30 replies to this topic

[ilovepolkadots]

(^) with that being said - i know you all want to help me with my Physics <3
maybe even help me understand how it works !

Spoiler

if that doesn't work out we will chalk it up to magic
and/or i will go monty python on your heretic ass

alright so lets get to it !

Quote

Lab 1 - Electric Fields and Potentials: PreLab

so the wee bit of a problem with getting started on this one is that i have no lab manual
therefore some of these cannot be answered yet O.O it is due tomorrow Jan 24th @ 1PM EST (USA)


manual for this section attached :3
so i am going to start trying to work these out
this post has now become super preemptive
but i am 100% sure i will need help before the end

Attached Files

•  manual.pdf   301.59K   4 downloads

Edited by ilovepolkadots, 23 January 2012 - 03:40 PM.

[Hydrogen]

What do you need help with? If you had a specific question or a specific area which you didn't understand, perhaps people could be more of a direct help.

[iargue]

Yeah. Can you provide what the experiment will be or anything? The questions all rely on some knowledge that we do not have of a project.

[ilovepolkadots]

that unfortunately is the main issue
the bookstore is closed so i have no access to a lab manual
=/ so i suppose this will have to wait

i was assuming i was just missing something
but apparently there is no progress to be made without some additional information

[Hydrogen]

Call/Email/IM/Text a friend? Acquire the lab manual somehow? I'm sure you can find something on the Internet to help you out until you can get to the book store

[ilovepolkadots]

prelab went great = 100%


now onto the homework D:

Edited by ilovepolkadots, 26 January 2012 - 11:16 AM.

[ilovepolkadots]

(

Quote

^) just these four that i need help on
i will also be trying to work them out - and then to the physics help center around 1pm (when it opens)

:3!

did one
xD now down to three!





(^) alright so this one is just geometry

E = - change in V / change in distance


the equipotential lines (shown) will be perpendicular to the electric field lines

blergh - the angle is 243 degrees counterclockwise from the +x axis
but i did not get that on my own and seriously cannot figure this one out x.X!

okay did the equation one
x.X
i was not using the diagram properly =/

now i am just down to those last two

done :3!
met up with some group members and i was able to become somewhat learned in the ways of physics
!


~(close til next time)~

Edited by ilovepolkadots, 26 January 2012 - 11:26 AM.

[ilovepolkadots]

another question on physics !



the electric field travels from more positive (12v) to more negative (0v)
so from the inner circle to outer-ring around it


(^) concentric circle if you need help
the voltz refer to the charge given to the ring/circle lines
inner = 12v and outer = 0v
:3!

[Sweeney]

Your optimism is remarkable!

[ilovepolkadots]

Sweeney, on 29 January 2012 - 08:41 AM, said:

Your optimism is remarkable!

;3
when it takes you an hour to do each problem its either: kill yourself or be crazily positive
and since i am happy with my existence, i chose the latter




(^) i am also unsure of this one =/

[ilovepolkadots]

another physics problem !
i am mainly having trouble with the conversions/units on this one
xP!

R = p * (l / A)

R = resistance
p = resistivity of metal (copper) = 16.7 nΩm
l = wire length
A = area of wire


LENGTH:: 6.1 mi = 9 816.9984 meters

(.55/2) millimeter = 0.000275 meters RADIUS
area of a circle = pi * r^2

R = (16.7 nΩm) x [9816.9984 meters / (pi * 0.000275 ^2)]
R = (16.7 nΩm) x 413202994.16 inverse meters
R = 6900490002.5nΩm^2

not the right units at all
o.0
and not surprisingly it was not the correct answer

Quote

Wiki:: The SI unit of electrical resistivity is the ohm⋅metre (Ω⋅m). It is commonly represented by the Greek letter ρ (rho).

Quote

nano (n) 10e-9

i did it !
for anyone who was wanting to check it out....



(1.68e-8Ωm)[9816.9984m / pi * (0.000275m)^2]
= 694.18 Ω = .69 kΩ

Edited by ilovepolkadots, 31 January 2012 - 03:26 PM.

[ilovepolkadots]

Quote

(^) okay this one is due in 3 hours
there are four guesses right
i literally cannot figure this one out
the first person to give me the right answer can have their choice of some NC items i am selling + some i have not listed because i am lazy
not into NC? we can talk about some items you might like :3

Quote

(^) this person had the same question but different numbers (still i cannot figure it out)
i also went to the physics help center where i wasted an hour of my time with a TA and accomplished nothing :C

{side note for the last question that is now shown}
VA-VB =I3*R3
(^) this is for me to hopefully be able to simply plug in later ;3

Quote

(^) another similar problem with different numbers
also cannot work it out =/

[ShadowLink64]

Using the principle of superposition for the first one..... Should take me no more than 10 minutes.

