Correct IDENTIFY: First use Ohm's law to find the resistance at 20.0° C; then calculate the resistivity from the resistance. Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance. SET UP: Ohm's law is R = V/I, R=pL/A, and the radius is one-half the diameter. EXECUTE: At 20.0°C, R = V/I= (13.0 V)/(18.4 A) = 0.707 2. Using R = pL/A and solving for p gives p=RA/L = RT(D/2)²/L = (0.707 )([(0.450 × 10-² m)/2]²/(1.30 m) = 8.64 × 10-6 N. m. ▾ Part B Find the temperature coefficient of resistivity at 20.0°C for the material of the rod. Express your answer in reciprocal degrees Celsius to two significant figures. 15. ΑΣΦ a= 9.780910 Qubmit Duerda pay ? (°C)-¹
Correct IDENTIFY: First use Ohm's law to find the resistance at 20.0° C; then calculate the resistivity from the resistance. Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance. SET UP: Ohm's law is R = V/I, R=pL/A, and the radius is one-half the diameter. EXECUTE: At 20.0°C, R = V/I= (13.0 V)/(18.4 A) = 0.707 2. Using R = pL/A and solving for p gives p=RA/L = RT(D/2)²/L = (0.707 )([(0.450 × 10-² m)/2]²/(1.30 m) = 8.64 × 10-6 N. m. ▾ Part B Find the temperature coefficient of resistivity at 20.0°C for the material of the rod. Express your answer in reciprocal degrees Celsius to two significant figures. 15. ΑΣΦ a= 9.780910 Qubmit Duerda pay ? (°C)-¹
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer the following question
![A 1.30 m cylindrical rod of diameter 0.450 cm is
connected to a power supply that maintains a
constant potential difference of 13.0 V across its
ends, while an ammeter measures the current
through it. You observe that at room temperature
(20.0°C) the ammeter reads 18.4 A, while at
92.0°C it reads 17.0 A. You can ignore any
thermal expansion of the rod.
For related problemsolving tips and strategies, you
may want to view a Video Tutor Solution of
Temperature dependence of resistance.
Submit
Correct
IDENTIFY: First use Ohm's law to find the resistance at 20.0° C; then calculate the resistivity from the resistance.
Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance.
SET UP: Ohm's law is R = V/I, R= pL/A, and the radius is one-half the diameter.
EXECUTE: At 20.0°C, R = V/I = (13.0 V)/(18.4 A) = 0.707 №.
Using R=pL/A and solving for p gives
-6
p=RA/L= Rπ(D/2)² /L = (0.707 №)π([(0.450 × 10−² m)/2]²/(1.30 m) = 8.64 × 10¯6 № · m.
Part B
Previous Answers
Find the temperature coefficient of resistivity at 20.0°C for the material of the rod.
Express your answer in reciprocal degrees Celsius to two significant figures.
IVE ΑΣΦ
α= 9.7809 10
Submit
4
Previous Answers Request Answer
?
X Incorrect; Try Again; One attempt remaining
(°C) -¹](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0655393b-8df3-4633-b13c-e0d6983d2306%2F4cc2a15b-26bd-4030-8aaa-b1dc7080de90%2F5nfun0r_processed.png&w=3840&q=75)
Transcribed Image Text:A 1.30 m cylindrical rod of diameter 0.450 cm is
connected to a power supply that maintains a
constant potential difference of 13.0 V across its
ends, while an ammeter measures the current
through it. You observe that at room temperature
(20.0°C) the ammeter reads 18.4 A, while at
92.0°C it reads 17.0 A. You can ignore any
thermal expansion of the rod.
For related problemsolving tips and strategies, you
may want to view a Video Tutor Solution of
Temperature dependence of resistance.
Submit
Correct
IDENTIFY: First use Ohm's law to find the resistance at 20.0° C; then calculate the resistivity from the resistance.
Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance.
SET UP: Ohm's law is R = V/I, R= pL/A, and the radius is one-half the diameter.
EXECUTE: At 20.0°C, R = V/I = (13.0 V)/(18.4 A) = 0.707 №.
Using R=pL/A and solving for p gives
-6
p=RA/L= Rπ(D/2)² /L = (0.707 №)π([(0.450 × 10−² m)/2]²/(1.30 m) = 8.64 × 10¯6 № · m.
Part B
Previous Answers
Find the temperature coefficient of resistivity at 20.0°C for the material of the rod.
Express your answer in reciprocal degrees Celsius to two significant figures.
IVE ΑΣΦ
α= 9.7809 10
Submit
4
Previous Answers Request Answer
?
X Incorrect; Try Again; One attempt remaining
(°C) -¹
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