Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Circuits: A certain material has a voltage of 120 V across 10 A and the length of the wire is 25 m, while the diameter is 5 mm. What it its resistivity?

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[V = IR\] \[P = IV\] \[R = \frac{ \delta L }{ A }\] i used the first one (R=V/I) is that right?
and you're not given the constant.
so first find the resistance

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yeah i got that 12 ohms
using the first formula right?
yes
that's it? but what are the length and diameter given for?
once you find R , use that and given diameter and length to find resistivity which is \[\delta\]
-_- grr. i thought the R was the resistivity.. fdjsalkfhdjkshfklhg got that question wrong on the quiz >:(
thanks tho. :)
R is resistance
oh okay. my teacher didn't explain clearly. :(
well, you know now, will help you in future exam

Not the answer you are looking for?

Search for more explanations.

Ask your own question