What is the change in length of a 15.748cm long column of mercury if its temperature changes from 14.7°C to 64.7°C, assuming the mercury is constrained to a cylinder but unconstrained in length (in cm)? Assume the dimensions of the cylinder do not change. The coefficient of volume expansion for mercury is 1.8×10-4°C-1 I need to look correct answer and show all work You AnsweredCorrect AnswerWhat is the change in length of a 15.748cm long column of mercuryif its temperature changes from 14.7°C to 64.7°C, assuming themercury is constrained to a cylinder but unconstrained in length (incm) ? Assume the dimensions of the cylinder do not change. Thecoefficient of volume expansion for mercury is 1.8x10-4°C-10.1417 margin of error +/- 1%

Answered: Image /qna-images/answer/c3fa5562-7de9-4131-9f22-1ff0f977ce91.jpg

Answered: What is the change in length of a…

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)...

See similar textbooks

Related questions

Question

What is the change in length of a 15.748cm long column of mercury if its temperature changes from 14.7°C to 64.7°C, assuming the mercury is constrained to a cylinder but unconstrained in length (in cm)? Assume the dimensions of the cylinder do not change. The coefficient of volume expansion for mercury is 1.8×10-4°C-1 I need to look correct answer and show all work

You Answered
Correct Answer
What is the change in length of a 15.748cm long column of mercury
if its temperature changes from 14.7°C to 64.7°C, assuming the
mercury is constrained to a cylinder but unconstrained in length (in
cm) ? Assume the dimensions of the cylinder do not change. The
coefficient of volume expansion for mercury is 1.8x10-4°C-1
0.1417 margin of error +/- 1%

Expert Solution

Step by step

Solved in 3 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions