

[Aznduud3]

YES!!! i got 0.02...m as well!!!
im assuming your smart so i must be corect too!!

i used a random x,y component from MY line of best fit, NOT the one they gave you in the table.

 

[chucknthem]

I think I screwed up, I got like 2.3x10^-4 metres lol
hopefully I'll only loose 1 mark for it since i showed my work :rofl:

 

[dunno04]

U are right chuck
hahah
=D