Recent content by sp1974

yeah its picky. I have put in 1.7, 1.71 1.705 and nothing. It does it everyonce in a while. I show teacher and all will be good. Thanks!

ok I figured this one out. I had my numbers flip flopped my numbers when subtracting

so if I am understanding 1/26 + 1/-1.6 = 1/f and I should come up with -1.705 cm? It looks right but thinking the computer is being picky like my other answer

Ok thanks. I know threwout the two semisters rounding has never been a factor but will try. Worst case if I use all my attempts I show teacher and get the 1 point for the problem. Thanks!

I also did it as 1/56 = 1/68 + 1/do and the computer didnt accept that on as well

I figured because the focal point was 56 and it says 68 behind it so I thought it would be f distance + the 68

The homework problem is "An object 26.0 cm in front of a certain lens is imaged 1.60 cm in front of that lens (on the same side as the object). What is the focal length (in centimeters) of this lens" 1/f = 1/di + 1/do I tried solving based on having distance of object (do) and...

The homework problem is "A sharp image is located 68.0 mm behind a 56.0 mm focal length converging lens. Calculate the object distance in millimeters" 1/f = 1/di + 1/do I tried solving based on having focal (f) and distance of image (di) so 1/56 mm = 1/(68 mm + 56 mm) + 1/do I...

The problem reads "What is the focal length (in meters) of a -4.2 diopter lens" The formula equation I have is = 1/f to get the diopter power -4.2 = 1/f and solve for f I get -.2381 This is a homework problem where the solutions are entered on line. I entered the solution above...

Yes I realize that now. I was over thinking the problem. It would be Vs/Vp=Ip/Is I believe. At least I know it would be 10/170 = .058A Thanks for coming back to clearify for me.

How about today? Can anyone offer some helpful hints to point me in the right direction?

All I know is the Voltage and that there is no power loss... I am looking for current

Ohm's Law?

I am having a problem with the following problem: A step-up transformer increases 10V to 170V. What is the current in Amperes in the secondary as compared to the primary? Assume 100% effiiency. The only formula I can see that has Amps in it is the turn ration...

sweet! 2.86V thanks!