Question

<Homework 5 Problem 25.75 A \( 1.50 \Omega \) wire is stretched uniformly to 1.20 times its original length. Part A What is its resistance now? \( R= \) \( \square \) \( \Omega \) Submit Request Answer < Return to Assignment Provide Feedback Search

          <Homework 5
Problem 25.75

A \( 1.50 \Omega \) wire is stretched uniformly to 1.20 times its original length.

Part A

What is its resistance now?
\( R= \) \( \square \) \( \Omega \)

Submit
Request Answer

< Return to Assignment

Provide Feedback

Search

<Homework 5
Problem 25.75

A 1.50 Ω wire is stretched uniformly to 1.20 times its original length.

Part A

What is its resistance now?
R= □ Ω

Submit
Request Answer

< Return to Assignment

Provide Feedback

Search

Added by Darnell C.

Physics for Scientists and Engineers with Modern Physics

Douglas C. Giancoli 4th Edition

Chapter 25

Instant Answer

Solved by Expert Mrinal R

Step 1

Step 1: Recall the formula for resistance of a wire: \[ R = \rho \frac{L}{A} \] where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is length, and \( A \) is cross-sectional area. Show more…

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<Homework 5 Problem 25.75 A \( 1.50 \Omega \) wire is stretched uniformly to 1.20 times its original length. Part A What is its resistance now? \( R= \) \( \square \) \( \Omega \) Submit Request Answer < Return to Assignment Provide Feedback Search