Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
OpenStudy (anonymous):
ik thought it be 1/2 but its not;/?
OpenStudy (anonymous):
did you use \[R=\rho l/A\]?
OpenStudy (anonymous):
no i looked at this
OpenStudy (anonymous):
i tried 1, 1 isnt it
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
its okay :/
OpenStudy (anonymous):
use the equation rajathsbhat gave
OpenStudy (anonymous):
A= pi r^2 = pi/4 d^2
OpenStudy (anonymous):
i dont get what to plug in? i know length doubles? so L=2? and diameter quadrouples so d=4?
OpenStudy (zehanz):
If the length doubles, the resistance doubles also.
If the diameter gets larger, there is more "room" for the current, so the resistance decreases.
In fact, you have to look at the area of the wire when you cut it: it becomes a 4 times as wide circle, so the area becomes 16 times bigger. This means the resistance is multiplied with 1/16.
Now you have to combine the two effects.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
so 2 x 1/16= 2/16!?
OpenStudy (anonymous):
@ZeHanz
OpenStudy (anonymous):
yelpppp
OpenStudy (anonymous):
y=h
OpenStudy (anonymous):
i only have one more guess left :( so i have to be positive its right
Still Need Help?
Join the QuestionCove community and study together with friends!