30 replies to this topic

[ilovepolkadots]

pre-lab
the TA posted it last night and i have been in class since 8am
(x.X!)

it is due in 1.5 hours
please help me D:
(NC compensation will once again the given for those interested, or NP)









(#1) straight line

Quote

If the I-V characteristic is a straight line, as in Fig. 1, then we say that the piece of conductor satisfies Ohm's law: V = IR, where R is a constant defined to be the resistance and has units of volts/ampere, or Ω (ohm).

(#2)
part a :: voltage - current - resistance

Quote

In the process you will learn how to use the multimeter to measure voltage, current, and resistance.

part b :: voltage - current - resistance
(based on title headings/lab procedure)

(#3) parallel

Quote

To measure the voltage across an object such as a resistor, the voltmeter is connected in parallel with the object as shown in Fig. 6b. Voltmeters have very large resistance so that only a small portion of the circuit's current will be diverted through the voltmeter.

(#4) series

Quote

To measure the current flowing through an object such as a resistor, the ammeter is connected in series with the object as shown in Fig. 6a.

though i am not sure the WHY for numbers 3 and 4
side note:: the why question is the same for #s 3 & 4


just need 5 & 6

did it
had to find the stupid accuracies for this instrument online since it is not provided in the lab manual

x.X!!!

Attached Files

•  Lab 2.pdf   322.41K   2 downloads

Edited by ilovepolkadots, 07 February 2012 - 08:38 AM.

[ilovepolkadots]

Lab !



(^) i need help with these

[ilovepolkadots]

(^)
for the first part i simply used the thin lens equation
1/p + 1/q = 1/f

q = distance of image from lens
p = distance of object from lens
f = focal length



it does not indicate if you have the number correct but the sig figs wrong sadly
and this program is insane with the sig fig rules so who really knows what is going on with that

[nymh]

I'm working on this problem dots, but in the meantime I just wanted to say that that is a curiously weak cornea
The average corneal power is 44 D
This cornea is only 13.3 D

Ok I have a kind of convoluted way of going about this problem because all of these fancy focal lengths and p's and q's make it hard for me to understand
So I just converted everything to diopters and went about it like we actually do in the optics world (admittedly probably not at all what they wanted you to do but whatever)

Cornea f = 7.5 cm = 13.3 diopters (D)
vertex = 0, Effective diopters (De) = 13.3

Lens f = 15 cm = 6.6 D
vertex = 6cm
De = 6.6/1+(.06)(6.6) = 10.93

Total power at cornea = 13.3 + 10.93 = 24.23 D
Then we can convert this back to focal length and just do your lens equation again
So 100/24.23 = f = 4.127
1/q + 1/33 = 1/4.127
1/q = .212
q = 4.717

That's what I get, maybe they want the q to be a negative number? The wording is kind of confusing

[ilovepolkadots]

the answer was 6
=/

and i am still not sure how that came about
x.X!

but i love you nymh


& there is another one like this but opposite
x.x that i need to post



got that one
it was 15cm


i just cannot draw the picture correctly
x.X!!

[ilovepolkadots]

what's this?!
another fun physics problem!!!





#25.


i only have two submissions left for this one
D:

and for the di i just got :: 8.3504
rounded to 8.4

can anyone verify this?



for the thin lens equation it is supposed to be inverse 1/di1
as the equation reads:: (1/f) = (1/do) + (1/di)

do = distance of object from lens
di = distance of image from lens
f = focal length

it was wrong =[

if anyone is interested i was putting the -8 cm on the wrong side
the do2/di1 was supposed to be at 8 + 5.1 thus making it 13.1cm
x.X!

Edited by ilovepolkadots, 20 March 2012 - 02:00 PM.