For a laboratory exercise, you determine the masses of two airtrack gliders as (*1 C 1) = (372.7+0.5) g and (2 Cm2)土= (314.2 +0.1) g. Find the precisions of the measurements (m1 ± Cm 1) and (2 ± Cm2).Select one:O a. 0.03%, 0.03%O b. 0.13%, 0.13%O C. 0.03%, 0.13%O d. 0.13%, 0.03%

For a laboratory exercise, you determine the masses of two airtrack gliders as ("1 C 1) = (372.7+0.5) g and (2 om2) = (314.2 +0.1) g. Find the precisions of the measurements (m1 ± dm1) and (2 ± dn2). Select one: O a. 0.03%, 0.03% Ob. 0.13%, 0.13% O C. 0.03%, 0.13% O d. 0.13%, 0.03%

For a laboratory exercise, you determine the masses of two airtrack gliders as ("1 C 1) = (372.7+0.5) g and (2 om2) = (314.2 +0.1) g. Find the precisions of the measurements (m1 ± dm1) and (2 ± dn2). Select one: O a. 0.03%, 0.03% Ob. 0.13%, 0.13% O C. 0.03%, 0.13% O d. 0.13%, 0.03%

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 25P: A sidewalk is to be constructed around a swimming pool that measures (10.0 0.1) m by (17.0 0.1) m....

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Question

For a laboratory exercise, you determine the masses of two airtrack gliders as (*1 C 1) = (372.7+0.5) g and (2 Cm2)
土
= (314.2 +0.1) g. Find the precisions of the measurements (m1 ± Cm 1) and (2 ± Cm2).
Select one:
O a. 0.03%, 0.03%
O b. 0.13%, 0.13%
O C. 0.03%, 0.13%
O d. 0.13%, 0.03%

Expert Solution

Step 1 :Introduction

Precision of an experiment is the measurement of how closely the measured values agree to each other . When relative error is used as a measurement of precision, , the precision is calculated by using the formula $P = \frac{E}{V} \times 100 %$ , where $E$ is the error value and $V$ is the measured value.

Step 2 :Calculation of the precision of first air track

Calculate the precision of first air track by using the formula $P_{1} = \frac{E_{1}}{V_{1}} \times 100 %$ , where $E_{1} = 0.5 g$ is the error value and $V_{1} = 372.7 g$ is the measured value.

$\begin{array}{rcl} P_{1} & = & \frac{E_{1}}{V_{1}} \times 100 % \\ = & \frac{0.5 g}{372.7 g} \times 100 % \\ \approx & 0.13 % \end{array}$

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