[ilovepolkadots]

ShadowLink64, on 02 February 2012 - 04:55 PM, said:

Using the principle of superposition for the first one..... Should take me no more than 10 minutes.

awesome
i will continue my futile effort to crunch numbers and await your post

[Kepster]

First off, we will use Kirchoffs rules on the two simple loops. We will set the sum of the rises and falls of Voltage equal to zero.
As a simplified explanation, let C1 and C2 refer to the currents from the left and right batteries respectively.

That is,

12 - 4.1(C_1) - 2.1(C_1+C_2) - 9C_1 = 0
which simplifies to
12 = 15.2*C_1 + 2.1*C_2

also,

12 - 6.1*C_2 - 2.1(C_1 + C_2) = 0
which simplifies to
12 = 8.2*C_2 + 2.1C_1

Now solve the second equation for C_1, getting
C_1 = 5.7143 - 3.905*C_2

Now substitute that value in for C_1 in the first equation:
12 = 15.2*(5.7143 - 3.905*C_2) + 2.1*C_2

Next solve for C_2 and find that C_2 =1.3074

Plug that back into our expression for C_1, and find C_1 = .6087

SO, the current across resistor 1 is going to be just that from the left battery: .6087
The current across resistor 2 should be just that from the right battery: 1.3074
The current across resistor 3 is the sum of these two: 1.9161
and the current across resistor 4 is just from the left battery: 1.3074

Edited by Kepster, 02 February 2012 - 05:00 PM.

[ShadowLink64]

Finished, but using mesh analysis. I believe that Kepster has the currents flipped, since they found C1 and C2 to be the same as what I did, but used left and right incorrectly in the end explanation, lol. << Kepster fixed it.

This is what I got:

Current from left battery = C1 = 0.6088 amps
Current from right battery = C2 = 1.30749 amps

Therfore, I1 = I4 = 0.6088 amps, I3 is the sum of these two (1.9163 amps), and I2 is 1.30749 amps.

To check this, the sum-of-voltages in a loop should equal zero. Therefore:

Left-branch:
12 - I1*R1 - I3*R3 - I4*R4 = 0
12 - (0.6088*4.1) - (1.9163*2.1) - (0.6088*9) = 0 (good)

Right-branch:
12 - I2*R2 - I3*R3 = 0
12 - (1.30749*6.1) - (1.9163*2.1) = 0 (good)

Ta-da. Give Kepster the NC, as I won't do shit with it. xD

[nymh]

Yay Slinky gets the NC

[ShadowLink64]

nymh, on 02 February 2012 - 05:11 PM, said:

Yay Slinky gets the NC

Nope, I'm giving it to Kepster, if our answers are right. I don't play Neopets.

[nymh]

ShadowLink64, on 02 February 2012 - 05:12 PM, said:

Nope, I'm giving it to Kepster, if our answers are right. I don't play Neopets.

Oh wait why did I totally overlook Kepster whoops

[Kepster]

Sorry, I didn't mean to steal any thunder. I saw your post only after I had posted mine ...

[nymh]

The fact that these were Kepster's first posts on the forum make me so happy

[ilovepolkadots]

Quote

.6087, 1.3, 1.9, 1.3

(^) wrong

[Kepster]

My guess is: It says specifically to express your answers in two significant figures.

That would make the answers:

.61, 1.3, 1.9, 1.3

EDIT: No, shoot. I was wrong, that's a relic from my first mistake. The fourth resistor is indeed from the left battery, therefore it should also be .61

Edited by Kepster, 02 February 2012 - 06:28 PM.

[ShadowLink64]

Kepster, on 02 February 2012 - 06:18 PM, said:

My guess is: It says specifically to express your answers in two significant figures.

That would make the answers:

.61, 1.3, 1.9, 1.3

EDIT: No, shoot. I was wrong, that's a relic from my first mistake. The fourth resistor is indeed from the left battery, therefore it should also be .61

Yes, I was about to say. xD Thanks.

0.61, 1.3, 1.9, 0.61

[ilovepolkadots]

correct !
it is a friends birthday so we are going to do wine
check out my thread pick out some stuff you like
and tomorrow i will screenie the other UFT NC i have and you can pick from that :3!


thanks again everyone :3